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s tan dard deviation(a) = 348 s tan dard deviation(o) = 112 Required probability is P x< 650 Required probability is P( X > 100 -() = P 650–p 100- = P( 650-789 100-70 Z > 348 112 = P(Z < -2. 08) = P(Z > 1.34) from Z-table = 0. 0901 from Z– table = 0.0188 The probability that the mean price is less than 650 is 0.0188. b) Given, sample size(n) = 42 mean(u) = 10 s tan dard deviation(6) = 13 Required probability is P < 15 = P -- ) - r(2- = P 10 < Z < 15- 15-10 = P(-0. 50 < Z < 2. 49) -0. 50 = 0. 9936 – 0. 3085 (from Z – table = 0. 6851 The probability that the mean no.of hours a week spent playing video games is between 9 and 15 is 0. 6851. II

s tan dard deviation(a) = 348 s tan dard deviation(o) = 112 Required probability is P x< 650 Required probability is P( X > 100 -() = P 650–p 100- = P( 650-789 100-70 Z > 348 112 = P(Z < -2. 08) = P(Z > 1.34) from Z-table = 0. 0901 from Z– table = 0.0188 The probability that the mean price is less than 650 is 0.0188. b) Given, sample size(n) = 42 mean(u) = 10 s tan dard deviation(6) = 13 Required probability is P < 15 = P -- ) - r(2- = P 10 < Z < 15- 15-10 = P(-0. 50 < Z < 2. 49) -0. 50 = 0. 9936 – 0. 3085 (from Z – table = 0. 6851 The probability that the mean no.of hours a week spent playing video games is between 9 and 15 is 0. 6851. II

A First Course in Probability (10th Edition)
A First Course in Probability (10th Edition)
10th Edition
ISBN:9780134753119
Author:Sheldon Ross
Publisher:Sheldon Ross
Chapter1: Combinatorial Analysis
Section: Chapter Questions
Problem 1.1P: a. How many different 7-place license plates are possible if the first 2 places are for letters and...
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READ THE INSTRUCTIONS. ILLUSTRATE MEANS TO DRAW THE GRAPH NOT ANSWER IT.I ALREADY HAVE THE SOLUTION I ONLY NEED THE GRAPH. UPVOTE FOR TYPEWRITTEN ONLY.
ONLY ILLUSTRATE THE DISTRIBUTION OF MEAN. DO THIS TYPEWRITTEN FOR UPVOTE. NO UPVOTE FOR HANDWRITTEN. THANK YOU a) Given, c) Given, sample size(n) = 27 mean(µ) = 789 s tan dard deviation(6) = 348 sample size(n) = 25 mean(u) = 70 s tan dard deviation(o) = 112 Required probability is P( x < 650 Required probability is P( x > 100 -() 650-u = P 100-p = P = P| Z < 650-789 348 100-70 112 = P( Z > = P(Z < -2. 08) = P(Z > 1.34) = 0. 0188 from Z – table = 0, 0901 from Z – table The probability that the mean price is less than 650 is 0. 0188. b) Given, sample size(n) = 42 теan(и) 3D 10 s tan dard deviation(6) = 13 Required probability is P 9 < i < 15 9-u 15-u = = P 9-10 <Z< 15-10 13 = P(-0.50 < Z < 2.49) = P Z < 2.49 - Z< -0. 50 = 0. 9936 – 0. 3085 (from Z – table = 0. = 0., 6851 The probability that the mean no.of hours a week spent playing video games is between 9 and 15 is 0. 6851.
Transcribed Image Text:ONLY ILLUSTRATE THE DISTRIBUTION OF MEAN. DO THIS TYPEWRITTEN FOR UPVOTE. NO UPVOTE FOR HANDWRITTEN. THANK YOU a) Given, c) Given, sample size(n) = 27 mean(µ) = 789 s tan dard deviation(6) = 348 sample size(n) = 25 mean(u) = 70 s tan dard deviation(o) = 112 Required probability is P( x < 650 Required probability is P( x > 100 -() 650-u = P 100-p = P = P| Z < 650-789 348 100-70 112 = P( Z > = P(Z < -2. 08) = P(Z > 1.34) = 0. 0188 from Z – table = 0, 0901 from Z – table The probability that the mean price is less than 650 is 0. 0188. b) Given, sample size(n) = 42 теan(и) 3D 10 s tan dard deviation(6) = 13 Required probability is P 9 < i < 15 9-u 15-u = = P 9-10 <Z< 15-10 13 = P(-0.50 < Z < 2.49) = P Z < 2.49 - Z< -0. 50 = 0. 9936 – 0. 3085 (from Z – table = 0. = 0., 6851 The probability that the mean no.of hours a week spent playing video games is between 9 and 15 is 0. 6851.
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