units in order to make the probability of underfilling approximately zero. Under these conditions, what is the expected profit per container? 4.115 Executive coaching and meeting effectiveness. Can executive coaching help improve business meeting effectiveness? This was the question of interest in an article published in Consulting Psychology Journal: Practice and Research (Vol. 61, 2009). The goal of executive coaching is to reduce content behaviors (e.g., seeking information, disagreeing/attacking) in favor of process behaviors (e.g., asking clarifying questions, summarizing). The study reported that prior to receiving executive coaching, the percentage of observed content behaviors of leaders had a mean of 75% with a standard deviation of 8.5%. In contrast, after receiving executive coaching, the percentage of observed content behaviors of leaders had a mean of 52% with a standard deviation of 7.5%. Assume that the percentage of observed content behaviors is approximately normally distributed for both leaders with and without executive coaching. Suppose you observe 70% content behaviors by the leader of a business meeting. Give your opinion on whether or not the leader has received executive coaching. 4.116 Box plots and the standard normal distribution. What relationship exists between the standard normal distribution and the box-plot methodology (Section 2.8) for describing distributions of data using quartiles? The answer depends on the true underlying probability distribution of the data. Assume for the remainder of this exercise that the distribution is normal.

a. Calculate the values of the standard normal random variable z, call them zL and zU, that correspond to the hinges of the box plot—that is, the lower and upper quartiles, QL and QU—of the probability distribution. M04_MCCL3396_14_GE_C04.indd 260 29/09/2021 14:01 SECTION 4.7 ● Descriptive Methods for Assessing Normality 261 The first two methods come directly from the properties of a normal distribution established in Section 4.6. Method 3 is based on the fact that for normal distributions, the z values corresponding to the 25th and 75th percentiles are -.67 and .67, respectively (see Example 4.17b). Because s = 1 for a standard normal distribution, IQR s = QU - QL s = .67 - 1 -.672 1 =