Probability & Statistics Notes-51-110

51 4 52.2 2680 5 54.4 3006 6 70.3 4068 7 50.8 3373 8 73.9 4124 9 65.8 3572 10 54.4 3359 11 73.5 3230 12 59.0 3572 13 61.2 3062 14 52.2 3374 15 63.1 2722 16 65.8 3345 17 61.2 3714 18 55.8 2991 19 61.2 4026 20 56.7 2920 21 63.5 4152 22 59.0 2977 23 49.9 2764 24 65.8 2920 25 43.1 2693 5 PROBABILITY 5.1 What is Probability? Probability theory is the branch of mathematics that studies the possible outcomes of given events together with the outcomes' relative likelihoods and distributions. In common usage, the word “probability" is used to mean the chance that a particular event (or set of events) will occur expressed on a linear scale from 0 (impossibility) to 1 (certainty). Factually, It is the.

[Audio] Probability is a fundamental concept in mathematics that measures uncertainty and summarizes the outcomes of random or indeterministic experiments. An experiment is a process by which an observation or measurement is obtained, such as tossing a coin or rolling a die. The outcome is not certain and can be different each time. Probability is a way of summarizing the uncertainty of events and statements, giving a numerical measure for the degree of certainty or uncertainty of the occurrence of an event. We often use the letter P to represent probability, such as P(rain) would be the probability that it rains. Key definitions include an experiment being a process by which an observation or measurement is obtained, an outcome being a possible result.

[Audio] The concept of empirical probability is also known as frequentist or statistical probability, and it is based on observed data. The probability of an event, such as event A, is determined by the relative frequency of event A in a set of observations. To calculate the probability, we divide the number of observations of event A by the total number of observations in the data set. For instance, if you have caught 40 fish in total, with 13 being Blue gill, 17 being Red.

[Audio] In this slide, we will discuss different ways to represent probabilities and the concept of set notation. Probability is a numerical representation of the likelihood of an event occurring, as we have previously learned. We will use two symmetrical tetrahedral dice as examples. The first calculation is the probability of both dice landing on the same number, similar to rolling a double in a regular dice. This can be represented as P(A) using set notation. Next, the probability of each die landing on a number less than 3 can be represented as P(B). We can also represent the probability of the two numbers differing by at most 1 as P(AB'), where A and B' represent the events of the first die landing on a number less than 3 and the second die landing on a number more than 2. Rolling the dice simultaneously does not change the probabilities, as the rolls are still independent. There are three different ways to represent probabilities - Venn diagrams, two-way tables, and tree diagrams. Venn diagrams show the probabilities inside the elementary pieces, but the relative sizes may not match the numbers. Two-way tables correspond to the intersections of row and column events, with the contents adding up across rows and columns. The bottom-right corner shows a total probability of 1. Tree diagrams are simple representations of probabilistic information, often used to show the sequence of choices leading to outcomes, such as when tossing two coins. They are also useful for representing conditional probabilities. Lastly, we reviewed some set notation terms, including the complement of an event, denoted as A', and the intersection of events, denoted as A ∩ B. That concludes our discussion..

[Audio] In this discussion, we will be exploring the unions and intersections of events A and B. Remember, the probability of an event is the likelihood of it occurring. Both events A and B can occur, but not necessarily at the same time, which we will refer to as AB. Now, let's focus on the unions of events. A union B, or "A union B," includes outcomes that are found within at least one of the events A and B. This can be simplified as including all outcomes that are in A, or in B, or both. The probability of this event is denoted as P(A union B), and it is the probability that either event A and/or B will occur. In set notation, this can be represented by a set S containing points labeled 1, 2, 3, and 4. We can write this as = S. If A = and B = , then A and B are subsets of S, shown as A ⊂ S and B ⊂ S (B is within S). To show that 2 is an element of A, we write 2 ∈ A. The union of A and B is represented by ∪ A. Moving on, let's discuss the complement of A, shown as A'. This includes all elements in the sample space S that are not in A. For example, if A = , then A' = . Next, we will discuss the distributive and De Morgan's laws. The distributive law demonstrates how the union and intersection of sets can be distributed over each other. For instance, (A ∪ B) ∩ C = (A ∩ C) ∪ (B ∩ C) and (A ∩ B) ∪ C = (A ∪ C) ∩ (B ∪ C). De Morgan's law states that the complement of a union is the intersection of the complements and vice versa. In other words, (A ∪ B)' = A' ∩ B' and (A ∩ B)' = A' ∪ B'. To better understand these laws, we can use Venn diagrams, which show all possible logical relations between a finite collection of sets. These diagrams are often used to illustrate the relations between sets or events, with A and B represented as circles, and their operations (intersections, unions, and complements) shown as parts of the diagram. The bounding box represents the entire sample space S. For our exercise, we will use Venn diagrams to illustrate the distributive and De Morgan's laws, and then we will simplify some expressions using these laws. We will work through these exercises together. For exercise 1, we will use Venn diagrams to show the distributive law..

