Question

Find the (a) mean, (b) variance, and (c) standard deviation of the binomial distribution for the given random variable, and (d) interpret the results. Sixty-three percent of U.S. mothers with school-age children choose fast food as a dining option for their families one to three times a week. You randomly select five U.S. mothers with school-age children and ask whether they choose fast food as a dining option for their families one to three times a week. The random variable represents the number of U.S. mothers who choose fast food as a dining option for their families one to three times a week.

Answer 1

a) The mean is 3.15.

b) The variance is 1.1655.

c) The standard deviation is 1.08.

d) The expected number of families in the sample who choose fast food as a dining option for their families one to three times a week is 3.15. The variance of 1.1655 and the standard deviation of 1.08 are the averages from which the sample results should diverge from the mean.

Explanation:

Binomial probability distribution

Probability of exactly x sucesses on n repeated trials, with p probability.

The expected value of the binomial distribution is:

`E(X) = np`

The variance of the binomial distribution is:

`V(X) = np(1-p)`

The standard deviation of the binomial distribution is:

`√(V(X)) = √(np(1-p))`

Sixty-three percent of U.S. mothers with school-age children choose fast food as a dining option for their families one to three times a week.

This means that
`p = 0.63`

You randomly select five U.S. mothers with school-age children and ask whether they choose fast food as a dining option for their families one to three times a week.

This means that
`n = 5`

(a) mean

`E(X) = np = 5*0.63 = 3.15`

The mean is 3.15.

(b) variance

`V(X) = np(1-p) = 5*0.63*0.37 = 1.1655`

The variance is 1.1655.

(c) standard deviation

`√(V(X)) = √(np(1-p)) = √(5*0.63*0.37) = 1.08`

The standard deviation is 1.08.

(d) interpret the results.

The expected number of families in the sample who choose fast food as a dining option for their families one to three times a week is 3.15. The variance of 1.1655 and the standard deviation of 1.08 are the averages from which the sample results should diverge from the mean.