Assume that a manhole is constructed to have a circular opening with a diameter of 22 in. men have shoulder breadths that are normally distributed with a mean of 18.2 in and a standard deviation of 1.0 in. a. assume that the connecticut light and power company employs 36 men who work in manholes. if 36 men are randomly selected, what is the probability that their mean shoulder breadth is less than 18.5 in? does this result suggest that money can be saved by making smaller manholes with a diameter of 18.5 in? why or why not

0.964 is the  probability that their mean shoulder breadth is less than 18.5 inch.    

Step-by-step explanation:

Given:

Mean, μ = 18.2 inch

Standard Deviation, σ = 1.0 inch

n = 36

We are given that the distribution of  shoulder breadths is a bell shaped distribution that is a normal distribution.

Formula:

a) P( mean shoulder breadth is less than 18.5 inch)

P(x < 18.5)

Calculation the value from standard normal z table, we have,  

Thus, 0.964 is the  probability that their mean shoulder breadth is less than 18.5 inch.

Yes, the result suggest that money can be saved by making smaller manholes with a diameter of 18.5 inch since 96.4% of the man holes have  their mean shoulder breadth  less than 18.5 inch.

56.6

step-by-step explanation:

the ratio of the perimeters is equal to the scale.

the ratio of the areas is equal to the square of the scale.

34 / 20 = k

a / 19.6 = k²

substituting:

a / 19.6 = (34 / 20)²

a = 19.6 (34 / 20)²

a ≈ 56.6