A-Level Normal distribution: worked solution

Question

Waiting times at a bakery are modelled as N(μ,σ²) minutes. On average 10% wait over 25 minutes and 10% wait under 5 minutes. (i) Find μ and σ to the nearest minute. (ii) Comment on whether the normal model is appropriate.

(i) Method: turn each probability statement into a statement about the standard normal Z, then solve. Get μ first from the symmetry, which avoids having to solve two equations at once. Let X be the waiting time, X ~ N(μ, σ²). Given: P(X > 25) = 0.10 and P(X < 5) = 0.10. Step 1: The two tail probabilities are equal, so 25 and 5 must be the same distance from the mean, one above and one below. The mean is the midpoint: μ = (25 + 5)/2 = 15 Step 2: Now use one of the tails to find σ. Standardise with Z = (X - μ)/σ: P(X > 25) = 0.10 P(Z > (25 - μ)/σ) = 0.10 From the normal tables, P(Z > 1.2816) = 0.10, so (25 - 15)/σ = 1.2816 10/σ = 1.2816 σ = 10/1.2816 = 7.803 Step 3: Round as asked, to the nearest minute: μ = 15 and σ = 8. (ii) Method: test the model against something we know for certain, namely that a waiting time cannot be negative. If the model gives a non-negligible chance of a negative wait, it is the wrong model. With μ = 15 and σ = 8: z = (0 - 15)/8 = -1.875 P(X < 0) = P(Z < -1.875) ≈ 0.0303 That is about 3%, roughly 1 customer in 33, predicted to wait a NEGATIVE length of time. That is not negligible, so the normal distribution is not an appropriate model for these waiting times.