8. On page 8 of the chapter 4 notes, an example found that P(x-2) = 0.31374. Find P(x=3) for this same example.
8. On page 8 of the chapter 4 notes, an example found that P(x-2) = 0.31374. Find P(x=3) for this same example.
Algebra & Trigonometry with Analytic Geometry
13th Edition
ISBN:9781133382119
Author:Swokowski
Publisher:Swokowski
Chapter3: Functions And Graphs
Section3.7: Operations On Functions
Problem 55E
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Hi, I added a picture for reference. Thanks
![8. On page 8 of the chapter 4 notes, an example found that
P(x-2) = 0.31374. Find P(x=3) for this same example.
O 0.31124
0.24999
O 0.31374
O 0.03112](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F85f92341-048d-46df-8884-f7af099bbb6d%2F42855090-4535-4d73-b288-322e919bbfb8%2Fu90kj3k_processed.png&w=3840&q=75)
Transcribed Image Text:8. On page 8 of the chapter 4 notes, an example found that
P(x-2) = 0.31374. Find P(x=3) for this same example.
O 0.31124
0.24999
O 0.31374
O 0.03112
![Solution
The random variable, x, represents the number of females out of this group.
We can apply the binomial probability formula to our original example.
Then n = 5, x = 2, p = 0.498, q = 0.502, and we get
=
P(x = 2) =
n!
x! (n − x)! p* q n-x
5!
2! (5-2)!
=
(.498)²(.502) 5-2
5.4 (3.2.1)
(2.1) (3.2.1)
-(.498)²(.502)5-2
10(.498)²(.502)5-2
= .31374
So... If we select 5 people randomly, there is a 31.4% chance that there will be exactly 2
females selected.
Notice that the answer we found above is the P(x) value in row 3 of the probability table for
counting females in a sample of 5 people.
We could find the other entries in that table by re-calculating each x-value from the table.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F85f92341-048d-46df-8884-f7af099bbb6d%2F42855090-4535-4d73-b288-322e919bbfb8%2Fopdfjy_processed.png&w=3840&q=75)
Transcribed Image Text:Solution
The random variable, x, represents the number of females out of this group.
We can apply the binomial probability formula to our original example.
Then n = 5, x = 2, p = 0.498, q = 0.502, and we get
=
P(x = 2) =
n!
x! (n − x)! p* q n-x
5!
2! (5-2)!
=
(.498)²(.502) 5-2
5.4 (3.2.1)
(2.1) (3.2.1)
-(.498)²(.502)5-2
10(.498)²(.502)5-2
= .31374
So... If we select 5 people randomly, there is a 31.4% chance that there will be exactly 2
females selected.
Notice that the answer we found above is the P(x) value in row 3 of the probability table for
counting females in a sample of 5 people.
We could find the other entries in that table by re-calculating each x-value from the table.
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