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
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ISBN:9781133382119
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Chapter3: Functions And Graphs
Section3.7: Operations On Functions
Problem 55E
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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
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.
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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