Answer:
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.
Step-by-step 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](https://tex.z-dn.net/?f=E%28X%29%20%3D%20np)
The variance of the binomial distribution is:
![V(X) = np(1-p)](https://tex.z-dn.net/?f=V%28X%29%20%3D%20np%281-p%29)
The standard deviation of the binomial distribution is:
![\sqrt{V(X)} = \sqrt{np(1-p)}](https://tex.z-dn.net/?f=%5Csqrt%7BV%28X%29%7D%20%3D%20%5Csqrt%7Bnp%281-p%29%7D)
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](https://tex.z-dn.net/?f=p%20%3D%200.63)
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](https://tex.z-dn.net/?f=n%20%3D%205)
(a) mean
![E(X) = np = 5*0.63 = 3.15](https://tex.z-dn.net/?f=E%28X%29%20%3D%20np%20%3D%205%2A0.63%20%3D%203.15)
The mean is 3.15.
(b) variance
![V(X) = np(1-p) = 5*0.63*0.37 = 1.1655](https://tex.z-dn.net/?f=V%28X%29%20%3D%20np%281-p%29%20%3D%205%2A0.63%2A0.37%20%3D%201.1655)
The variance is 1.1655.
(c) standard deviation
![\sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{5*0.63*0.37} = 1.08](https://tex.z-dn.net/?f=%5Csqrt%7BV%28X%29%7D%20%3D%20%5Csqrt%7Bnp%281-p%29%7D%20%3D%20%5Csqrt%7B5%2A0.63%2A0.37%7D%20%3D%201.08)
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.