Week 5 practice quiz 3

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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, 12:34 PM Quizzes - MGMT 650 9046 Statistics for Managerial Decision Making (2242) - UMGC Learning Management System Attempt Score 10/ 11 -90.91 % Overall Grade (Highest Attempt) 10/ 11 - 90.91 % Question 1 1 /1 point If the population mean is 57 and the population standard deviation is 7, what is the z value for 63?7 Round to three decimal places. Answer: 0.857 v/ Ww Hide question 1 feedback Z = (number - mean) / standard deviation Question 2 1 /1 point What is the Excel function used to find the area from the left side of a standard normal graph to the Z value? ) NORM.DIST() ) NORM.INV() ) NORM.S.INV() v ) NORM.S.DIST() Question 3 1 /1 point Kyoko gave Misa a box of marbles for her birthday. If the average number of marbles in the brand that Kyoko gave to Misa is 75 with a standard deviation of 4, what is the probability that Misa received more than 77 marbles? ) 6877 v ) .3085 ) .6915 ~).3195 W Hide question 3 feedback https://onlinestatbook.com/2/calculators/normal.html https://learn.umgc.edu/d2l/Ims/quizzing/user/attempt/quiz_start_frame_auto.d21?o0u=930725&isprv=&qi=2119154 &cfql=0&dnb=0&fromQB=0&inProgre... 1/9

2/13/24, 12:34 PM Quizzes - MGMT 650 9046 Statistics for Managerial Decision Making (2242) - UMGC Learning Management System 8 67 7 75 70 83 §7 * Area from a value (Use to compute p from Z) ~ Value from an area (Use to compute Z for confidence intervals) Specify Parameters: Mean (75 | sofe ] ® Bbove |77 ' Below 18 . Between 1056 and |1.96 l Outside |-1.68 and |1.98 [ Results: Area (probability) = =1 - NORM.DIST(77,75,4,1) After finding the z value=(77-75) / 4, probability is =1-NORM.S.DIST(z,1) to find the probability. Question 4 1 /1 point Kyoko gave Misa a box of marbles for her birthday. If the average number of marbles in the brand that Kyoko gave to Misa is 75 with a standard deviation of 4, what is the probability that Misa received less than 79 marbles? O 85.66% ) 17.53% v ) .8413 ) .1587 W Hide question 4 feedback https://onlinestatbook.com/2/calculators/normal.html https://learn.umgc.edu/d2l/Ims/quizzing/user/attempt/quiz_start_frame_auto.d21?o0u=930725&isprv=&qi=2119154 &cfql=0&dnb=0&fromQB=0&inProgre... 2/9

2/13/24, 12:34 PM Quizzes - MGMT 650 9046 Statistics for Managerial Decision Making (2242) - UMGC Learning Management System 63 67 1 75 79 83 87 ® Area from a value (Use to compute p from Z) ) Value from an area (Use to compute Z for confidence intervals) Specify Parameters: Mean |75 SD |4 () Above |77 | ® Below 79 () Between |-1.96 and |1.96 () Outside [-1.96 and [1.96 | Results: Area (probability) = [0.8413 | | Recalculate | =NORM.DIST(79, 75, 4, 1) Question 5 1 /1 point Kyoko gave Misa a box of marbles for her birthday. If the average number of marbles in the brand that Kyoko gave to Misa is 75 with a standard deviation of 4, what is the probability that Misa received more than 78 marbles or less than 74 marbles? () .4013 () .2266 v ) 6279 () .372] W Hide question 5 feedback https://onlinestatbook.com/2/calculators/normal.html https://learn.umgc.edu/d2l/Ims/quizzing/user/attempt/quiz_start_frame_auto.d21?o0u=930725&isprv=&qi=2119154&cfql=0&dnb=0&fromQB=0&inProgre... 3/9

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2/13/24, 12:34 PM Quizzes - MGMT 650 9046 Statistics for Managerial Decision Making (2242) - UMGC Learning Management System A \& 63 67 71 75 79 83 87 ® Area from a value (Use to compute p from Z) () Value from an area (Use to compute Z for confidence intervals) Specify Parameters: Mean |75 | SD |4 | @ O O) O Above | 1.96 | Below 1.96 Between |-1.96 | and |1.96 [ Outside |74 | and |78 | Results: Area (probability) = |0.6279 liRecalculateﬂi P((X < 74) or (X > 78) Simple Addition for mutually exclusive events = PX < 74)) + P(X > 78) =NORM.DIST(74, 75, 4, 1) + (1-NORM.DIST(78, 75, 4, 1)) Question 6 1 /1 point Kyoko gave Misa a box of marbles for her birthday. If the average number of marbles in the brand that Kyoko gave to Misa is 75 with a standard deviation of 4, what is the approximate probability that Misa received between 71 and 79 marbles? v ) 0.68 . 0.95 ) 0.997 W Hide question 6 feedback https://onlinestatbook.com/2/calculators/normal.html https://learn.umgc.edu/d2l/Ims/quizzing/user/attempt/quiz_start_frame_auto.d21?o0u=930725&isprv=&qi=2119154&cfql=0&dnb=0&fromQB=0&inProgre... 4/9