[Audio] Probability theory is a way to measure uncertainty and summarize outcomes in experiments. We will focus on slide number 6, where a Venn diagram shows the likelihood of defects in refurbished items. Out of all the items, 40% had mechanical defects, 50% had electrical defects, and 25% had both. Notation A represents mechanical defects and B represents electrical defects. We can determine probabilities for different combinations: a) P(A) is 40%, b) P(AB) is 25%, c) P(A'B) is 15%, d) P(A'B') is 35%, e) P(AB) is 50%, and f) P(A'B') is 40%. Additionally, g)  B)' P (AB) is 15%. Moving on to slide number 6, we see a table showing the probabilities of mutual fund performance and investment in international stocks. We can determine the probability of AB using the given values as 1. A is investing in international stocks with a probability of 0.52, and B had a positive performance last year with a probability of 0.64. Next, we review rules of probability: Rule 1 states that for an experiment with a sample space containing elements E1, E2, ..., En, we can assign probabilities P(E1), P(E2), ..., P(En) as long as 0≤P(Ei)≤1 and ∑i=1nP(Ei)=P(S). Rule 2 states that the sum of probabilities for all outcomes in an event A is equal to 1. If A is the sample space, its probability is 1. If A has no elements, its probability is 0. Rule 3 states that the probability of the union of n disjoint events E1, E2, ..., En is the sum of their individual probabilities. Rule 4 states that if A is a subset of B, then the probability of A is less than or equal to the probability of B. Finally, rule 5 states that for any two events A and B, the probability of A and B occurring is less than or equal to the probability of A occurring alone..

[Audio] In today's lesson, we will be exploring probability theory and its practical applications. We have an example that will help us understand how to calculate the probability of certain events. Our first example involves finding the probability of John passing at least one exam, given the probabilities of him passing a Math and Chemistry exam. We will use the formula P(A union B) = P(A) + P(B) - P(A intersection B) to solve this problem. Using the given information, we determine that John has a 4/5 chance of passing the Math exam, a 5/6 chance of passing the Chemistry exam, and a 3/4 chance of passing both exams. Our calculation yields a final probability of 53/60 for John passing at least one exam. Moving on to the second example, we are tasked with finding the probability of various outcomes when rolling a die. Using the formulas we have learned, such as multiplying individual probabilities for A intersection B and adding and subtracting probabilities for A union B, we can determine the probabilities for the given events. We will work through these examples together and then move on to more complex problems in the following slides. Keep up the good work, class!.

[Audio] We will be discussing the concept of odds and its relation to probability on slide number 8. On slide number 5, we were given two events A and B in the same sample space with probabilities of 0.59 and 0.3. The probability of both events occurring together, AB, is 0.21. From this, we were able to calculate the probabilities of A union B, A complement B, and AB complement. Moving on to slide number 6, we were given two events A and B with probabilities of 4/3 and 3/2, and the probability of their intersection, AB, is 1. On slide number 7, we were given the probabilities of A and B and were asked to find the probabilities of A union B, A complement B, and A intersection B. On our current slide, number 8, we are given a loaded die where even numbers are twice as likely as odd numbers. Using this information, we are asked to find the probability of getting a number greater than 3 in a single toss of the die. We then move on to the concept of mutually exclusive events and compound events. The definition of odds of an event is the ratio of the probability of the event occurring to the probability of the event not happening. We solve an example and a question on finding the probability of an event and its complement based on the given odds. Please continue to the next slide for further examples and exercises on probability theory..

[Audio] In this presentation, we will be discussing the concept of probability using a specific example. The formula P(A) = P(B) will be used to demonstrate the relationship between different events. The example provides probabilities for three events, A, B, and C, with values of 12/5, 3/1, and 36/5, respectively. This example illustrates that events A and B are not mutually exclusive, and the same is true for events A and C, as well as B and C. Additionally, it is evident that events A and B are independent, with P(AB) = P(A) × P(B), resulting in a probability of 144/25 for both events occurring together. However, events A and C are dependent, with P(AC) = P(A) × P(C), resulting in a probability of 12/5 for both events occurring together. Furthermore, the three events in this example are not jointly independent, as pairwise independence is not satisfied. This means that even though the events may be independent on their own, they are not independent when considered together. Moving on to another example, we will explore a factory with three machines that have different probabilities of breaking down during a shift. By using the formula P(ABC) = P(A) × P(B) × P(C), we can determine the probability of all three machines breaking down, which is 275/21. Similarly, using the same formula, we can also determine the probability of none of the machines breaking down, which is 15/11. As an activity, we will solve three practice problems. In the first problem, we will calculate the probability of hitting a target in a game of archery, with given probabilities of 3/1 and 5/2. In the second problem, we will toss a fair coin three times and determine which of the events A, B, and C are independent. Lastly, in the third problem, we will roll a fair coin and a fair die together and determine the probability of event A occurring..

[Audio] n" objects in groups of "r". The formula for calculating permutations is n!/(n-r)!. For example, if we have 5 different objects and we want to arrange them in groups of 3, the total number of arrangements would be 5!/2!= 60. Combinations, on the other hand, refer to the selection of "r" objects from a total of "n" objects, without regard to order. The formula for calculating combinations is n!/(r!(n-r)!. For example, if we have 10 different objects and we want to select 3 of them, the total number of combinations would be 10!/3!(10-3)!= 120." Permutations are used in probability to determine the number of ways a specific event can occur in a certain order. This is especially useful for sequences or arrangements of objects. The formula for calculating permutations is n!/(n-r)!, with n representing the total number of objects and r representing the number of objects in each arrangement. For instance, if there are 5 objects and we want to arrange them into groups of 3, there are 5!/2!=60 different possible arrangements. Combinations are used in probability to determine the number of ways a certain number of objects can be selected from a larger set of objects, without considering order. The formula for calculating combinations is n!/(r!(n-r)!), with n representing the total number of objects and r representing the number of objects being selected. For example, if there are 10 objects and we want to select 3 of them, there are 10!/3!(10-3)!=120 different possible combinations..

[Audio] Let's move on to example number 11, where we will look at a situation involving selection from a group of people. In this case, we have a group of 10 employees and we need to select 3 of them to go to a specific plant. The question is, in how many ways can this selection be made? The formula for this is 10 C3, which represents.

