Mathematics, 28.06.2021 22:20 Officaljazz18

A pizza company runs a marketing campaign based on their delivery time for pizzas. They claim that they will deliver a pizza within 30 minutes of ordering or it is free. In practice the time it takes to prepare a pizza and it being delivered is normally distributed with mean 25 minutes and standard deviation 3 minutes. What is the probability a pizza is delivered for free?On a particular Sunday, 40 pizzas were ordered. What is the probability that more than 2 were delivered for free?If the company wants to reduce the proportion of pizzas that are delivered free to 1%, what should the delivery time be advertised as?

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Answer from: meganwintergirl

0.0475 = 4.75% probability a pizza is delivered for free.

0.2955 = 29.55% probability that more than 2 were delivered for free.

The delivery time should be advertised as 32 minutes.

Step-by-step explanation:

To solve this question, we need to understand the binomial distribution and the normal distribution.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

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

And p is the probability of X happening.

Normal Probability Distribution:

Problems of normal distributions can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the z-score of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.

Normally distributed with mean 25 minutes and standard deviation 3 minutes.

This means that $\mu = 25, \sigma = 3$

What is the probability a pizza is delivered for free?

More than 30 minutes, which is 1 subtracted by the p-value of Z when X = 30.

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{30 - 25}{3}$

Z = 1.67

Z = 1.67 has a p-value of 0.9525

1 - 0.9525 = 0.0475

0.0475 = 4.75% probability a pizza is delivered for free

What is the probability that more than 2 were delivered for free?

Multiple pizzas, so the binomial probability distribution is used.

0.0475 probability a pizza is delivered for free, which means that p = 0.0475

40 pizzas, which means that n = 40

This probability is:

$P(X 2) = 1 - P(X \leq 2)$

In which

$P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2)$

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 0) = C_{40,0}.(0.0475)^{0}.(0.9525)^{40} = 0.1428$

$P(X = 1) = C_{40,1}.(0.0475)^{1}.(0.9525)^{39} = 0.2848$

$P(X = 2) = C_{40,2}.(0.0475)^{2}.(0.9525)^{38} = 0.2769$

Then

$P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2) = 0.1428 + 0.2848 + 0.2769 = 0.7045$

$P(X 2) = 1 - P(X \leq 2) = 1 - 0.7045 = 0.2955$

0.2955 = 29.55% probability that more than 2 were delivered for free.

If the company wants to reduce the proportion of pizzas that are delivered free to 1%, what should the delivery time be advertised as?

The 99th percentile, which is X when Z has a p-value of 0.99, so X when Z = 2.327.

$Z = \frac{X - \mu}{\sigma}$

$2.327 = \frac{X - 25}{3}$

X - 25 = 2.327*3

X = 32

The delivery time should be advertised as 32 minutes.

Answer from: Quest

step-by-step explanation:

Answer from: Quest

im here

step-by-step explanation:

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