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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 backs You select 7 statistics students at random, with replacement. What is the probability that exactly 3 sleep on their backs? Hint: It is probably easiest to use Minitab or Geogebra to solve this. A P(x=3) = 0.003375 B P(x=3) = 0.0000017 P(x=3) = 0.15 P(x=3) = 0.061662

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 exactly 3 sleep on their backs? Hint: It is probably easiest to use Minitab or Geogebra to solve this. A P(x=3) = 0.003375 B P(x=3) = 0.0000017 P(x=3) = 0.15 P(x=3) = 0.061662

MATLAB: An Introduction with Applications
MATLAB: An Introduction with Applications
6th Edition
ISBN:9781119256830
Author:Amos Gilat
Publisher:Amos Gilat
Chapter1: Starting With Matlab
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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 backs. You select 7 statistics students at random, with replacement. What is the probability that exactly 3 sleep on their backs? Hint: It is probably easiest to use Minitab or Geogebra to solve this. A P(x-3) = 0.003375 P(x=3) = 0.0000017 !3D %3D P(x-3) = 0.15 %3D C P(x-3) =D 0.061662
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 exactly 3 sleep on their backs? Hint: It is probably easiest to use Minitab or Geogebra to solve this. A P(x-3) = 0.003375 P(x=3) = 0.0000017 !3D %3D P(x-3) = 0.15 %3D C P(x-3) =D 0.061662
Expert Solution
Step 1

Solution:

Given information:

31 statistics students sleep on their sides. 

20 statistics students sleep on their stomachs 

9 statistics students sleep on their backs. 

Total statistics students= 60

Let X be the random variable that the number of statistics students sleep on their backs. 

The probability that the  statistics students sleep on their backs is p=960=0.15

n= 7 random sample of statistics students . 

The random variable X has binomial distribution with parameters n= 7 and p = 0.15

X~B(n=7,p=0.15)

The probability mass function of X is 

P(X=x)=nxpx×(1-p)n-x    x=0,1,....,n=7

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