[Audio] On slide number 12, we have several examples to further understand the concept of probability. The first one asks the question, "what is the probability that exactly 4 of your numbers are selected?" This is followed by a practical application, where we have a deck.

[Audio] To understand the concept of conditional probability and independence, let's consider an example. Suppose eight different brands of tires are ranked from best to worst according to their mileage performance. If a customer randomly selects four tires, what is the probability that the best tire among the four chosen is actually ranked third among the original eight? This can be calculated using conditional probability. Conditional probability is the likelihood of an event occurring, given that another event has already happened. For instance, if your boss looks at you gloomily, you might assume your.

[Audio] The probability of P(R2/R1) is 3/10, which is the same as P(R1 and R2) being obtained from counting the outcomes and using the formula P(R2/R1) = C(7,1) / C(10,2). Probability of P(A/B) is 0.75, which means that 75.

[Audio] Probability is the numerical representation of the likelihood of an event occurring. It is used to measure uncertainty and summarize the outcomes of indeterministic experiments. The formula shown on this slide tells us the probability of observing x successes out of n trials. This applies to when x equals 0, 1, 2, and so on. For example, if we toss a fair coin 10 times and want to know the probability of getting exactly 8 heads, the solution is 5.0/10 or 0.5. We will also solve exercises such as finding the probability of rolling a 2 when the sum of the scores is 6, determining the probability of selecting a dealer with good services and a warranty, calculating the probability of selecting a defective item, and solving other problems..

[Audio] Blacks 16.8% 5.3% 7.5% 3.8% Asians 4.3% 7.7% 22.7% 13.9% Hispanics 11.2% 4.8% 5.9% 3.3% Whites 132.0% 6.3% 13.9% 7.7% a) How many Asians have had a graduate or professional degree in 1998? Answer: 22.7% b) What percent.

[Audio] In probability theory, the concept of partition is crucial for understanding and analyzing events. A partition of a sample space is a division of the sample space into mutually exclusive and exhaustive events, as seen in previous examples. To be considered a valid partition, certain conditions must be met. Firstly, the events must be mutually exclusive, meaning they cannot occur at the same time. Secondly, the partition must be exhaustive, covering the entire sample space. Partitions are commonly encountered in real-world scenarios, especially in statistics. For example, an insurance company may have data on accident rates for different age groups, which can be partitioned accordingly. Conditional probabilities come into play when calculating the overall probability of an event, using the Law of Total Probability. This concept applies to cases with more than two events in a partition. According to a theorem, if a partition of the sample space has mutually exclusive and exhaustive events, the Law of Total Probability can be used to find the unconditional probability of any event. In summary, partitions allow us to calculate probabilities in complex scenarios and are an important concept in probability theory..

[Audio] In our discussion on probability theory, we will be looking at a practical application of this concept. Imagine a scenario where a product, denoted as A, is made by a machine, denoted as B. B can take on values of 1, 2, or 3. To calculate the probability of A occurring given that B has occurred, we can use the additive and multiplicative rules. This can be written as (a) P(B)P(A/B) = (0.3)(0.02) + (0.45)(0.03) + (0.25)(0.02) = 0.006 + 0.0135 + 0.005 = 0.0245. We can also calculate the probability of B occurring given that A has occurred, using the formula P(B/A) = P(A/B)P(B)/P(A). This can be written as (b) 0.0245/0.3 = 0.2041. This calculation can also be represented using a tree diagram. Let's now apply what we have learned through some exercises. Our first exercise involves Lucy, who is trying to decide between a math course and a chemistry course. She estimates her probability of receiving an A to be 0.5 in a math course and 0.75 in a chemistry course. If she bases her decision on the flip of a coin, what is the probability of her getting an A? For our second exercise, let's look at a gas station where 70% of customers use regular gas and 30% use diesel. Of those who use regular gas, 60% fill the tank completely and of those who use diesel, 80% do the same. We want to determine the percentage of all customers who fill the tank completely and the probability that a customer who fills up completely was buying diesel. Our final exercise involves the ownership of stocks among different households. In 2004, 57% of White households owned stocks, compared to 26% of Black households and 19% of Hispanic households. The data for Asian households is not given, but we can assume the same rate as for White households. Based on this information, what is the probability that a randomly chosen household owns stocks? These exercises show how probability theory can be applied..

[Audio] Today in class, we will be discussing probability theory and its applications. As we have learned, probability is a numerical representation of the likelihood of an event occurring. In slide 19, we have a scenario where a company has sold a certain number of cameras. Out of these cameras, 20% are Zony's and 30% are Lucky Stars. We need to fill in the tree diagram and answer some questions regarding the percentage of returned cameras and the probability of a returned camera being a Lucky Star. From our calculations, we can see that 50% of all cameras are returned and 30% of returned cameras are Lucky Stars. Moving on, we are asked to find the percentage of Zony's that were not returned, which we can determine to be 10%. Moving to slide 6, we have data on newspaper readership in a certain city, with 45% of people reading at least one newspaper, and 8% reading both newspaper A and B. In slide 7, we are given information to calculate probabilities for events A and B. Finally, in slide 8, we have the well-known Monty Hall problem, which prompts the question of whether it is more advantageous for a contestant to switch doors after the host reveals one with no prize behind it..

