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If the amount of time to prepare and deliver a pizza from MOD PIZZA is normally distributed with a mean of 20 minutes and standard deviation of 5 minutes.
Determine the percent of pizzas that were prepared and delivered:
(a) Between 20 and 25 minutes (b) In less than 18 minutes
( c ) If MOD PIZZA advertises that the Pizza is free if it takes more than 30 minutes to deliver, what percent of the pizza will be free?

I need to answer a,b and c - it is not mutiple choice. I also need to show all relevant steps and plus I really want to understand how to do this! I am completely stuck on this one and its due tonight. Please help!! :) Ill give badges, brainliest, thanks, all of that for whoever can help me!

Sagot :

Using the normal distribution, we have that:

a) 34.13% of pizzas were prepared and delivered between 20 and 25 minutes.

b) 34.46% of pizzas were prepared and delivered in less than 18 minutes.

c) 2.28% of the pizza will be free.

Normal Probability Distribution

The z-score of a measure X of a normally distributed variable with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex] is given by:

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

  • The z-score measures how many standard deviations the measure is above or below the mean.
  • Looking at the z-score table, the p-value associated with this z-score is found, which is the percentile of X.

The mean and the standard deviation are given, respectively, by:

[tex]\mu = 20, \sigma = 5[/tex]

For item a, the probability is the p-value of Z when X = 25 subtracted by the p-value of Z when X = 20, hence:

X = 25:

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

[tex]Z = \frac{25 - 20}{5}[/tex]

Z = 1

Z = 1 has a p-value of 0.8413.

X = 20:

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

[tex]Z = \frac{20 - 20}{5}[/tex]

Z = 0

Z = 0 has a p-value of 0.5.

0.8413 - 0.5 = 0.3413.

Hence 34.13% of pizzas were prepared and delivered between 20 and 25 minutes.

For item b, the probability is the p-value of Z when X = 18, hence:

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

[tex]Z = \frac{18 - 20}{5}[/tex]

Z = -0.4

Z = -0.4 has a p-value of 0.3446.

34.46% of pizzas were prepared and delivered in less than 18 minutes.

For item c, the proportion is one subtracted by the p-value of Z when X = 30, hence:

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

[tex]Z = \frac{30 - 20}{5}[/tex]

Z = 2

Z = 2 has a p-value of 0.9772.

1 - 0.9772 = 0.0228, hence 2.28% of the pizza will be free.

More can be learned about the normal distribution at https://brainly.com/question/28135235

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