Exam 2



Exam 2

DS-203

Fall 2004

Name___________________

Show all your work.

1. In a sodium dietary study, the initial systolic blood pressure (in mm Hg) for all female participants is found to have a mean value of 120.5 mm and a standard deviation of 12.4 mm. Let x denote the initial systolic blood pressure for a female participant in this study.

a. If the distribution of x is normal and a sample of n = 16 female subjects is randomly selected, what is the probability that the sample mean (x is between 115.5 mm and 123.5 mm?

P (11.5 < [pic] < 123.5) = P (115.5-120.5 < 123.5-120.5)

12.4 / √16 12.4 / √16

= P (-1.61 < z < .97) = .8340 - .0537 = .78

b. What is the probability that the sample mean is at most 125.5 mm?

P ( [pic] ≤ 125.5 ) = P( z < 125.5 -120.5) = P ( z < 1.61)

12.4 / √16

= .9463

c. What is the initial blood pressure (in mmHg) of the upper 5% of all the female participants?

.95

5% 5%

[pic] ___________________________ ______________________ z

120.5 0

From z-table

Z .95 = 1.645

Z = x-µ

[pic]

1.645 = x - 120.5

12.4

x = (1.645)( 12.4) + 120.5

x = 140.898

2. You are planning a survey of starting salaries for recent business major graduates from your college. From a pilot study you estimate that the standard deviation is about $8000. What sample size do you need to have a margin of error equal to $500 with 95% confidence?

[pic] = $ 8000 Margin of error = Z* __[pic]__

[pic]

Z * For 95% Confidence = 1.96

8000

500 = 1.96 [pic]

500 [pic]= 1.96 (8000)

[pic]= 15680 = 31.36

500

n = 983.45 [pic]984

3. You want to rent an unfurnished one-bedroom apartment for next semester. The mean monthly rent for a random sample of 100 apartments advertised in the local newspaper is $540. Assume that the standard deviation is $80. Find a 98% confidence interval for the mean monthly rent for unfurnished one-bedroom apartments available for rent in this community.

n = 100 Z* for 98% confidence interval = 2.33

[pic]= 540

[pic]= 80

[pic] [pic] Z* __[pic]__

[pic]

__80__

540 [pic] 2.33 [pic]

540 [pic] (2.33) (8)

540 [pic]18.64

(521.36 558.64)

4. A large manufacturing plant has averaged 7 “reported accidents’ per month. Suppose that accident counts over time follow a poison distribution with mean 7 per month.

a) What is the probability of 7 accidents in a month?

P( x = k) = e[pic]µ[pic] P( x = 7) = e[pic]7[pic] = (0.0009119)(823543)

k! 7! 5040

= 750.974 = .149 5040

b) What is the probability of at leas 2 accidents in a month?

P ( x [pic] 2) = 1- [ P (x = 0) + P (x = 1)]

= 1 – [e[pic]7[pic] + e[pic]7[pic]] = 1 – (.0009119 + .0063832)

0! 1!

= 1 - .0072951 = .9927

5. You have two scales for measuring weight. Both scales give answers that vary a bit in repeated weightings of the same item. If the true weight of an item is 2 grams (g), the first scale produces readings X that have mean 2.000 g and standard deviation .002 g. The second scale’s readings Y have mean 2.001g and standard deviation .001 g.

a) What are the mean and standard deviation of the difference Y-X between the readings? (The readings X and Y are independent.)

[pic]= [pic]-[pic]= 2.001-2 = .001

[pic] [pic][pic] = [pic] [pic][pic]+ [pic] [pic][pic] = (.001)[pic]+ (.002)[pic] = .000005

[pic] [pic]= [pic] = .0022

b) You measure once with each scale and average the readings. Your result is Z = (X+Y)/2. What are (z and (z?

[pic] = [pic] [pic]+ [pic]= [pic]( 2 + 2.001) = 2.0005

[pic] [pic][pic] = ([pic])[pic][pic][pic][pic] = ([pic])[pic] ( .0022) [pic]

[pic] [pic] = [pic] ( .0022) = .0011

c) Is the average Z more or less variable than the reading Y of the less variable scale?

More variable .0011 vs. .001

Multiple Choice Questions

Select the best answer.

1. A sample of size 5 is to be selected from a population whose distribution is normal.

a. The (x sampling distribution will be normal. A

b. We can’t assume that the(x sampling distribution will be normal.

c. None of the above.

2. A sample of size 50 is to be selected from a population whose distribution is positively skewed.

a. The (x sampling distribution will be approximately normal. A

b. We can’t assume that the(x sampling distribution will be normal.

c. None of the above.

3. For which of the following sample sizes will the standard deviation of the (x distribution be the smallest?

a. n = 5 C

b. n = 10

c. n = 20

d. It will be the same for all.

4. If a population has a standard deviation (, then the standard deviation of mean of 100 randomly selected items from this population is

a) ( C

b) 100(

c) (/10

d) (/100

e) 0.1

Use the following problem for questions 5-7.

Let X denotes the number of siblings for a college professor at a particular university. Suppose that X has the following probability distribution.

X 0 1 2 3 4 5 6

P(X) .2 .3 .15 .13 .12 .08 .02

5. What proportion of professors at this university are only children?

a) .5 b) .2 c) .65 d) .73 e) None of these B

6. What is the probability that a randomly selected professor has two or fewer siblings?

a) .5 b) .2 c) .65 d) .73 e) None of these C

7. What is the probability that a randomly selected professor comes from a family with 2 or fewer children?

a) .5 b) .2 c) .65 d) .73 e) None of these A

Use the following problem for questions 8 and 9.

Parking violations in a particular city result in a fine of either $5, $10, or $25, depending on the type of violation. Suppose that X = amount of fine has the following probability distribution.

X 5 10 25

P(X) .2 .70 .10

8. The mean value of X is. B

a) 5.22 b) 10.50 c) 27.25 d) 6300 e) None of these

9. The standard deviation of X is.

a) 5.22 b) 10.50 c) 27.25 d) 6300 e) None of these A

10. Which of the following variables is not discrete?

a) The number of students absent from a class. D

b) The number in a group of 20 people who have college degrees.

c) The number of students in a class who earned an A on the midterm exam.

d) The dept of a crack in dry clay.

e) None of the above.

11. A random sample is to be selected from a population with mean ( = 100 and standard deviation ( = 20. Determine the mean of the (x distribution for a sample of size 16.

a. ( = 100 A

b. ( = 100/16

c. ( = 100/4

d. None of the above

12. A random sample is to be selected from a population with mean ( = 100 and standard deviation ( = 20. Determine the standard deviation of the (x distribution for a sample of size 16.

a. ( = 20 C

b. ( = 20/16

c. ( =20/4

d. None of the above

[pic] = [pic]P (x) = 5 (.2) + 10 (.7) + 25 (.1) = 10.5

[pic]

[pic] = [pic]= [pic]

= [pic] = 5.22

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