The Post Office has established a record in a major midwestern city for delivering 90 percent of its local mail the next working day. Answer the following questions based on this information:a. If you mail eight local letters, what is the probability that all of them will be delivered the next day?b. If you mail eight local letters, what is the average number you expect to be delivered the next day?c. Calculate the standard deviation of the number delivered when 8 local letters are mailed.d. When there are 8 local letters mailed, what is the probability that the number delivered will be within 2 standard deviations of the mean?

The binomial distribution is a "DISCRETE probability distribution that summarizes the probability that a value will take one of two independent values under a given set of parameters. The assumptions for the binomial distribution are that there is only one outcome for each trial, each trial has the same probability of success, and each trial is mutually exclusive, or independent of each other".

The probability mass function for the Binomial distribution is given as:

[tex]P(X)=(nCx)(p)^x (1-p)^{n-x}[/tex]

Where (nCx) means combinatory and it's given by this formula:

[tex]nCx=\frac{n!}{(n-x)! x!}[/tex]

Let X the random variable of interest, on this case we now that:

[tex]X \sim Binom(n=8, p=0.9)[/tex]

Solution to the problem

Part a

For this case we want to find the following probability:

[tex] P(X=8) [/tex]

And we can use the pmf function and we got:

[tex]P(X=8)=(8C8)(0.9)^8 (1-0.9)^{8-8}=0.43[/tex]

Part b

The expected valie for the binomial distribution is given by:

[tex] E(X) = np = 8*0.9 =7.2[/tex]

Part c

For this case the variance is given by:

[tex]Var(X) = np(1-p) =8*0.9*(1-0.9) = 0.72[/tex]

And the standard deviation would be just the square root of the variance and we got:

[tex] Sd(X) = \sqrt{Var(X)}= \sqrt{0.72}=0.85[/tex]

Part d

For this case two deviations from the mean represent [tex] 2\sigma = 2*0.85= 1.7[/tex]

For this case we can use the binomial approximation to the normal distribution since the value of p is greater than 0.5, we have that np>5.

So we can approximate the distribution of X with

[tex] X \sim N(7.2, 0.85)[/tex]

And we want this probability:

[tex] P(7.2 -2*0.85 \leq X \leq 7.2 +2*0.85)= P(5.5 \leq X \leq 8.9)[/tex]

And we apply the continuity correction and we got:

[tex]P(5.5 \leq X \leq 8.9)= P(X \leq 8.9+0.5) -P(X\leq 5.5+0.5) =P(X \leq 9.4) -P(X\leq 6) [/tex]

And using the z score formula given by:

[tex] z = \frac{x -\mu}{\sigma}[/tex]

We find the two z scores for thw two values and we have:

[tex] z= \frac{9.4-7.2}{0.85}= 2.59[/tex]

[tex] z= \frac{6-7.2}{0.85}= -1.41[/tex]

[tex]P(Z<2.59)-P(Z<-1.41) = 0.9952-0.0793=0.9159[/tex]