Mathematics, 07.04.2020 20:20 Chrien222

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

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Answer from: graciewyatt6833

a) $z = \frac{0.88-0.85}{0.0357}= 0.840$

$z = \frac{0.81-0.85}{0.0357}= -1.120$

And we can find the probability like this:

P(0.81< z< 0.88) =P(-1.120

b) P(p

$z = \frac{0.87-0.85}{0.0357}= 0.560$

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

P(p

Step-by-step explanation:

Previous concepts

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:

$X \sim Binom(n=100, p=0.85)$

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

$P(X)=(nCx)(p)^x (1-p)^{n-x}$

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

$nCx=\frac{n!}{(n-x)! x!}$

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

np = 100*0.85=8510

n(1-p) = 100*(1-0.85) = 1510

So then we satisfy the two conditions and we can use the normal approximation for the distribution of $\hat p$

$\hat p \sim N(p , \sqrt{\frac{p(1-p)}{n}})$

With:

$\mu_{\hat p} = 0.85$

$\sigma_{\hat p}=\sqrt{\frac{0.85*(1-0.85)}{100}}= 0.0357$

Part a

For this case we want to find this probability:

P(0.81< p< 0.88)

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

$z = \frac{p\\mu_{p}}{\sigma}$

And replacing we got:

$z = \frac{0.88-0.85}{0.0357}= 0.840$

$z = \frac{0.81-0.85}{0.0357}= -1.120$

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

P(0.81< z< 0.88) =P(-1.120

Part b

P(p

$z = \frac{0.87-0.85}{0.0357}= 0.560$

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

P(p

Answer from: Quest

answer: b.

step-by-step explanation:

the y coordinate for point b is 0, and answer b is the only one with that y coordinate.

Asap this is a quiz and i cannot redo it

Answer from: Quest

both graphs increase on their domain

neither graph has a y-intercept

step-by-step explanation:

the following points can be visualized easily from the graph

there is a hike in both the graphs so the graphs are increasing on their domain.x-intercept of log x and log 6x are differentthere is a vertical asymptote of both the graphs so there is no y-interceptas both the graphs are increasing so there is no horizontal asymptote

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Dartmouth distribution warehouse makes deliveries of a large number of products to its customers. it...