Statistics Fall2013 Class 05 - Instructional Material for Assignment 03 and 04

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Page 1: Statistics Fall2013 Class 05 - Instructional Material for Assignment 03 and 04

8/13/2019 Statistics Fall2013 Class 05 - Instructional Material for Assignment 03 and 04

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Vietnam National University – HCMC Academic Year 2013 – 2014, Semester 01International University Date: Wednesday, October 23, 2013

Page 1 of 3 

COURSE: STATISTICS FOR BUSINESS (BA080IU)

CLASS 05

INSTRUCTIONAL MATERIAL

FOR INDIVIDUAL ASSIGNMENT 03 AND 04

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SOLUTION FOR PROBLEM 3-29:

Let   be the annual income of hedge fund manager without withholding. 

 ′ be the annual income of hedge fund manager with withholding.

Hence,

   =

  −[300+(

  −300)(0.05)] 

Solution 01:

′ (million) () $1,330 0.2

$1,140 0.2

$855 0.3

$665 0.1

$475 0.1

$285 0.05

$95 0.05

By using the calculator, we have:

The standard deviation of hedge fund managers’ income with withholding:

( ) ≈  $351.7567  

The variance of hedge fund managers’ income with withholding:

( ) = [( )] ≈  $123,732.75  

Solution 02: 

By using the calculator, we can find:

The standard deviation of hedge fund managers’ income without withholding:

( ) ≈  $370.2702  

The variance of hedge fund managers’ income without withholding:

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Vietnam National University – HCMC Academic Year 2013 – 2014, Semester 01International University Date: Wednesday, October 23, 2013

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( ) = [( )] ≈  $137,100  

   =  − [300+(  − 300)(0.05)] = 0.95  +285 

The variance of hedge fund managers’ income with withholding:

( ) =( +) = ( ) = (0.95)(137,100) ≈  $123,732.75  

The standard deviation of hedge fund managers’ income with withholding:

( ′) = ( ) ≈  $351.7567  

SOLUTION FOR PROBLEM 3-40:

Let

 be the number of computers in the lab.

 be the number of enrollment ceiling.

 be the probability of reliability of each computer.

a. The probability that each of the 30 students will get a computer in working condition:

(30≤ ≤ 33) =   33  (0.9)(0.1) ≈ 0.5769

 

 b.i. Decreasing the enrollment ceiling.

( ≤ ≤  33) =   33  (0.9)(0.1) ≥ 0.95

 

By using the calculator, we can find that ≈  27. In order to improve the probability that each of

the 30 students will get a computer in working condition, we can need to decrease 3 enrollment

ceiling.

 b.ii. Increasing the number of computers in the lab.

(30≤ ≤ ) =   (0.9)(0.1) ≥  0.95

 

By using the calculator, we can find that ≈ 37. In order to improve the probability that each of

the 30 students will get a computer in working condition, we can need to increase 4 number of

computers in the lab.

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 b.iii. Increasing the reliability of all the computers.

(30≤ ≤ 33) =   33  ()(1− ) ≥  0.95

 

By using the calculator, we can find that ≈ 0.9576. In order to improve the probability that

each of the 30 students will get a computer in working condition, we can need to increase 5.76%

reliability of all the computers.

SOLUTION FOR PROBLEM 3-74:

a. Assuming the advertisement claim is true,  = 2/ 5=0.4 

The probability of the observed event that only 2 of 20 doctors recommended the product:

(   = 2) = 202

(0.4)(0.6) ≈  0.0031 

 b. Assuming the advertisement claim is true

The probability of observing two or fewer successes:

(  ≤ 2) = 20  (0.4)(0.6) ≈  0.0036

 

c. Given the sampling results, the advertisement should not be believed because the observedevent that only 2 of 20 doctors recommended the product would occur with the probability of

2/20=0.1  that would be much lower than the probability of 2/5=0.4  the advertisement

claims.

d. The expected number of successes in a sample of 20 is (20)(0.4) = 8.

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