Dartmouth distribution warehouse makes deliveries of a large number of products to its customers. it is known that 85% of all the orders it receives from its customers are delivered on time. let p be the proportion of orders in a random sample of 100 that are delivered on time. find the probabiltiy that the value of p will be:-

a) between .81 and .88

b) less than .87

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".

Solution to the problem

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

[tex]X \sim Binom(n=100, p=0.85)[/tex]

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]

For this case we need to check if we can use the normal approximation with the following two conditions:

[tex] np = 100*0.85=85>10[/tex]

[tex] n(1-p) = 100*(1-0.85) = 15>10 [/tex]

So then we satisfy the two conditions and we can use the normal approximation for the distribution of [tex]\hat p[/tex]

[tex] \hat p \sim N(p , \sqrt{\frac{p(1-p)}{n}}) [/tex]

With:

[tex] \mu_{\hat p} = 0.85[/tex]

[tex]\sigma_{\hat p}=\sqrt{\frac{0.85*(1-0.85)}{100}}= 0.0357[/tex]

Part a

For this case we want to find this probability:

[tex] P(0.81< p< 0.88)[/tex]

And we can find the z score for the limits given by:

[tex] z = \frac{p\\mu_{p}}{\sigma}[/tex]

And replacing we got:

[tex] z = \frac{0.88-0.85}{0.0357}= 0.840[/tex]

[tex] z = \frac{0.81-0.85}{0.0357}= -1.120[/tex]

And we can find the probability with the normal standard table or excel and we got:

[tex] P(0.81< z< 0.88) =P(-1.120<z<0.840)= P(Z<0.840)-P(Z<-1.120) = 0.7995 -0.1314= 0.6682[/tex]

Part b

[tex] P(p<0.87)[/tex]

[tex] z = \frac{0.87-0.85}{0.0357}= 0.560[/tex]

And we can find the probability with the normal standard table or excel and we got:

[tex]P(p<0.87) = P(z<0.560) = 0.712[/tex]