Question

Five males with an​ X-linked genetic disorder have one child each. The random variable x is the number of children among the five who inherit the​ X-linked genetic disorder. Determine whether a probability distribution is given. If a probability distribution is​ given, find its mean and standard deviation. If a probability distribution is not​ given, identify the requirements that are not satisfied.
x ​P(x)
0 0.029
1 0.147
2 0.324
3 0.324
4 0.147
5 0.029
a. Does the table show a probability distribution?
b. Find the mean of the random variable x.

Answer 1

Answer and Explanation:

Given : Five males with an​ X-linked genetic disorder have one child each. The random variable x is the number of children among the five who inherit the​ X-linked genetic disorder.

To find :

a. Does the table show a probability distribution?

b. Find the mean and standard deviation of the random variable x.

Solution :

a) To determine that table shows a probability distribution we add up all six probabilities if the sum is 1 then it is a valid distribution.

$\sum P(X)=0.029+0.147+0.324+0.324+0.147+0.029$

$\sum P(X)=1$

Yes it is a probability distribution.

b) First we create the table as per requirements,

x      P(x)         xP(x)           x²            x²P(x)

0    0.029         0              0                0

1     0.147        0.147           1              0.147

2    0.324       0.648         4              1.296

3    0.324       0.972         9              2.916

4    0.147        0.588        16              2.352

5    0.029       0.145         25            0.725

   ∑P(x)=1      ∑xP(x)=2.5               ∑x²P(x)=7.436

The mean of the random variable is

$\mu=\sum xP(x)=2.5$

The standard deviation of the random sample is

$Variance=\sigma^2$

$\sigma^2=\sum x^2P(x)-\mu^2$

$\sigma^2=7.436-(2.5)^2$

$\sigma^2=7.436-6.25$

$\sigma^2=1.186$

$\sigma=\sqrt{1.186}$

$\sigma=1.08$

Therefore, The mean is 2.5 and the standard deviation is 1.08.