Corporate sustainability of CPA firms. Refer to the Business and Society (March 2011) study on the sustainability behaviors of CPA corporations, Exercise 2.23 (p. 83). Recall that the level of support for corporate sustainability (measured on a quantitative scale ranging from 0 to 160 points) was obtained for each of 992 senior managers at CPA firms. The accompanying Minitab printout gives the mean and standard deviation for the level of support variable. It can be shown that level of support is approximately normally distributed.

a. Find the probability that the level of support for corporate sustainability of a randomly selected senior manager is less than 40 points.

b. Find the probability that the level of support for corporate sustainability of a randomly selected senior manager is between 40 and 120 points.

c. Find the probability that the level of support for corporate sustainability of a randomly selected senior manager is greater than 120 points.

d. One-fourth of the 992 senior managers indicated a level of support for corporate sustainability below what value?

Descriptive Statistics: Support

Variables	N	Mean	StDev	Variance	Minimum	Maximum	Range
Support	992	67.755	26.871	722.036	0.000	155.000	155.000

Referring to exercise 2.23, in sustainability behaviors of CPA corporations, the level of support for corporate sustainability was obtained by 992 senior managers at CPA firms.

The level of support for sustainability x has a mean of 67.755 and a standard deviation of 26.871

Assume that x is normally distributed.

Here the mean and standard deviation of the random variable x is given by,

$$\begin{gathered} \mu =67.755and\sigma =26.871 \\ x=40 \end{gathered}$$

The z-score is,

$$\begin{gathered} z=\frac{x-\mu }{\sigma } \\ =\frac{40-67.755}{26.871} \\ =-1.0329 \\ P(x<40)=P(x<40) \\ =P(z<1.0329)) \\ =1-0.8485 \\ =0.1515 \\ P(x<40)=0.1515 \end{gathered}$$

Thus, the required probability is 0.1515.

Here the mean and standard deviation of the random variable x is given by,

$$\begin{gathered} \mu =67.755and\sigma =26.871 \\ x=40 \end{gathered}$$

The z-score is,

$$\begin{gathered} =\frac{40-67.755}{26.871} \\ =-1.0329 \end{gathered}$$

Again,

$$\begin{gathered} x=120 \\ z=\frac{x-\mu }{\sigma } \\ =\frac{120-67.755}{26.871} \\ =1.9443 \\ P(40<x<120)=P(x<120)-P(x<40) \\ =P(z<1.9443)-P(z<1.0329) \\ =P(z<1.9443)-(1-P(z<1.0329\left)\right) \\ =0.9738-1+0.8485 \\ =0.8223 \end{gathered}$$

Thus, the required probability is 0.8223.

Here the mean and standard deviation of the random variable x is given by,

$$\begin{gathered} \mu =67.755and\sigma =26.871 \\ x=120 \\ z=\frac{x-\mu }{\sigma } \\ =\frac{120-67.755}{26.871} \\ =1.9443 \\ P(x>120)=(1-P(x<120\left)\right) \\ =(1-P(z<1.9443\left)\right) \\ =1-0.9738 \\ =0.0262 \\ P(x>120)=0.0262 \end{gathered}$$

Thus, the required probability is 0.0262.

Let a be the support vale,

$$\begin{gathered} Px<a)=\frac{1}{4} \\ P\left(\frac{x-\mu }{\sigma }<\frac{x-\mu }{\sigma }\right)=0.25 \\ P\left(z<\frac{a-67.755}{26.871}\right)=0.25 \\ \Phi \left(\frac{a-67.755}{26.871}\right)=0.25 \\ \frac{a-67.755}{26.871}={\Phi }^{-1}(0.25) \\ a=67.755+26.871\times {\Phi }^{-1}(0.25) \\ a=67.755+26.871\times (-0.67449) \\ a=49.63077921 \\ a=49.6 \\ a=49.6 \end{gathered}$$

So, the value of a is 49.6

Therefore, one-fourth of the 992 senior managers with corporate sustainability below is 49.6.