Respuesta :
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:
[tex]E(X) = np[/tex]
The variance of the binomial distribution is:
[tex]V(X) = np(1-p)[/tex]
The standard deviation of the binomial distribution is:
[tex]\sqrt{V(X)} = \sqrt{np(1-p)}[/tex]
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 [tex]p = 0.63[/tex]
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 [tex]n = 5[/tex]
(a) mean
[tex]E(X) = np = 5*0.63 = 3.15[/tex]
The mean is 3.15.
(b) variance
[tex]V(X) = np(1-p) = 5*0.63*0.37 = 1.1655[/tex]
The variance is 1.1655.
(c) standard deviation
[tex]\sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{5*0.63*0.37} = 1.08[/tex]
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.
