91Ó°ÊÓ

We are given the mean duration of commercials (µ) as 75 seconds and the standard deviation (σ) as 20 seconds. We need to find the z-score of the given durations using the formula: z = (X - µ) / σ where X is the duration of the commercial.

For part a, we need to find the probability of a commercial lasting less than 35 seconds. So, X = 35. Calculate the z-score: z = (35 - 75) / 20 z = -40 / 20 z = -2 For part b, we need to find the probability of a commercial lasting longer than 55 seconds. So, X = 55. Calculate the z-score: z = (55 - 75) / 20 z = -20 / 20 z = -1

Now that we have the z-scores, we can use the z-table to find the corresponding probabilities. For part a, the z-score is -2. Looking at the z-table, we find the probability P(Z < -2) to be approximately 0.0228. So, there is an approximate 2.28% chance that a commercial will last less than 35 seconds. For part b, the z-score is -1. Looking at the z-table, we find the probability P(Z < -1) to be approximately 0.1587. However, we need the probability of a commercial lasting longer than 55 seconds, so we need to find the probability P(Z > -1). Since the total probability is 1, we can find this value as: P(Z > -1) = 1 - P(Z < -1) P(Z > -1) = 1 - 0.1587 P(Z > -1) = 0.8413 So, there is an approximate 84.13% chance that a commercial will last longer than 55 seconds.