MGMT650 2021 July Week 5 Quiz Latest

MGMT650 Statistics for Managerial Decision Making

Week 5 Quiz

Question 1According to Investment Digest ("Diversification and the Risk/Reward Relationship", Winter 1994, 1-3), the mean of the annual return for common stocks from 1926 to 1992 was 16.5%, and the standard deviation of the annual return was 19%.

What is the probability that the stock returns are less than 18.5%

.468

.510

.542

.490

Question 2According to Investment Digest ("Diversification and the Risk/Reward Relationship", Winter 1994, 1-3), the mean of the annual return for common stocks from 1926 to 1992 was 16.5%, and the standard deviation of the annual return was 19%.

What is the probability that the stock returns are greater than 17%?

.490

.468

.531

.510

Question 3According to Investment Digest ("Diversification and the Risk/Reward Relationship", Winter 1994, 1-3), the mean of the annual return for common stocks from 1926 to 1992 was 16.5%, and the standard deviation of the annual return was 19%.

a. What is the probability that the stock returns are greater than 0%?

b. What is the probability that the stock returns are less than 18%?

What is the area between the mean and our actual score?

0.308 in part a, 0.039 in part b

0.408 in part a, 0.319 in part b

0.308 in part a, 0.032 in part b

0.408 in part a, 0.039 in part b

Question 4According to Investment Digest ("Diversification and the Risk/Reward Relationship", Winter 1994, 1-3), the mean of the annual return for common stocks from 1926 to 1992 was 16.5%, and the standard deviation of the annual return was 19%.

a. What is the probability that the stock returns are greater than 0%?

b. What is the probability that the stock returns are less than 18%?

P(X > 0) = 29.7% ; P(X<18) =53.2%

P(X > 0) = 19.2%; P(X<18) =46.8%

P(X > 0) = 80.7%; P(X<18) = 53.2%

P(X > 0) = 86.7%; P(X<18) = 53.2%

Question 5Match the term to its definition.

Population

Stratum

Cluster

1.A category composed of people with certain similarities, such as gender, race, religion, or even grade level. Each person belongs to only one group. A simple random sample of equal size is taken from each group.

2.A division of the population into separate physical units such as districts, neighborhoods, or streets.  Each person belongs to only one unit.  A simple random sample is taken from the units.  All people within the selected units are surveyed.

3.A group of units (persons, objects, or other items) enumerated in a census or from which a sample is drawn.

Question 6We can eliminate sampling error by selecting an unbiased sample

                True

                False

Question 7It is better to try to match the characteristics of the sample to the population rather then relying on randomization

                True

                False

Question 8According to the 68-95-99.7 rule what percent of the population are more than 2 standard deviations away from the mean?

5

68

2.5

95

Question 9According to the 68-95-99.7 rule what percent of the population are between 1 standard deviation below the mean and 2 standard deviations above the mean?

82

95

79

68

Question 10        The sampling distribution of the mean becomes approximately normally distributed when which of the following conditions is met?

The population Distribution is not symmetric.

The sample size is large.

The standard deviation of the population is large.

A single random sample is drawn from the population.

Question 11        You select a sample of 100 and find a mean of 60 and a standard deviation of 12.  What is the standard error of the sampling distribution of sample means?

1.2

0.6

12

0.5

Question 12What is the Z score where 92% of the normal curve is below that z score and 8% of the normal curve lies above that same z score?

1.41

1.96

-1.96

-1.41

Question 13What are the Z values of the limits of the area covering the middle 25% of the area under the normal curve?

From -0.674 to 0.674

From -0.319 to 0.319

From -0.250 to 0.250

From 0 to 0.250

Question 14The Cheebles cookie factory changed their recipe.  The inspectors took a sample of the new cookies and found that the mean weight of the sample was 42 grams with a standard deviation of 4 grams.  The Cheebles CEO specifically asked the inspectors to use these statistics to find the lower and upper boundary weights of 50% of their cookies

What are the Z values of the limits of the area covering the middle half of the area under the normal curve that the inspectors would use to find this information for the CEO of Cheebles?

From 0 to 1.00

From -0.674 to 0.674

From -0.680 to 0.680

From -1.00 to 1.00