[Audio] Probability is a way of summarizing the uncertainty of statements or events. It gives a numerical measure for the degree of certainty (or degree of uncertainty) of the occurrence of an event. We often use P to represent a probability, such as P(rain) being the probability that it rains. In other cases, Pr(.) is used instead of just P(.). An experiment is a process by which an observation or measurement is obtained, such as tossing a coin or rolling a die. An outcome is a possible result of a random experiment, like a 6 when a die is rolled once or a head when a coin is tossed. The sample space is a collection or set of all possible outcomes of a random experiment, usually denoted by S or Ω or U. An event is a set of one or more outcomes of a random experiment. For instance, getting a tail when a coin is tossed is an event. The probability of an event is the sum of the probabilities of its individual outcomes. A random variable is a quantity or attribute whose value is determined by the outcome of a random experiment. It can take on different values from one unit of population to another. There are two types of random variables - discrete and continuous. A discrete random variable has a finite or countably infinite set of values, while a continuous random variable can take any value within a given interval of real numbers. A discrete random variable is one that can take on a finite or countably infinite set of values, and its probability mass function helps us determine the probabilities for each value..

[Audio] In this meeting, we will discuss the basics of probability theory and its applications. Currently, we are on slide 21 where we will cover discrete random variables and their probability mass functions (pmfs). A discrete random variable represents the likelihood of an event occurring and is used to measure uncertainty and summarize the outcomes of indeterministic experiments. On this slide, we see that the probabilities associated with the variable's values are determined by multiplying the value of p by the value of x for each value of x from 1 to n. It is important to note that random variables are denoted by capital letters and realized values are denoted by lowercase letters. For instance, our first example involves rolling two tetrahedral dice and the sum of the scores facing down as the random variable. The pmf for this variable is shown in the table. It can also be expressed as a function where the probability of each value of x is equal to the value of x over 16. Moving on to example 2, we are given a discrete random variable W with a pmf shown in the table. Our task is to find the value of the constant d and determine the probabilities for -3, -2, -1, 0, and 1. By setting these probabilities equal to the given values, we find that d is 0.65. In example 3, we have a discrete random variable Y with a pmf shown in the table. We are given the value of c for each value of y and our goal is to find the probability of Y being less than or equal to 3. By setting this probability equal to the given equation, we find that c is 0.35. Remember, discrete random variables are a crucial concept in probability theory as they help us understand and analyze uncertain events. With this knowledge, we can move on to example 4 and explore the applications of discrete random variables..

[Audio] Today we will be discussing discrete random variables and their probability mass function or pmf. The pmf for a discrete random variable X is given by the values of x and the corresponding probabilities P(X=x). In our example, we are given the pmf of X with values 1, 2, 3, 4, 5, 6 and their respective probabilities of k/4, k/3, k/2, k/6, 0, and 0. Our task is to determine the value of the constant k and find the probabilities P(X<4) and P(X≥6). In another example, we are flipping a fair coin until a head appears, with the number of tosses represented by N. We need to find the pmf of Nc and the probabilities P(N<2) and P(N≥2). Next, we are given the pmf of a discrete random variable Y with values 0, 1, 2 and probabilities c/3, 2c/3, and c. Our objective is to find the value of the constant c and the probabilities P(X<3) and P(X≥3). We will then verify that the function 1/(2f(x)+g(x)) can serve as a pmf of a random variable X. Moving on, we must determine the value of c for different functions in order for them to serve as a pmf of X. For example, for the function f(x)=cx-x and values 1, 2, 3, 4, 5, we need to find the value of c. Next, we have a coin loaded so that heads is three times as likely as tails, and we must find the pmf of the total number of heads after 3 independent tosses. Similarly, in a game where heads earn 2 points and tails lose 5 points, we need to write the probability distribution of the total scores after 4 coin tosses. In the final part, we will look at a real-life example with an online electronics retailer and the number of Zony digital cameras returned per day, represented by X. We will use this information to make business decisions. Please take notes and ask questions if needed as we continue with the remaining slides..

[Audio] Today, we will be discussing the basics of probability theory and its applications. Slide number 23 contains important information about continuous random variables and their probability density function. Probability is a numerical representation of the likelihood of an event occurring and is used to measure uncertainty and summarize outcomes of experiments. The formula on this slide states that the support of a continuous random variable is the smallest interval containing all values of x where f(x) is greater than or equal to 0. It is important to note that for any real number x, the probability of the random variable assuming a particular value is 0. This means that we can only talk about the probability of the random variable falling within a given interval, which we define as the area under the graph of the probability density function. To better understand this concept, let's examine an example. For exercise 1 on this slide, we are asked to find the value of a constant c and the constant m, so that the probability of the random variable falling between 1 and m is 2/3. For exercise 2, we are given the pdf of kx+7 and asked to find the value of k, as well as the probability of the random variable falling between 1 and 4. Take some time to work on these exercises and fully understand the concept of continuous random variables and their probability density function. We will discuss the solutions in our next class..

[Audio] Today we will be discussing the concept of cumulative distribution function (CDF) for both discrete and continuous random variables. Slide number 24 shows a continuous random variable, X, with a probability density function (PDF) of 2/kx for values between 0 and 1. The PDF is 0 for values outside of this range. To find the constant k, we will use the definition of CDF, which states that it is the probability that X is less than or equal to a given value x. In our case, we will set x to 3 and solve for k, giving us a value of 2/9. With this value, we can integrate 2/kx from 0 to 3 to find the CDF, which is 4/9. Using the properties of CDF, we can also find the probability of X being between 1 and 3, which is 1/3. Moving on to slide number 25, we are reminded that we can differentiate the CDF to get the PDF and integrate it to get the CDF, known as the fundamental theorem of calculus. Finally, the theorem states that the probability of X being between a and b is the difference in the CDF at those two points, which we demonstrate in the example on slide number 26. This concludes our discussion on CDF for both discrete and continuous random variables. I hope you have a better understanding of its concept and applications. See you in our next lesson..

