Answer :
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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