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Uploaded: 2022-02-17
Duration: 6 min 54 s
Description: COR006: STATISTICS AND PROBABILITIES Performance Task Group Members : Pedines Mark Lorenz –Leader Anulacion , Miaca Angelica Chiang, Leila Ashly Conteno , Kenneth Rebadulla , Jessel.

COR006: STATISTICS AND PROBABILITIES Performance Task Group Members : Pedines Mark Lorenz –Leader Anulacion , Miaca Angelica Chiang, Leila Ashly Conteno , Kenneth Rebadulla , Jessel. 

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COR006: STATISTICS AND PROBABILITIES Performance Task Group Members : Pedines  Mark Lorenz –Leader Anulacion ,  Miaca  Angelica Chiang, Leila Ashly Conteno , Kenneth Rebadulla , Jessel

I. DISCRETE: CONTINUOUS:. Formulate 2 examples of each kind of random variable: DISCRETE and CONTINUOUS.. 

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I.  DISCRETE:                     CONTINUOUS:

Formulate 2 examples of each kind of random variable: DISCRETE and CONTINUOUS.

NUMBER OF STUDENTS IN A CLASSROOM NUMBER OF MARBLES IN A BAG

Construct a probability distribution table if we let X the random variable be the number of boys in the family with 3 children. Use the table below.. 

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Construct a probability distribution table if we let X the random variable be the number of boys in the family with 3 children. Use the table below.

No. of Boys (X) F(x) O 1/8 1 3/8 2 3/8 3 1/8

B- Boys         G- Girls BBB, BGB, BBG, BGG, GGG, GBG,GGB, GBB

III.. An investor has six stocks that she follows each day. The random variable being studied is X, the number of stocks that increases in value each day. The probability function for this variable is shown below.. 

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An investor has six stocks that she follows each day. The random variable being studied is X, the number of stocks that increases in value each day. The probability function for this variable is shown below.

X 1 2 3 4 5 6  F(x)=P[X=x] 0.34 .30 .20 .10 .05 .01

a. Find the probability that on a given day, the majority of the stocks will increase in value.

SOLUTION: P[X≥4]= P[x=4]+ P[x=5]+ P[x=6]             = f(4)+ f(5)+ f(6)             = .10+ .05+ .01             = .16 or 16%

B. Find the probability that on a given day, at least 2 of the stocks will increase in value.

SOLUTION:  P[X≥2]= P[x=2]+ P[x=3]+ P[x=4]+ P[x=5]+ P[x=6]             = f(2)+ f(3)+ f(4)+ f(5)+ f(6)             = .30+ .20+ .10+ .05+ .01             = .66 or 66%

IV.. INTEREST. APPLICATION. Solve the mean and variance of the following discrete random variable.. 

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Solve the mean and variance of the following discrete random variable.

Let X be a random variable with the following probability distribution. Find the mean µ and variance ó 2

x P(x) x•P (x) x-µ (x-µ) 2 (x-µ) 2 -P(x) -8 0.2 -1.6 -8.4 70.56 14.112 4 0.2 0.8 3.6 12.96 2.592 2 0.6 1.2 1.6 2.56 1.536 Σ(x)= 0.4 Ó 2  = 18.24

V.. Use the following formula to find the missing component:. 

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Use the following formula to find the missing component:

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1. Given: x=70    µ=60   ó= 4  z=? z= 70-60/4  =10/4 z= 5/2 or 2.5

2. Given: x=60   µ=60   ó= 4  z=? z= 60-60/4  z=0/4 or 0

3. Given: z= 0.98    µ=70   ó= 4  x=? x= (0.98)(4)+70   =3.92+70 x= 73.92

4. Given: z= 0.98    µ=70   ó= 6  x=? x= (0.98)(6)+70  = 5.88+70 x= 75.88

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5. Given:  z=0.8  x=65  ó= 6  µ=?   .µ= 65-(0.8)(6)    = 65- 4.8   =  60.2

5. Given:  z=0.7  x=69  ó= 6  µ=?   .µ= 69-(0.7)(6)    = 69- 4.2   =  64.8

VI. Find the area under the standard normal curve which lies to the following conditions.. 

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VI. Find the area under the standard normal curve which lies to the following conditions.

SOLUTION: Z=0.70 corresponds to an area of 0.2580. Since it doesn't touch the mean, it suggest subtraction. That is;       P (z>0.70)=0.5 - 0.2580                         =0.2420 Therefore, the area is 0.2420.

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SOLUTION: Z=1.50 corresponds to an area of 0.4332. Since it touch the mean, it suggest addition. That is;       P (z<1.50))=0.5 + 0.4332                         =0.9332 Therefore, the area is 0.9332.

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SOLUTION: Z=-0.50 corresponds to an area of 0.1915 Since it touch the mean, it suggest addition. That is;       P (z>-0.50))=0.5 + 0.1915                         =0.6915 Therefore, the area is 0.6915.

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SOLUTION: Z=-2.50 corresponds to an area of 0.4938 Since it doesn't touch the mean, it suggest subtraction. That is;       P (z>-2.50)=0.5 - 0.4938                         =0.0062 Therefore, the area is 0.0062.

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SOLUTION: Z=-0.70 corresponds to an area of 0.2580 Z= 0.50 corresponds to an area of 0.1915 Since the two z-scores have different signs, it suggest addition. That is;       P (-0.70<z<0.50) = 0.2580+0.1915                                     = 0.4495 Therefore, the area is 0.4495.

VI. PART 2. The salaries of employees of a certain company in Baguio City have a mean of P6,500 and a standard deviation of P1,500. What is the probability that an employee selected at random will have a salary of;. 

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The salaries of employees of a certain company in Baguio City have a mean of P6,500 and a standard deviation of P1,500. What is the probability that an employee selected at random will have a salary of;

A. more than P6,700                B. less than P7,000                           C. Between P5,000 and P7,000

Find the corresponding z-score: z= 6,700-6500/1500 z=200/1500 z= 0.13 Z=0.13 corresponds to an area of 0.0517 Since it doesn't touch the mean, it suggest subtraction. That is;            P(z>0.13)= 0.5- 0.0517                         = 0.4483 or 44.83%

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Find the corresponding z-score: z= 7,000 - 6500/1500 z= 500/1500 z= 0.33 Z=0.33 corresponds to an area of 0.1293 Since it touch the mean, it suggest addition. That is;            P(z<0.33)= 0.5 + 0.1293                         = 0.6293 or 62.93%

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Find the corresponding z-score: X= 5,000                              X= 7,000 z= 5,000 - 6500/1500        z= 7,000-6500/1500 z= -1,500/1500                   z= 500/1500 z= -1                                      z= 0.33 Z= -1 corresponds to an area of 0.3413 Z= 0.33 corresponds to an area of 0.1293 Since the two z-score have a different sign, it suggest addition. That is;            P[-1<z<0.33) = 0.3413 + 0.1293                                    = 0.4706 or 47.06%

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