Respuesta :
Answer:
(a) The mean is 7.5.
The variance is 1.875.
The standard deviation is 1.369.
(b) The event is not unusual.
Step-by-step 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:
[tex]\mu=E(X)\\[/tex]
[tex]=np\\=10\times 0.75\\=7.5[/tex]
The mean is 7.5.
Compute the variance of the Binomial distribution as follows:
[tex]\sigma^{2}=V(X)[/tex]
[tex]=np(1-p)\\=10\times 0.75\times (1-0.75)\\=1.875[/tex]
The variance is 1.875.
Compute the standard deviation of the Binomial distribution as follows:
[tex]\sigma=\sqrt{V(X)}[/tex]
[tex]=\sqrt{1.875}\\=1.369[/tex]
The standard deviation is 1.369.
(b)
The probability mass function of a Binomial distribution is:
[tex]P(X=x)={10\choose x}0.75^{x}(1-0.75)^{10-x};\ x=0,1,2,3...[/tex]
Compute the value of P (X = 10) as follows:
[tex]P(X=10)={10\choose 10}\ 0.75^{10}(1-0.75)^{10-10}[/tex]
[tex]=1\times 0.056314\times 1\\=0.0563[/tex]
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.
