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 0.032
1 0.152
2 0.316
3 0.316
4 0.152
5 0.032

Answer 1

I believe the correct answer is: Mean = 2.5 and S.D = 1.1145

Explanation:

Since we are given the probability distribution as below, we will go ahead and calculate the mean and standard deviation using the following formula:

E(X)= ∑xP(X=x)

Standard deviation = √Variance(x)

Var(x) = E(x²) - [E(x)]²

X 0 1 2 3 4 5

p(x=x) 0.032 0.152 0.316 0.316 0.152 0.032

First we have to find E(x)

E(x) = (0x0.032) + (1x0.152) + (2x0.316) + (3x0.316) + (0.152) + (5x0.032)

The answer therefore the mean E(x) = 2.5

To find variance, we use Var(x) = E(x²) - [E(x)]²

But we have to calculate E(x²) first since we already have found E(x)

So E(x²) = (0²x0.032) + (1²x0.152) + (2²x0.316) + (3²x0.316) + (4²x0.152) + (5²x0.032)

therefore E(x²) = 7.492

Now we have all the values we can calculate variance:

Var(x) = E(x²) - [E(x)]²

=7.492 - [2.5]²

= 1.242

With variance now we can find the standard deviation

S.D = √Var(x)

= √1.242

Answer = 1.1145

Answer 2

Adding up all 6 probabilities gives a total of 1, which means that this is a valid distribution function. Observing that P(0) = P(5), P(1) = P(4), and P(2) = P(3), this is a symmetric distribution.

Its mean will be in the middle, which is at x = 2.5.

The standard deviation is calculated by multiplying each P(x) by the square of deviation (x - average)^2: (0.032)(0 - 2.5)^2 + (0.152)(1 - 2.5)^2 + (0.316)(2 - 2.5)^2 + (0.316)(3 - 2.5)^2 + (0.152)(4 - 2.5)^2 + (0.032)(5 - 2.5)^2 = 1.242, then taking the square root: sqrt(1.242) = 1.1145.