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 DISTRIBUTIONOF MEAN. DO THIS TYPEWRITTEN FORUPVOTE. NO UPVOTE FORHANDWRITTEN. THANK YOUa) Given,c)Given,sample size(n) = 27mean(µ) = 789s tan dard deviation(6) = 348sample size(n) = 25mean(u) = 70s tan dard deviation(o) = 112Required probability is P( x < 650Required probability is P( x > 100-()650-u= P100-p= P= P| Z <650-789348100-70112= P( Z >= P(Z < -2. 08)= P(Z > 1.34)= 0. 0188from Z – table= 0, 0901from Z – tableThe probability that the mean price is less than 650is0. 0188.b) Given,sample size(n) = 42теan(и) 3D 10s tan dard deviation(6) = 13Required probability is P9 < i < 159-u15-u== P 9-10

It is given that the sample size, mean and standard deviation of each part. Here, need to draw the…

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)

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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