Question

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.

Short answer

The opinion is the leader has received executive coaching.

Step-by-step solution

Given information

The process behavior of leaders before coaching has a mean of 75% and a standard deviation of 8.5%

$i.e,\mu =0.75and\sigma 0.085$

The process behavior of leaders after coaching has a mean of 52% and a standard deviation of 7.5%

$So,\mu =0.52and\sigma =0.075$

Calculating the Probability

Here the observed content behaviors is$\mathbf{of}$70% by the leader of a business meeting.

The probability of 70% content behaviors before coaching the leader is

role="math" localid="1660288326895" $$\begin{gathered} P\left(x\le 0.70\right)=P\left(\frac{x-0.75}{0.085}\le \frac{0.70-0.75}{0.085}\right) \\ =P\left(z\le -0.58824\right) \\ =0.278187 \\ 𝆏0.2782 \end{gathered}$$

Here the probability is 0.2782

The probability of 70% behaviors after coaching the leader is,

$$\begin{gathered} P\left(x\le 0.70\right)=P\left(\frac{x-0.75}{0.075}\le \frac{0.70-0.52}{0.075}\right) \\ =P\left(z\le 2.4\right) \\ =0.991802 \\ 𝆏0.9918 \end{gathered}$$

Here the probability is 0.9918

Here for the second case, the probability is high.

The opinion is the leader has received executive coaching.