A pizza delivery shop averages 20 minutes per delivery with a standard deviation of 4 minutes. What is the probability that a pizza takes less than 15 minutes to be delivered?

We have the mean, μ = 20 and the standard deviation, σ = 4

We want P (X<15)

Standardizing gives:

P (X<15) = P (Z< (15-20) / 4)

P (X<15) = P (Z< - 1.25) ⇒ When the value of the z-score is negative, we will read the value of z<1.25 on the z-table then subtract the answer from 1

P (X<15) = 1 - P (Z<1.25)

P (X<15) = 1 - 0.8944

P (X<15) = 0.1056

the probability that pizza takes less than 15 minutes to be delivered is 0.1056 = 10.56%

Round answer as needed