Week 5 practice quiz 2

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Liberty University *

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650

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Management

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Feb 20, 2024

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2/13/24, 11:13 AM Quizzes - MGMT 650 9046 Statistics for Managerial Decision Making (2242) - UMGC Learning Management System Question 1 (Mandatory) Attempt Score 49 /93 -52.69% Overall Grade (Highest Attempt) 49 / 93 - 52.69 % / | 7 points According 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% ).490 ) 510 () .542 () .468 W Hide question 1 feedback https://onlinestatbook.com/2/calculators/normal.html -40.5 Specify Parameters: rﬂum!vwr—-—wmmwmnr—m—'"m 16.5 355 545 715 ® Area from a value (Use to compute p from Z) ) Value from an area (Use to compute Z for confidence intervals) Mean |16.5 SD (19 () Above |60 | ® Below [185 ) Between |71 | and |79 | () Outside [-1.96 | and [1.96 | Results: Area (probability) = |0.5419 | | Recalculate | https://learn.umgc.edu/d2l/Ims/quizzing/user/attempt/quiz_start_frame_auto.d21?o0u=930725&isprv=&qi=2119155&cfql=0&dnb=0&fromQB=0&inProgre... 1/7

2/13/24, 11:13 AM Quizzes - MGMT 650 9046 Statistics for Managerial Decision Making (2242) - UMGC Learning Management System =NORM.DIST(18.5, 16.5, 19, 1). Question 2 (Mandatory) 7 | 7 points According 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%? () .531 () .468 () .490 W Hide question 2 feedback https://onlinestatbook.com/2/calculators/normal.html 405 215 25 16.5 35.5 545 735 ® Area from a value (Use to compute p from Z) O Value from an area (Use to compute Z for confidence intervals) Specify Parameters: Mean (165 SD |19 ® Above [17 | ) Below 18.5 () Between |71 and |79 | () Outside [-1.96 | and [1.96 ] Results: Area (probability) = |0.4895 | | Recalculate | =1 - NORM.DIST(17, 16.5, 19, 1) Question 3 (Mandatory) According 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 https://learn.umgc.edu/d2l/Ims/quizzing/user/attempt/quiz_start_frame_auto.d21?ou=930725&isprv=&qi=2119155&cfql=0&dnb=0&fromQB=0&inProgre... / | 7 points 27

2/13/24, 11:13 AM Quizzes - MGMT 650 9046 Statistics for Managerial Decision Making (2242) - UMGC Learning Management System 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%? :) PX > 0) = 29.7% ; P(X<18) =53.2% () PX'> 0) = 86.7%; 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% W Hide question 3 feedback a =1 - NORM.DIST(O, 16.5, 19, 1) b = NORM.DIST(18, 16.5, 19, 1) Question 4 (Mandatory) Match the term to its definition. __2__ Cluster 3 Population ] Stratum Question 5 (Mandatory) /7 |/ 7 points 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. 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. A group of units (persons, objects, or other items) enumerated in a census or from which a sample is drawn. O / 7 points We can eliminate sampling error by selecting an unbiased sample () True () False W Hide question 5 feedback Sampling error cannot be avoided https://learn.umgc.edu/d2l/Ims/quizzing/user/attempt/quiz_start_frame_auto.d21?ou=930725&isprv=&qi=2119155&cfql=0&dnb=0&fromQB=0&inProgre... 3/7

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2/13/24, 11:13 AM Quizzes - MGMT 650 9046 Statistics for Managerial Decision Making (2242) - UMGC Learning Management System Question 6 (Mandatory) 0 / 7 points It is better to try to match the characteristics of the sample to the population rather then relying on randomization ) True ") False W Hide question 6 feedback Randomization will match the characteristics in an unbiased way. Question 7 (Mandatory) 0 / 7 points According to the 68-95-99.7 rule what percent of the population is more than 2 standard deviations away from the mean? _ )68 () 2.5 ()5 O 3> Question 8 (Mandatory) 0 / 7 points According 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? ) 68 ()95 82 ()79 W Hide question 8 feedback = (1/2 x 68.26) + (1/2 x 95.44) Question 9 (Mandatory) 7 | 7 points The sampling distribution of the mean becomes approximately normally distributed when which of the following conditions is met? () The sample size is large. () A'single random sample is drawn from the population. () The population Distribution is not symmetric. () The standard deviation of the population is large. Question 10 (Mandatory) 7 | 7 points https://learn.umgc.edu/d2l/Ims/quizzing/user/attempt/quiz_start_frame_auto.d21?o0u=930725&isprv=&qi=2119155&cfql=0&dnb=0&fromQB=0&inProgre... 4/7

