SOLUTION: Suppose that 20,000 married adults in a country were randomly surveyed as to the number of children they have. The results are compiled and are used as theoretical probabilities. L
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Question 1206559: Suppose that 20,000 married adults in a country were randomly surveyed as to the number of children they have. The results are compiled and are used as theoretical probabilities. Let X = the number of children married people have.
x P(x) xP(x)
0 0.15
1 0.20
2 0.35
3
4 0.10
5 0.05
6 (or more) 0.05
(a) Find the probability that a married adult has three children. (Enter your answer to two decimal places.)
(c) Find the expected value. (Enter your answer to two decimal place.)
Answer by math_tutor2020(3817) (Show Source): You can put this solution on YOUR website!
X = number of children married people have
x | P(x) |
0 | 0.15 |
1 | 0.20 |
2 | 0.35 |
3 | ???? |
4 | 0.10 |
5 | 0.05 |
6 (or more) | 0.05 |
The two key rules for a probability distribution are:- Each P(x) item must be between 0 and 1.
- All of the P(x) values must add to 1.
So,
P(0)+P(1)+P(2)+P(3)+P(4)+P(5)+P(6) = 1
0.15+0.20+0.35+P(3)+0.10+0.05+0.05 = 1
0.9+P(3) = 1
P(3) = 1 - 0.9
P(3) = 0.1
The probability a married adult has exactly 3 kids is 0.1
This is what the completed table will look like
x | P(x) |
0 | 0.15 |
1 | 0.20 |
2 | 0.35 |
3 | 0.10 |
4 | 0.10 |
5 | 0.05 |
6 (or more) | 0.05 |
--------------------------------------------------------------------------
Now to fill out the column labeled x*P(x).
As the name of this column suggests, we'll multiply each paired x and P(x) value.
Eg: 0*0.15 = 0 and 1*0.20 = 0.20
x | P(x) | x*P(x) |
0 | 0.15 | 0 |
1 | 0.20 | 0.20 |
2 | 0.35 | 0.70 |
3 | 0.10 | 0.30 |
4 | 0.10 | 0.40 |
5 | 0.05 | 0.25 |
6 (or more) | 0.05 | 0.30 |
The "6 (or more)" entry at the bottom is treated simply as x = 6. I wasn't sure how to handle that portion.
Spreadsheet software is recommended. But since the table is fairly small, doing it by hand isn't that bad.
Add up the values in the x*P(x) column to arrive at the expected value 2.15
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