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In a group of 60 statistics students, 31 sleep on their sides, 20 sleep on their stomachs, and 9 sleep on their bac: You select 7 statistics students at random, with replacement. What is the probability that more than 3 sleep on th backs? Hint: It is probably easiest to use Minitab/Geogebra to solve this. P(x<3) = 0.92623 P(x>3) = 0.01210 P(x<=3) = 0.9879 P(x>=3) = 0.07377

In a group of 60 statistics students, 31 sleep on their sides, 20 sleep on their stomachs, and 9 sleep on their bac: You select 7 statistics students at random, with replacement. What is the probability that more than 3 sleep on th backs? Hint: It is probably easiest to use Minitab/Geogebra to solve this. P(x<3) = 0.92623 P(x>3) = 0.01210 P(x<=3) = 0.9879 P(x>=3) = 0.07377

MATLAB: An Introduction with Applications
MATLAB: An Introduction with Applications
6th Edition
ISBN:9781119256830
Author:Amos Gilat
Publisher:Amos Gilat
Chapter1: Starting With Matlab
Section: Chapter Questions
Problem 1P
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Please see the attachment below. I need help with this please and thank you. 

In a group of 60 statistics students, 31 sleep on their sides, 20 sleep on their stomachs, and 9 sleep on their backs. You select 7 statistics students at random, with replacement. What is the probability that more than 3 sleep on their backs? Hint: It is probably easiest to use Minitab/Geogebra to solve this. P(x<3) = 0.92623 P(x>3) = 0.01210 %3D C P(x<=3) = 0.9879 P(x>=3)% 3D 0.07377 %3D
Transcribed Image Text:In a group of 60 statistics students, 31 sleep on their sides, 20 sleep on their stomachs, and 9 sleep on their backs. You select 7 statistics students at random, with replacement. What is the probability that more than 3 sleep on their backs? Hint: It is probably easiest to use Minitab/Geogebra to solve this. P(x<3) = 0.92623 P(x>3) = 0.01210 %3D C P(x<=3) = 0.9879 P(x>=3)% 3D 0.07377 %3D
Expert Solution
Step 1

Solution:

Given information:

Total statistics students =60

31 statistics students sleep on their sides

20 statistics students sleep on their stomachs. 

and 9 statistics students sleep on their backs. 

To find the probability of statistics students sleep on their backs. 

Probability = p=Number of statistics students sleep on their backsTotal number of statistics students Probability = p=960=0.15

The probability of statistics students sleep on their backs is 0.15 

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