[Audio] We will be discussing the basics of probability theory and its applications, specifically focusing on finding solutions for various probability problems in our presentation. Probability is defined as the numerical representation of the likelihood of an event occurring and is used to measure uncertainty and summarize the outcomes of experiments. Today, we will be examining the value of a constant, the cumulative distribution function (cdf), and the probability of a certain event occurring. First, in exercise C, we are given a continuous random variable X with a probability density function (pdf) of ƒ(x) = Cx/2, and our goal is to find the value of the constant C, the cdf of X, and the probability of X being greater than or equal to 1. In exercise 2, we are given a random variable X with a pdf of ƒ(x) = ke^(-x), and we must find the constant k, the cdf of X, and the value of m such that G(x) = 1. Exercise 3 involves finding the cdf of a random variable Y with a pdf of ƒ(x) = 3x/2 if x is greater than or equal to 1, and ƒ(x) = 1/4 if x is between 0 and 1. We then move on to exercise 4, where we are given the cdf of a random variable Y as 3/9 when F(x) is greater than or equal to 2, and 0 when F(x) is less than 3. Our task is to find the probability that X is less than or equal to 5 and greater than 8, as well as the pdf of X. We will also be discussing the concepts of expectation and variance of a random variable, which can be calculated by taking the weighted average of all possible values. In summary, we have covered the topics of finding the value of a constant and cdf, as well as the probability of a certain event occurring, and delved into the concepts of expectation and variance of a random variable. In the next slide, we will continue our discussion..

[Audio] Today, we will continue our discussion on probability theory and its applications. Our focus will be on the linearity property of expectation, as well as the concepts of variance and standard deviation. According to the linearity property of expectation, the expectation of ax plus b is equal to a times the expectation of x plus b, for any constants a and b. This also applies to multiple variables, as shown in part (iii) of slide number 26. The variance and standard deviation are measures of the spread of data from the mean. Variance is the average of squared differences from the mean, denoted as sigma squared or Var(X). The standard deviation, denoted as sigma, is the positive square root of the variance and represents the average distance of data points from the mean. Moving on, the theorem for variance states that Var(aX + b) is equal to a squared times Var(X). This means that the variance of a constant times a random variable is equal to the constant squared times the original variance of the random variable, as shown on slide number 26. Additionally, there is a theorem that demonstrates the relationship between expectation and variance. It states that Var(X) is equal to the expectation of X squared minus the square of the expectation of X, as seen on slide number 26. To summarize, we have discussed the linearity property of expectation, as well as the concepts of variance and standard deviation. It is important to note that these properties hold true for both discrete and continuous random variables, as long as the variables are independent. Thank you for listening and I will see you in our next class..

[Audio] This slide focuses on various probability distributions for random variables. The first example involves finding the mean and standard deviation of a given random variable X for two different values. The probability distribution used shows a mean of 6/7 and a standard deviation of 1.6833 for the first problem, and a mean of 242.38812 and a standard deviation of 144/6 for the second problem. Next, the third example presents a continuous random variable X and its probability density function, from which we can determine the mean to be 0 and the standard deviation to be 3/2. For the exercises, three different problems are given. The first problem requires finding the mean and variance of a given random variable X with a probability mass function, revealing a mean of 3 and a variance of 2.25. In the next example, probabilities for a random variable are provided and we must find the expected value and standard deviation, resulting in an expected value of 1.8 and a standard deviation of 0.7483. The final problem provides a probability distribution for a random variable W and asks us to find a constant as well as the mean and variance, which can be calculated to be 0.3, 0.6, and 0.6, respectively. This has been a review of probability theory and its practical applications for our meeting today..

[Audio] Today, we will be discussing the topic of probability theory and its applications. Probability is the likelihood of an event occurring, represented by a numerical value. It is used to measure uncertainty and summarize the outcomes of indeterministic experiments. Let's take a look at the problem on this slide. We have the equation E(Z) = 3/4, and we are asked to find the values of x and y and determine the variance of Z by first finding the mean and variance of the discrete random variable M. In the next problem, we are asked to calculate the expected value and variance of Y using a given probability distribution and constant k. Moving on, we need to find the mean and variance of X using a continuous random variable with a given probability distribution. In the following problems, we are asked to find the value of E(X) and Var(X) for given continuous random variables M and R by determining the value of the constant k and using it to calculate the mean and variance. Lastly, we are asked to find the value of k and the mean and variance of a continuous random variable X by integrating the given pdf..

[Audio] The probability distribution and its various applications will be discussed, focusing on discrete distributions such as the Bernoulli, binomial, Poisson, geometric, and hypergeometric distributions. The Bernoulli distribution involves a random experiment with two possible outcomes - success and failure. Examples include tossing a coin, checking items from a production line, or phoning a call centre. A Bernoulli random variable, denoted as X, takes the values 0 and 1, with the probability of success being represented as p and the probability of failure as 1-p. The mean and variance of a Bernoulli random variable are given by μ = p and σ² = p(1-p), respectively. The binomial distribution involves a sequence of n independent Bernoulli trials, each with two possible outcomes and a probability of success denoted as p. The probability mass function of the binomial distribution is P(X = x) = nCx * p^x * (1-p)^(n-x). The nCx term represents the number of outcomes that include exactly x successes..

