ONLY ILLUSTRATE THE DISTRIBUTION OF MEAN. DO THIS TYPEWRITTEN FOR UPVOTE, NO UPVOTE FOR HANDWRITTEN, THANK YOU Given, Given, a) sample size(n) = 27 mean(u) = 789 s tan dard deviation(a) = 348 sample size(n) = 25 mean(u) = 70 s tan dard deviation(o) = 112 Required probability is P( x < 650 Required probability is X > 100 -) 650-p = P 100- = P = P Z < 650–789 100-70 = P Z > = 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( 9
ONLY ILLUSTRATE THE DISTRIBUTION OF MEAN. DO THIS TYPEWRITTEN FOR UPVOTE, NO UPVOTE FOR HANDWRITTEN, THANK YOU Given, Given, a) sample size(n) = 27 mean(u) = 789 s tan dard deviation(a) = 348 sample size(n) = 25 mean(u) = 70 s tan dard deviation(o) = 112 Required probability is P( x < 650 Required probability is X > 100 -) 650-p = P 100- = P = P Z < 650–789 100-70 = P Z > = 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( 9
MATLAB: An Introduction with Applications
6th Edition
ISBN:9781119256830
Author:Amos Gilat
Publisher:Amos Gilat
Chapter1: Starting With Matlab
Section: Chapter Questions
Problem 1P
Related questions
Question
TYPEWRITTEN ONLY PLEASE UPVOTE. DOWNVOTE FOR HANDWRITTEN. DO NOT ANSWER IF YOU ALREADY ANSWERED THIS
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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