Question

2) Seventy-five 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 ten U.S. mothers with

school-aged 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.

(a) Find the mean, variance, and standard deviation for the binomial distribution. (6 points)

(b) If ten mothers chose fast food as a dining option for their families one to three times a week,

would this be considered an unusual event? Be sure to justify your answer using mathematics.

(4 points)

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Answer 1

(a) The mean is 7.5.

The variance is 1.875.

The standard deviation is 1.369.

(b) The event is not unusual.

Explanation:

The random variable X can be defined as the number of U.S. mothers with school-aged children who choose fast food as a dining option for their families one to three times a week.

The proportion of the random variable X is, p = 0.75.

A random sample of n = 10 U.S. mothers with school-aged children are selected and were asked if they choose fast food as a dining option for their families one to three times a week.

The event of a US mother choosing fast food as a dining option for their families one to three times a week is independent of the other mothers.

The random variable X thus follows a Binomial distribution with parameter n = 10 and p = 0.75.

(a)

Compute the mean of the Binomial distribution as follows:

`\mu=E(X)\\`

`=np\\=10* 0.75\\=7.5`

The mean is 7.5.

Compute the variance of the Binomial distribution as follows:

`\sigma^(2)=V(X)`

`=np(1-p)\\=10* 0.75* (1-0.75)\\=1.875`

The variance is 1.875.

Compute the standard deviation of the Binomial distribution as follows:

`\sigma=√(V(X))`

`=√(1.875)\\=1.369`

The standard deviation is 1.369.

(b)

The probability mass function of a Binomial distribution is:

`P(X=x)={10\choose x}0.75^(x)(1-0.75)^(10-x);\ x=0,1,2,3...`

Compute the value of P (X = 10) as follows:

`P(X=10)={10\choose 10}\ 0.75^(10)(1-0.75)^(10-10)`

`=1* 0.056314* 1\\=0.0563`

The probability of ten mothers choosing fast food as a dining option for their families one to three times a week is 0.0563.

Unusual events are those events that have a very low probability of happening. The probability of unusual event is generally less than 0.05.

In this case the value of P (X = 10) is 0.0563, approximately 0.06.

So, P (X = 10) ≈ 0.06 > 0.05.

Thus, the event is not unusual.