BAN 530 - Sam Houston State University



BAN 530

Homework # 3

PLEASE NOTE: The solutions in this handout are presented using the Standardized Normal Distribution (Z) Table found in Appendix 1, page A3 in the back of the Heizer and Render text AND on page 310 of the Evans text. The solutions shown in the handout given in class last Thursday used the cumulative standardized normal distribution table from the texts. The solutions will be the same, however, the approach is different. I apologize for any inconvenience this may have caused. V. Muehsam

1. A sample of eight runs from a production run of 1000 items yielded the following number of defective items: 6, 12, 14, 8, 10, 12, 7, 7. Find the:

a) mean

b) variance

c) standard deviation

d) z-score for a production run with 10 defective items. Interpret.

2. A sample of six new graduates produced the following starting salaries (in $1,000s): 38.5, 42.1, 29.5, 30.1, 45.7, 27.3. Find the:

a) mean

b) variance

c) standard deviation

d) z-score for a starting salary of $32,000. Interpret.

3. A sample of returns for ten portfolios managed by a particular financial manager produced the following returns (in %): 6.5, 12.2, 3.5, 15.6, 10.4, 8.5, 7.8, 0.9, 11.1, 17.3. Find the:

a) mean

b) variance

c) standard deviation

d) z-score for a portfolio with a return of 8.4%. Interpret.

4. Find the z-score with the following area in the left tail:

a) 0.01

b) 0.03

c) 0.05

d) 0.12

5. Find the z-score with the following area in the right tail:

a) 0.025

b) 0.06

c) 0.085

d) 0.15

1. A sample of eight runs from a production run of 1000 items yielded the following number of defective items: 6, 12, 14, 8, 10, 12, 7, 7. Find the:

a. mean

[pic]

b. variance

|x |x2 |[pic] |[pic] |

|6 |36 |-3.5 |12.25 |

|12 |144 |2.5 |6.25 |

|14 |196 |4.5 |20.25 |

|8 |64 |-1.5 |2.25 |

|10 |100 |0.5 |0.25 |

|12 |144 |2.5 |6.25 |

|7 |49 |-2.5 |6.25 |

|7 |49 |-2.5 |6.25 |

|[pic]76 |[pic] |[pic] |[pic] |

[pic]8.57

Using the computational formula the sample variance is:

[pic]

[pic]

c. standard deviation

The standard deviation is: [pic]

d. z-score for a production run with 10 defective items. Interpret.

[pic]

A production run with 10 defective items is 0.17 standard deviations above the mean.

2. A sample of six new graduates produced the following starting salaries (in $1,000s): 38.5, 42.1, 29.5, 30.1, 45.7, 27.3. Find the:

a. mean

[pic]

c. variance

|x |x2 |[pic] |[pic] |

|38.5 |1482.25 |2.97 |8.8209 |

|42.1 |1772.41 |6.57 |43.1649 |

|29.5 |870.25 |-6.03 |36.3609 |

|30.1 |906.01 |-5.43 |29.4849 |

|45.7 |2088.49 |10.17 |103.4289 |

|27.3 |745.29 |-8.23 |67.7329 |

|[pic]213.2 |[pic] |[pic] |[pic] |

[pic]57.80

Using the computational formula the sample variance is:

[pic][pic]

c. standard deviation

The standard deviation is: [pic]

d. z-score for a starting salary of $32,000. Interpret.

[pic]

A starting salary of $32,000 is 0.46 standard deviations below the mean starting salary.

3. A sample of returns for ten portfolios managed by a particular financial manager produced the following returns (in %): 6.5, 12.2, 3.5, 15.6, 10.4, 8.5, 7.8, 0.9, 11.1, 17.3. Find the:

a. mean

[pic]

c. variance

|x |x2 |[pic] |[pic] |

|6.5 |42.25 |-3.72 |13.8384 |

|12.2 |148.84 |1.98 |3.9204 |

|3.5 |12.25 |-6.72 |45.1584 |

|15.6 |243.36 |5.38 |28.9444 |

|10.4 |108.16 |0.18 |0.0324 |

|8.5 |72.25 |-1.72 |2.9584 |

|7.8 |60.84 |-2.42 |5.8564 |

|9.3 |86.49 |-0.92 |0.8464 |

|11.1 |123.21 |0.88 |0.7744 |

|17.3 |299.29 |7.08 |50.1264 |

|[pic]102.2 |[pic] |[pic] |[pic] |

[pic]16.94

Using the computational formula the sample variance is:

[pic][pic]

c. standard deviation

The standard deviation is: [pic]

d. z-score for a portfolio with a return of 8.4%. Interpret.

