A-Level Normal distribution: worked solution

Question

On a morning with 20 incoming calls, n are complex. If the probability that the mean resolution time of these n complex calls exceeds 24 minutes is at most 0.01, find the set of values of n.

Step 1 - write down the distribution of the SAMPLE MEAN, not of a single call. Let C be the resolution time of one complex call. From the earlier part of the question, C ~ N(20, 6²). For a sample of n such calls the sample mean Cbar is also normal, with the SAME mean but with the variance divided by n: Cbar ~ N(20, 6²/n), that is Cbar ~ N(20, 36/n) Its standard deviation is therefore 6/sqrt(n). Step 2 - write the condition in the question as a probability statement. "the probability that the mean resolution time exceeds 24 minutes is at most 0.01" means P(Cbar > 24) <= 0.01 Step 3 - standardise, using Z = (Cbar - mean)/(standard deviation). P( Z > (24 - 20)/(6/sqrt(n)) ) <= 0.01 (24 - 20)/(6/sqrt(n)) = 4 sqrt(n)/6 = (2/3)sqrt(n) P( Z > (2/3)sqrt(n) ) <= 0.01 Step 4 - convert the tail probability into a critical z-value. The upper 1% point of the standard normal is z = 2.3263 (since P(Z > 2.3263) = 0.01). The right-hand tail shrinks as the cut-off moves further right, so for the tail area to be AT MOST 0.01 the cut-off must be AT LEAST 2.3263: (2/3)sqrt(n) >= 2.3263 Step 5 - solve for n. sqrt(n) >= (3/2)(2.3263) = 3.4894 Square both sides (both sides are positive, so the inequality direction is unchanged): n >= 3.4894² = 12.18 Step 6 - apply the practical restrictions. n counts calls, so it must be a positive whole number: n >= 13. Only 20 calls came in that morning, so n cannot exceed 20: n <= 20. Set of values of n: {n in Z+ : 13 <= n <= 20}