Executive networking and firm performance. Refer to the Journal of Accounting Public Policy (Vol. 34, 2015) study of the impact of executive networking on firm performance, Exercise 2.101 (p. 117). Recall that firm performance was measured as annual return on equity (ROE), recorded as a percentage. The mean ROE for the firms studied was 13.93%, and the standard deviation was 21.65%. Assume that these values represent m and s for the population ROE distribution and that this distribution is normal. What value of ROE will be exceeded by 80% of the firms?

Referring to the Journal of Accounting Public Policy (Vol. 34, 2015) study impact of executive networking on firm performance, exercise 2.101,

The firm performance is measured as annual return on equity (ROE), recorded as a percentage, i.e., x follows a normal distribution with a mean of 13.93 and a standard deviation of 21.65.

x follows a normal distribution with a mean of 13.93 and a standard deviation of 21.65, i.e.$\mu =13.93 and \sigma =21.65$

The value of return on equality (ROE), i.e., a is,

$\begin{array}{rcl}P\left(x<a\right)& =& 0.8\\ P\left(\frac{x-\mu }{\sigma }<\frac{a-\mu }{\sigma }\right)& =& 0.8\\ P\left(z<\frac{a-13.93}{21.65}\right)& =& 0.8\\ \Phi \left(\frac{a-13.93}{21.65}\right)& =& 0.8\\ & & \end{array}$

$\begin{array}{rcl}\frac{a-13.93}{21.65}& =& {\Phi }^{-1}\left(0.8\right)\\ a& =& 13.93+\left(21.65\times {\Phi }^{-1}\left(0.8\right)\right)\\ a& =& 13.93+\left(21.65\times 0.7881\right) \left[From S\tan dard Normal Table\right]\\ a& =& 13.93+17.0624\\ a& =& 30.9924\\ & & \end{array}$

$a=30.9924$

So, the value of a is 30.9924.

Therefore, the return on equality (ROE) value that will be exceeded by 80% of the firms is 30.9924.