2/13/24, 12:34 PM Quizzes - MGMT 650 9046 Statistics for Managerial Decision Making (2242) - UMGC Learning Management System 63 67 7 75 79 83 87 ® Area from a value (Use to compute p from Z) ) Value from an area (Use to compute Z for confidence intervals) Specify Parameters: Mean |75 SD |4 ) Above |77 | () Below 79 ® Between |71 and |79 | () Outside [-196 | and [1.96 | Results: Area (probability) = |0.6827 | | Recalculate | P(71 <X <79)=PX <79 -PX<71) = NORM.DIST(79, 75, 4, 1) - NORM.DIST(71, 75, 4, 1) Question 7 1 /1 point A pizza delivery vendor has a goal of delivering all orders within 50 minutes. A sample was taken and found a mean of 40 minutes and a sample standard deviation of 6 minutes. Assuming the distribution is normal, what percent of the population did not meet the goal of 50 minutes? . 1.80% () 2.39% () 9.56% v ) 4.78% W Hide question 7 feedback https://onlinestatbook.com/2/calculators/normal.html https://learn.umgc.edu/d2l/Ims/quizzing/user/attempt/quiz_start_frame_auto.d21?o0u=930725&isprv=&qi=2119154&cfql=0&dnb=0&fromQB=0&inProgre... 5/9

2/13/24, 12:34 PM Quizzes - MGMT 650 9046 Statistics for Managerial Decision Making (2242) - UMGC Learning Management System ey 22 Mean SD 28 34 40 46 ® Area from a value (Use to compute p from Z) ) Value from an area (Use to compute Z for confidence intervals) Specify Parameters: 40 6 ® Above (U Below () Between () Outside |-1.96 | and [1.96 150 | 79 71 52 38 and |79 | Results: Area (probability) = [0.0478 | | Recalculate | P(not(X < 50))=1-P(X < 50) =1 - NORM.DIST(50, 40, 6, 1) Question 8 1 /1 point A company sampled the driving distance of employees from their homes to the office. The sample found a mean of 35 minutes with a standard deviation of 20 minutes. What percent of the employees drive more than one hour (60 minutes)? ) 5.84% \/Q 10.56% ©)11.80% O 12.28% W Hide question 8 feedback https://onlinestatbook.com/2/calculators/normal.html https://learn.umgc.edu/d2l/Ims/quizzing/user/attempt/quiz_start_frame_auto.d21?o0u=930725&isprv=&qi=2119154&cfql=0&dnb=0&fromQB=0&inProgre... 6/9

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2/13/24, 12:34 PM Quizzes - MGMT 650 9046 Statistics for Managerial Decision Making (2242) - UMGC Learning Management System | -25 15 Mean 35 SD 20 ® ZAbove () Below (' Between () Outside [-1.96 35 55 Specify Parameters: 60 | 79 71 ® Area from a value (Use to compute p from Z) ) Value from an area (Use to compute Z for confidence intervals) and |79 | and 196 Resul ts: | Recalculate | Area (probability) = |0.1056 | P(X>60)=1-PX <60)=1-NORM.DIST(60, 35, 20, 1) Question 9 1 /1 point A fast food restaurant is trying to lure customers to wait in line even if the line is long by offering a 50% discount if you wait more than a specified number of minutes. A sample found that customers waited a mean time of 12 minutes with s sample standard deviation of 3 minutes. If they want to set the specified number of minutes such that only 4% of the population will wait more than that number, what number should they set it to (round to a whole number). () 12 minutes () 14 minutes v/ ) 17 minutes () 19 minutes Ww Hide question 9 feedback Easy Way https://onlinestatbook.com/2/calculators/normal.html https://learn.umgc.edu/d2l/Ims/quizzing/user/attempt/quiz_start_frame_auto.d21?o0u=930725&isprv=&qi=2119154&cfql=0&dnb=0&fromQB=0&inProgre... 7/9

2/13/24, 12:34 PM Quizzes - MGMT 650 9046 Statistics for Managerial Decision Making (2242) - UMGC Learning Management System 3 6 9 12 15 18 21 ) Area from a value (Use to compute p from Z) ® Value from an area (Use to compute Z for confidence intervals) Specify Parameters: Area |04 | Mean |12 | SD 3 ] Results: | Recalculate | @® Above [17.252 ) Below | ) Between | | () Outside | ] Still Easy P(X > wait time limit) = 4% then P(X < wait time limit) = 96% and X = NORM.INV(0.96,12,3) = 17.2521 Not as Easy P(X < wait time limit) = 96% P(Z < (wait time-mean)/sd) = 96% What is the Z-score for 96%? Z = NORM.S.INV(.96) = 1.7507 z=1.7507 = (x-mean)/sd = (x-12)/3 3(1.7507) = x - 12 x = 3(1.7507) + 12 Question 10 0/ 1 point A survey of 113 houses in a neighborhood found that 0.6 of them have a basement. What is the standard error of the sampling distribution? Use three decimal digits. Answer: 0.056 X (0.046) W Hide question 10 feedback With n the sample size and p the sample proportion, the standard error is = SQRT( p(1-p)/n) https://learn.umgc.edu/d2l/Ims/quizzing/user/attempt/quiz_start_frame_auto.d21?o0u=930725&isprv=&qi=2119154&cfql=0&dnb=0&fromQB=0&inProgre... 8/9

2/13/24, 12:34 PM Quizzes - MGMT 650 9046 Statistics for Managerial Decision Making (2242) - UMGC Learning Management System Question 11 1 /1 point A company is interested in how many miles their employees drive to the office. A sample of 160 employees found a mean of 33 minutes and a sample standard deviation of 10 minutes. What is the standard error of the mean of the sampling distribution? Use three decimal digits. Answer: 0.791 v W Hide question 11 feedback With n the sample size and s the sample standard deviation, the standard error = s/SQRT(n) Done https://learn.umgc.edu/d2l/Ims/quizzing/user/attempt/quiz_start_frame_auto.d21?o0u=930725&isprv=&qi=2119154&cfql=0&dnb=0&fromQB=0&inProgre... 9/9

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Week 5 practice quiz 3

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