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
College Algebra (MindTap Course List)
12th Edition
ISBN:9781305652231
Author:R. David Gustafson, Jeff Hughes
Publisher:R. David Gustafson, Jeff Hughes
Chapter8: Sequences, Series, And Probability
Section8.7: Probability
Problem 66E: Determine if the statement is true or false. If the statement is false, then correct it and make it...
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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](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F0c576a2f-357a-470c-9c40-a9828a0be514%2F207b4d22-61fa-403d-b8e8-b34759caae79%2Fv3kq2sj_processed.jpeg&w=3840&q=75)
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
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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
n= 7 random sample of statistics students .
The random variable X has binomial distribution with parameters n= 7 and p = 0.15
The probability mass function of X is
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