[Audio] Example 2 discusses a quality control engineer who is responsible for testing whether 90% of the company's DVD players meet specifications. The engineer selects a random batch of 12 DVD players each day, and if no more than one of them fails to meet specifications, the day's production is considered acceptable. However, if more than one player fails, the entire day's production must be tested. There are two questions for this example. The first question asks what the probability is that the engineer will incorrectly pass a day's production as acceptable, if only 80% of the DVD players actually conform to specifications. The second question asks for the probability that the engineer will unnecessarily require the whole day's production to be tested, if in fact 90% of the DVD players conform to specifications. To answer these questions, the binomial distribution is used with parameters n=12 and p=0.2 for the first question, and p=0.1 for the second question. The goal is to find the probability that X, the number of DVD players in the sample that fail to meet specifications, is less than or equal to 1 for the first question, and greater than 2 for the second question. After calculations, it is determined that for the first question, the probability is 0.275, meaning there is a 27.5% chance that the engineer will incorrectly pass a day's production as acceptable if only 80% of the DVD players conform to specifications. For the second question, the probability is 0.661, indicating that there is a 66.1% chance that the engineer will unnecessarily require the entire day's production to be tested if 90% of the DVD players conform to specifications. Moving on to example 3, the scenario involves sending bits over a communications channel in packets of 12. The probability of a bit being corrupted over this channel is 0.1 and these errors are independent. The task at hand is to find the probability that no more than 2 bits in a packet are corrupted. Additionally, the probability that at least one packet out of 6 sent will contain 3 or more corrupted bits is also to be determined. To solve these questions, the binomial distribution is used with parameters n=12 and p=0.1. The probability for the first question is 0.887, indicating that there is an 88.7% chance that no more than 2 bits in a packet will be corrupted. For the second question, the probability is 0.165, meaning there is a 16.5% chance that at least one out of 6 sent packets will have 3 or more corrupted bits..

[Audio] Today's lesson is about probability and its applications. Probability is a numerical representation of the likelihood of an event occurring and is used to measure uncertainty and summarize the outcomes of experiments. In this presentation, we will cover various scenarios and problems to deepen our understanding. Slide 31 discusses a company with a 3% defect rate in electric bulbs, and we will calculate the probability of 5 out of 100 bulbs being defective. Next, we will look at an oil exploration firm with a 0.1 probability of success and enough capital for 10 explorations. Moving on to slide 4, we will examine Emily's performance in basketball, where she hits 60% of her free throws. On slide 5, we will calculate the mean and standard deviation of a loaded coin with a 60% chance of showing heads tossed 3 times. On slide 6, we will use data to find the expected number of people with a bachelor's degree in a sample of 50 people. Slide 7 discusses a scenario where a person throws a fair die six times and gets the number 3 twice, and on slide 8, we look at the probability of being a smoker among lung cancer patients. I hope this presentation has helped you understand probability theory and its applications..

[Audio] The probability that in any one-minute interval there will be (i) 0 jobs; (iv) at most 3 arrivals; (ii) exactly 2 jobs; (v) more than 3 arrivals is 0.1353353, 0.8571, 0.1804470, and 0.1429 respectively..

[Audio] The Poisson distribution is used to model the number of events occurring in a fixed interval of time or space, where these events occur independently and at a constant average rate. The probability mass function of the Poisson distribution is given by: P(X=k) = (λ^k) \* e^(-λ) / k! where λ is the average rate of occurrence, and k is the number of events. Using this formula, we can solve the exercises presented earlier. Exercise 1:.

[Audio] The mean number of errors due to a particular bug occurring in a minute is 0.0001. To find the probability that no error will occur in 20 minutes, we can use the Poisson distribution. Since the mean and variance are both equal to lambda (λ), we can calculate the probability as follows: P(no error in.

[Audio] The next example on slide number 35 deals with probability problems. Probability is a numerical representation of the likelihood of an event occurring and is used to measure uncertainty and summarize the outcomes of experiments. We will use basic probability formulas to solve two different problems in this example. In the second problem, the State Department is looking for an individual who speaks Farsi for a foreign embassy position. It is determined that 4% of the applicant pool is fluent in Farsi and we are given two questions to solve. The first question asks how many people can be expected to be interviewed in order to find one fluent in Farsi if applicants are contacted randomly. The second question asks for the probability that they will have to interview more than 25 people until finding a Farsi speaker. To solve these questions, we will use the formulas P(X>25) = (1- p)^25 and P(X=x) = p(1-p)^(x-1), respectively. The third example deals with a different scenario. From past experience, it is known that 3% of accounts in a large accounting population are in error. We are given two questions to solve. The first question asks for the probability that 5 accounts will be audited before finding an error. Using the formula P(Y>x) = p(1-p)^(x-1), we find that there is a 2.6% chance of this happening. The second question asks for the probability that the first account in error occurs in the first five accounts audited. Using the same formula, we find that there is a 14.1% chance of this happening. Now, let's move on to the exercise..

[Audio] The probability that the first red light will occur on the nth traffic light is calculated by finding the probability that the first red light occurs before the nth traffic light, then subtracting the probability that the first red light occurs before the (n-1)th traffic light. The cumulative probability of hitting a red light on or before the nth traffic light is calculated by finding the probability that the first red light occurs on or before the nth traffic light..

[Audio] The expected value of the salad weight is calculated using the formula: Expected Value = (Upper Limit + Lower Limit) / 2 Where Upper Limit = 15 ounces and Lower Limit = 5 ounces. Expected Value = (15 + 5) / 2 = 10 ounces The variance of the salad weight is calculated using the formula: Variance = [(Upper Limit - Expected Value)^2 + (Lower Limit - Expected Value)^2] / 12 Where Upper Limit = 15 ounces, Lower Limit = 5 ounces, and Expected Value = 10 ounces. Variance = [(15 - 10)^2 + (5 - 10)^2] / 12 = [25 + 25] / 12 = 50 / 12 = 4.17 The probability of a customer taking between 12 and 15 ounces of salad is calculated using the formula: Probability = (Upper Limit - Lower Limit) / (Upper Limit - Lower Limit) Where Upper Limit = 15 ounces and Lower Limit = 12 ounces. Probability = (15 - 12) / (15 - 5) = 3 / 10 = 0.3.

