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
Glencoe Algebra 1, Student Edition, 9780079039897, 0079039898, 2018
18th Edition
ISBN:9780079039897
Author:Carter
Publisher:Carter
Chapter10: Statistics
Section10.4: Distributions Of Data
Problem 19PFA
Related questions
Question
100%
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.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F052d18b6-9bbc-49dd-8c86-267b553be472%2F8af58b36-3dd1-4b5d-b6d7-d45f49133918%2Fkn0e9pa_processed.jpeg&w=3840&q=75)
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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