Question

The level of support for corporate sustainability (measured on a quantitative scale ranging from 0 to 160 points) was obtained for 980 senior managers at CPA firms. The accompanying table gives the mean and standard deviation for the level of support variable. It can be shown that level of support is approximately normally distributed Complete parts a through d StDev 684 157 000 a Find the probability that the level of support for corporate sustainability of a randomly selected senior manager is less than 39 points The probability that a randomly selected manager gets less than 39 points is 0 823 Round to three decimal places as needed)

Answer 1

This is a normal distribution question with
$\\Mean (\mu)= 66.684 \\Standard\;Deviation (\sigma)= 25.608$
Since we know that
$z_{ score } = \frac{x-\mu}{\sigma}$

a) x = 39
P(x < 39.0)=?
$z = \frac {39.0-66.684}{25.608}$
z = -1.0811
This implies that
P(x < 39.0) = P(z < -1.0811) = 0.1398
PS: you have to refer z score table to find the final probabilities.
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