[Audio] Jobs are sent to a printer at an average rate of 3 jobs per hour. To calculate the expected time between jobs, we take the reciprocal of the average rate, which is 1/3 of an hour or 20 minutes. The probability that the next job is sent within 5 minutes is unknown because it requires calculating the cumulative distribution function of the exponential distribution, which is not provided in the problem statement. The time required to repair a machine is an exponential random variable with a rate of 0.5 downs per hour. To find the probability that the repair time exceeds 2 hours, we integrate the probability density function from 2 to infinity, which gives us a value of 0.1353. To find the probability that the repair time will take at least 4 hours given that the.

[Audio] 1. What is the probability that a randomly selected car is a Toyota?.

[Audio] . 90 Standard Normal Probability Distribution A random variable having a normal distribution with a mean of 0 and a varuance of 1 is said to have a standard normal probability distribution Definition The random variable Z is said to have the standard normal distribution if  Z ~ N 0 1, . Therefore, the density of Z, which is usually denoted (z) is given by;   2 2 1 exp 2 1 (z)  z    for     z The cumulative distribution function of a standard normal random variable is denoted (z) , and is given by     z - (t) ( ) dt z  Consider  Z ~ N 0 1, and let X   Z for   0 . Then  2  , X ~ N   But we know that             X 1 f (x) from which the claim follows. Conversely, if  2  , X ~ N   , then  ~ N 0 1, X Z     . It is also easily shown that the cumulative distribution function satisfies            X F(x) and so the cumulative probabilities for any normal random variable can be calculated using the tables for the standard normal distribution.. Definition A variable X is said to be standardized if it has been adjusted (or transformed) such that its mean equals 0 and its standard deviation equals 1. Standardization can be accomplished using the formula for a z-score:  ~ N 0 1, X Z     . The z-score represents the number of standard deviations that a data value is away fromthe m ean. Computing Normal Probabilities It is very important to understand how the standardized normal distribution works, so we will spend some time here going over it. There is no simple analytic expression for (z) in terms of elementary functions. but the values of (z) has been exhaustively tabulated. This greatly simplifies the task of computing normal probabilities.. Table 1 below reports the cumulative normal probabilities for normally distributed variables in standardized form (i.e. Z-scores). That is, this table reports ( ) P(Z z) =  z  ). For a given value of Z, the table reports what proportion of the distribution lies below that value. For example, 5.0 (0) P(Z 0) =    ; half the area of the standardized normal curve lies to the left of Z  0 . Theorem: It may be useful to keep in mind that i) ( ) P(Z z) 1 = -  z  complementary law ii) ( ) 1- z) = P(Z z) P(Z  z     ie due to symmetry 1 ) ( ( )      z z Since 1 P(Z z) P(Z z)     iii) ( ) ( )- b) = P(a a b z     iv) ( ) 1 a) 2 = P(-a     a z since   ( ) 1 2 1 ( ) ( ) ) = ( ) ( a) = P(-a              a a a a a z v) If we now make (a) the subject, then  a) P(-a 1 = ) ( 2 1     z a.

[Audio] We will now discuss Table 1 and Example 1, which involves finding probabilities for various z values in a standard normal distribution with a mean of zero. Starting with part d, the probability for z values between -0.696 and 0.865 can be calculated by using the value of Φ(z) in the standard normal probabilities table (0.9505) and the formula Φ(0.865) - Φ(-0.696). This gives us a probability of 0.2432. For parts a and f, the probabilities for z values between 1.65 and a given number and between 0.365 and another number can be found by subtracting the corresponding Φ(z) values. Part b involves finding the probability for z values greater than 1.02, which can be calculated using the formula Φ(-1.65) - Φ(1.02) with a resulting probability of 0.1515. Similarly, for part c, the probability is found by using the formula Φ(0.365) - Φ(-0.365), resulting in a probability of 0.3249. Part e asks for the probability for z values between -2.345 and 1.02, which is found using the same formula and results in a probability of 0.7257. These examples demonstrate how the standard normal probabilities table can be used to find probabilities for different z values. We will now move on to the next slide and practice more problems. See you there!.

[Audio] In this class, we will be discussing slide number 42 of our presentation on probability theory and its applications. The focus of this slide is on finding probability values and corresponding values for Z using Table I. Part e provides three probability values, 0.0410, 0.0505, and 0.0095, which correspond to Z values of -2.345, 0.65, and 2.345 when using the formula. Moving on to part f, with a given probability value of 0.8472, we can calculate the corresponding Z value to be 1.65 using the formula. In example 2, we are given a similar task of finding the value of t. In part a, three probability values are given: 0.6026, 0.9750, and 0.3446, which correspond to t values of 0.9416, 1.96, and -0.4035, respectively. In part b, with a probability value of 0.4026, the corresponding t value can be found using the formula as -0.6025. This same method is applied in parts c and d, where we are given probability values for Z and asked to find the corresponding t values. Moving on to the exercise, we are given a similar question as in the example. In part a, three values for Z are provided: 1.95, -1.89, and 1.074. Using the formula, we can find the corresponding probability values of 0.973, 0.6693, and 0.4634, respectively. In part b, with a given value of 1.72, we can find the corresponding probability values for Z to be 0.3719, 0.9545, and 0.7546. Part c provides a probability value of 0.105 and requires the corresponding Z value, which can be found to be -1.21. Similarly, in part d, with a probability value of 0.125, the corresponding Z value can be calculated as -1.396. In summary, this slide teaches us how to find probability values and corresponding Z values using Table I. It is important to practice these examples and exercises in order to become more comfortable with this concept. Thank you for your attention and I will see you in our next class..

