ANSWER NEEDED PLEASE

Question

CAN'T THINK ANYMORE!!!

 

Let the random variable x represent the number of girls in a family with three children. Assume the probability of a child being a girl is 0.39. The table on the right describes the probability of having x number of girls. Determine whether the table describes a probability distribution. If it does, find the mean and standard deviation. Is it unusual for a family of three children to consist of three girls?
x, P(x)
0, 0.227
1 0.435
2, 0.278
3, 0.060
μ = ?
(Round to two decimal places as needed.)
B.
The table is not a probability distribution.

Answer 1

It is a probability distribution since the following two conditions are met.

1. The probability of each value is between 0 and 1.

2. The sum of the probabilities =1

 

Mean = 0 x 0.227 + 1 x 0.435 + 2 x 0.278 + 3 x 0.060 =1.17

 

Variance = (1 x 0.435 ^2 + 2 x 0.278 ^2 + 3 x 0.06 ^2 - 6 x 1.17 ^2)/6

=[1 x (1.17-.435) ^2 + 2 x (1.17-0.278) ^2 + 3 x (1.17-0.06) ^2]/6 = 0.971

 

Standard deviation = square root (0.971) = 0.99

 

The probability of having three girls in the family is low --- 0.06 = 6% which implies that it is unusual.