[pic]

A return of 8.4% is 0.44 standard deviations below the average return.

4. Find the z-score with the following area in the left tail:

e) 0.01 Z = -2.33

f) 0.03 Z = -1.89

g) 0.05 Z = -1.65

h) 0.12 Z = -1.18

5. Find the z-score with the following area in the right tail:

i) 0.025 Z = 1.96

j) 0.06 Z = 1.56

k) 0.085 Z = 1.38

l) 0.15 Z = 1.04

BAN 530

Normal Probability Distribution Examples

Suppose that the amount owed to the IRS (after withholding) at the end of the year is normally distributed with a mean refund of $911 and a standard deviation of $525.

a) What proportion of tax returns show a refund of more than $1500?

b) What proportion of tax returns show a refund of between $100 and $500?

c) What proportion of the tax returns show that the taxpayer owes money to the government?

SOLUTIONS:

X ~ N(911, 5252)

For a) What proportion of tax returns show a refund of more than $1500?

P(X > 1500) = P(Z > 1.12) = .5 - .3686 = .1314

[pic]

For b) What proportion of tax returns show a refund of between $100 and $500?

P(100 < X < 500) = P(-1.54 < Z < -0.78) = .4382 - .2823 = .1559

[pic] [pic]

For c) What proportion of the tax returns show that the taxpayer owes money to the government?

P(X < 0) = P(Z < -1.74) = .5 - .4591 = .0409

[pic]

Ban 530

Normal Probability Distribution Examples

Toolworkers are subject to work-related injuries. One disorder, caused by strains to the hands and wrists, is called carpel tunnel syndrome. This disorder strikes as many as 23,000 workers per year. The U. S. Labor Department estimates that the average cost of this disorder to employees and insurers is approximately normally distributed with a mean of $30,000 and a standard deviation of $9,000.

a) What proportion of the costs fall between $18,000 and $27,000?

b) What cost is exceeded by 12.5% of the costs?

SOLUTIONS:

X ~ N (30,000, 9,0002)

For a) P(18,000 < X < 27,000) = P(-1.33 < Z < -0.33) = .4082 - .1293 = .2789

[pic] [pic]

For b) P(X > X0) = 0.1250

P(Z > Z0) = P(Z > 1.16) = 0.1250

To find the Z value that has 0.1250 to the right—look up .5 – 0.1250 = .3750 and find the Z value that corresponds to this area (Z = 1.16)

[pic] so x = 30,000 + 1.16(9,000) = $40,440

(This solution is different from the handout I gave in class last Thursday. Since we had agreed to always use the Z value further from zero—we should use 1.16 instead of 1.15.

Ban 530

Normal Probability Distribution Examples

Suppose that the amount of money spent by university students at Padre Island during Spring Break is normally distributed with a mean of $63.50 per day and a standard deviation of $12.75.

a) What proportion of the students spend more than $55.00 per day?

b) What proportion of the students spend less than $61.50 per day?

c) What proportion of the students spend between $65.00 and $85.00 per day?

SOLUTIONS:

X ~ N (63.50, 12.752)

For a) P(X > 55.00) = P(Z > -0.67) = .5 + .2486 = .7486

[pic]

For b) P(X < 61.50) = P(Z < -0.16) = .5 -.0636 = .4364

[pic]

For c) P(65.00 < X < 85.00) = P(0.12 < Z < 1.69) = .4545 - .0478 = .4067

[pic] [pic]

BAN 530

Normal Probability Distribution Examples

As the work force is requiring a greater need for computer literacy, human resource managers are using standardized exams to measure a potential employee’s computer ability. Suppose that on one such standardized exams, the scores are normally distributed with a mean of 73 and a standard deviation of 8.

a) What proportion of the scores will be less than 78?

b) A particular human resource manager will offer jobs to only those job seekers who score in the top 10%. What is the minimum score needed before this human resource manager will offer someone a job?

SOLUTIONS:

X ~ N (73, 82)

For a) P(X < 78) = P(Z < 0.63) = .5 + .2357 = .7357

[pic]

For b) P(X < X0) = 0.9000

P(Z < Z0) = P(Z < 1.29) = 0.9000

To find the Z value that has 0.1000 to the right (0.9000 to the left)—look up .5 – 0.1000 = .4000 and find the Z value that corresponds to this area (Z = 1.29)

[pic] so x = 73 + 1.29(8) = 83.32

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