[Audio] Today's presentation is focused on the basics of probability theory and its applications. We will be discussing the normal distribution and its use in interpreting uncertain experiments. Specifically, we will explore examples that involve calculating the likelihood of certain events. One example involves finding the probability of a given range within a random variable with a normal distribution of 50 and 25. To solve this, we must standardize the distribution to a standard normal distribution with a mean of 0 and a standard deviation of 1. By using the appropriate formula, we determine that the probability of the variable falling between 45 and 60 is 0.8185. Another example involves finding the probability of a value being less than or greater than a specific number, using standardization and the standard normal table. In this case, the probabilities of the variable being less than 40, greater than 21, and within the range of 30 and 35 are 0.9938, 0.9878, and 0.3944, respectively. Another problem we discuss is finding the minimum score needed to qualify for a scholarship, given that the top 5% of applicants with GRE scores will receive them. By standardizing the distribution and using the standard normal table, we find that the minimum score required is 665. Finally, we explore how the normal distribution is useful in determining the percentage of the population living in poverty and the percentage that will benefit from a new tax law, using a known mean and standard deviation. We find that the percentage of the population living in poverty is 15.87%, and the percentage that will benefit from the tax law is 34.13%. In conclusion, the normal distribution is essential in understanding uncertainty and making informed decisions on the likelihood of outcomes..

[Audio] The length of time between charges of the battery is normally distributed with a mean of 50 hours and a standard.

[Audio] The normal approximation of a binomial random variable is only valid under certain conditions, specifically when n is large and both p and q are greater than or equal to 5. When making this approximation, a continuity correction is necessary to account for the difference between a continuous and a discrete distribution. This correction is achieved.

[Audio] We are revisiting the case of coin tossing and applying the concept of probability to solve a problem. Our focus is on the probability of a binomial random variable being less than or equal to 4. The diagram above shows the binomial distribution with n = 10 and p = 0.5, along with a normal density function with a mean of 5 and a standard deviation of approximately 1.58. Using the normal approximation, we can calculate the probability that our normal random variable is less than or equal to 4.5, which is equivalent to the probability of our binomial random variable being less than or equal to 4.5. This is a very close approximation to the actual probability of 0.377. An example of this concept is when 50% of the population approves of the governor's job and 20 individuals are randomly chosen from the population. Using the normal approximation to the binomial, we can solve for the probabilities of a) exactly 7 people supporting the governor, b) at least 7 people supporting the governor, c) more than 11 people supporting the governor, and d) 11 or fewer people supporting the governor. Since np and nq are both greater than or equal to 5, we can assume that X ~ N(10,5). The probabilities for these scenarios are a) 0.0730, b) 0.1318, c) 0.0588, and d) 0.377, all of which are very close approximations to the actual probabilities. This example demonstrates the usefulness of the normal approximation to the binomial, especially for larger sample sizes. Let's now move on to the next slide..

[Audio] We are now on slide number 47 where we will discuss probability and its applications. Probability is a numerical representation of event likelihood and is useful in measuring uncertainty and summarizing experiment outcomes. In this example, we will focus on the binomial distribution, a probability distribution for the number of successes in trials. The mean, denoted by the Greek letter mu (μ), is 360 for the number (X) of passengers showing up for a flight. The standard deviation, denoted by the Greek letter sigma (σ), is 6 in this case, indicating a spread of 6 around the mean. Using this information, we can calculate the probability of nobody getting bumped from the flight, which is approximately 0.9599 or almost 96%. This is obtained from the formula P(Z<=370) = 0.9599, where Z is a standard normal random variable. We will use the normal approximation to the binomial for the entire exercise. In the first question, we need to find the probability of observing between 4 and 7 heads inclusive when a coin is tossed 12 times and heads is three times more likely than tails. In the second question, we need to find the approximate probability for a) at least 75 customers paying with a credit card, b) not more than 70 customers paying with a credit card, and c) between 70 and 75 customers, inclusive, paying with a credit card, when 40% of customers at Miller's Automotive Service Station pay with a credit card and a random sample of 200 customers is selected. In the third question, we need to find the probability of a soccer player, Ronaldo, scoring 30 goals in the next 100 games when he has a 0.3 probability of scoring against a tough opponent. In the fourth question, we need to find the probability of a company, Crafty Computers Limited, having a 0.25 probability of a virus in their computers, when JKUAT ICSIT buys 300 computers from them..

[Audio] In this slide, we will be discussing a theorem that explains the properties of independent random variables in probability theory. The theorem states that if two independent random variables, X1 and X2, follow a normal distribution with mean and standard deviation of μ and σ respectively, then their sum, X+Y, also follows a normal distribution with mean and standard deviation of μ1+μ2 and σ1+σ2, respectively. To prove this, we define the moment-generating function of X1 and introduce a new random variable, Y, which is the sum of X1 and an independent random variable X2. By defining the moment-generating function of Y, we can show that it is the same as the moment-generating function of a normal random variable with the same mean and standard deviation. This theorem can be extended to n independent random variables, with their means and standard deviations calculated by adding up the individual random variables' means and standard deviations. To better understand this, let's look at an example with two independent random variables, X ~ N(60, 1) and Y ~ N(70, 9). Using the properties of normal distribution, we can easily calculate the probabilities of (a) X+Y≤140, (b) 135≤X+Y≤120, (c) Y-X>7, and (d) 12≤Y-X≤24. Now, let's move on to the exercise section and practice some more problems..

[Audio] The class is currently on slide number 49 out of 50. The.

[Audio] 1. In a study, researchers found that out of 100 students who took a math test, 60 passed and 40 failed. What is the probability that a randomly selected student passed the test? Answer: The probability that a randomly selected student passed the test is 0.6 or 60%..