If 75% of a budget is $270, what is a full budget
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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<
)
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%
X ≈ N(20, 4²)
We want P(X<15)
Standardising gives
P(X<15) = P(Z<
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%