2/13/24, 11:13 AM Quizzes - MGMT 650 9046 Statistics for Managerial Decision Making (2242) - UMGC Learning Management System 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? 0.6 (1.2 0.5 (12 W Hide question 10 feedback standard error = 12/SQRT(100) =12/10 = 1.2 Question 11 (Mandatory) 0 / 7 points What 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 () -1.41 () 1.96 W Hide question 11 feedback https://onlinestatbook.com/2/calculators/normal.html ) Area from a value (Use to compute p from Z) ® Value from an area (Use to compute Z for confidence intervals) Specify Parameters: Area |92 | Mean |0 ] sD [i | Results: | Recalculate | () Above | ® Below [1.405 ) Between | | ) Outside | | =NORM.S.INV(.92) https://learn.umgc.edu/d2l/Ims/quizzing/user/attempt/quiz_start_frame_auto.d21?o0u=930725&isprv=&qi=2119155&cfql=0&dnb=0&fromQB=0&inProgre... 5/7

2/13/24, 11:13 AM Quizzes - MGMT 650 9046 Statistics for Managerial Decision Making (2242) - UMGC Learning Management System Question 12 (Mandatory) 7 | 7 points What are the Z values of the limits of the area covering the middle 25% of the area under the normal curve? . )From -0.319 t0 0.319 . ) From -0.250 to 0.250 ~ ) From -0.674 to0 0.674 _ ) From 0 to 0.250 W Hide question 12 feedback https://onlinestatbook.com/2/calculators/normal.html — 1 T ' T JR— ) Area from a value (Use to compute p from Z) ® Value from an area (Use to compute Z for confidence intervals) Specify Parameters: Area |25 | Mean |0 | sD [1 | Results: | Recalculate | () Above | | ) Below | | ® Between|-0.319 and 0.319 | (' Outside | | The normal distribution is symmetric. 0.5 of the probability is below the mean and 0.5 is above the mean. For the middle 0.25, 0.25/2 is below the mean and 0.25/2 is above the mean. So, the lower limit = NORM.S.INV(0.5-.25/2) and the upper limit =NORM.S.INV(0.5+.25/2) for the upper limit. Question 13 (Mandatory) 0 / 9 points The 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 asked the inspector to use these statistics to find the lower and upper boundary weights of the middle 50% of their cookies. What are the Z values of the middle 50% of the population? https://learn.umgc.edu/d2l/Ims/quizzing/user/attempt/quiz_start_frame_auto.d21?o0u=930725&isprv=&qi=2119155&cfql=0&dnb=0&fromQB=0&inProgre... 6/7

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2/13/24, 11:13 AM Quizzes - MGMT 650 9046 Statistics for Managerial Decision Making (2242) - UMGC Learning Management System () From 0 to 1.00 () From -0.674 to 0.674 () From -1.00 to 1.00 . ) From -0.680 to 0.680 W Hide question 13 feedback https://onlinestatbook.com/2/calculators/normal.html () Area from a value (Use to compute p from Z) ® Value from an area (Use to compute Z for confidence intervals) Specify Parameters: Area |0.50 ] Mean |0 | SD |1 | Results: [ Recalculate ] ) Above | | ) Below| | @ Between |-0.674 and 0.674 | ) Outside | | The normal distribution is symmetric. 0.5 of the probability is below the mean and 0.5 is above the mean. For the middle 0.50, 0.50/2 is below the mean and 0.50/2 is above the mean. So, the lower limit = NORM.S.INV(0.5-.50/2) and the upper limit =NORM.S.INV(0.5+.50/2) for the upper limit. Done https://learn.umgc.edu/d2l/Ims/quizzing/user/attempt/quiz_start_frame_auto.d21?o0u=930725&isprv=&qi=2119155&cfql=0&dnb=0&fromQB=0&inProgre... 7/7

Week 5 practice quiz 2

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