Answer:
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.
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
In which [tex]C_{n,x}[/tex] is the number of different combinations of x objects from a set of n elements, given by the following formula.
[tex]C_{n,x} = \frac{n!}{x!(n-x)!}[/tex]
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 [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the z-score of a measure X is given by:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
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 [tex]\mu = 25, \sigma = 3[/tex]
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.
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{30 - 25}{3}[/tex]
[tex]Z = 1.67[/tex]
[tex]Z = 1.67[/tex] 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 [tex]p = 0.0475[/tex]
40 pizzas, which means that [tex]n = 40[/tex]
This probability is:
[tex]P(X > 2) = 1 - P(X \leq 2)[/tex]
In which
[tex]P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2)[/tex]
So
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
[tex]P(X = 0) = C_{40,0}.(0.0475)^{0}.(0.9525)^{40} = 0.1428[/tex]
[tex]P(X = 1) = C_{40,1}.(0.0475)^{1}.(0.9525)^{39} = 0.2848[/tex]
[tex]P(X = 2) = C_{40,2}.(0.0475)^{2}.(0.9525)^{38} = 0.2769[/tex]
Then
[tex]P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2) = 0.1428 + 0.2848 + 0.2769 = 0.7045[/tex]
[tex]P(X > 2) = 1 - P(X \leq 2) = 1 - 0.7045 = 0.2955[/tex]
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.
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]2.327 = \frac{X - 25}{3}[/tex]
[tex]X - 25 = 2.327*3[/tex]
[tex]X = 32[/tex]
The delivery time should be advertised as 32 minutes.