Question

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?

Answer 1

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

X ≈ N(20, 4²)

We want P(X<15)

Standardising gives

P(X<15) = P(Z<$\frac{15-20}{4}$)
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

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