Solved: Genetics. In Exercises 1518, refer to the accompanying table, which describes

Chapter 5, Problem 15

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QUESTION:

Genetics. In Exercises 1518, refer to the accompanying table, which describes results from groups of 10 births from 10 different sets of parents. The random variable x represents the number of girls among 10 children. Number of Girls x P(x) 0 0.001 1 0.010 2 0.044 3 0.117 4 0.205 5 0.246 6 0.205 7 0.117 8 0.044 9 0.010 10 0.001 Mean and Standard Deviation Find the mean and standard deviation for the numbers of girls in 10 births.

Questions & Answers

QUESTION:

ANSWER:

Problem 15

Genetics. In Exercises 15-18, refer to the accompanying table, which describes results from groups of 10 births from 10 different sets of parents. The random variable x represents the number of girls among 10 children.

Mean and Standard Deviation. Find the mean and standard deviation for the numbers of girls in 10 births.

Step by Step Solution

Step 1 of 2

Mean for a probability distribution is,

Mean,

Construct a table as shown below:

x	P(x)	x.P(x)
0	0.001	0 x 0.001 = 0
1	0.01	1 x 0.01 = 0.01
2	0.044	2 x 0.044 = 0.088
3	0.117	3 x 0.117 = 0.351
4	0.205	4 x 0.205 = 0.82
5	0.246	5 x 0.246 = 1.23
6	0.205	6 x 0.205 = 1.23
7	0.117	7 x 0.117 = 0.819
8	0.044	8 x 0.044 = 0.352
9	0.01	9 x 0.01 = 0.09
10	0.001	10 x 0.001 = 0.01