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.

75 tickets have been sold so far.

step-by-step explanation:

in this question it is clearly mentioned that 45 children's tickets have been sold and this number is 60% of the total tickets sold.

now let's assume the total number of tickets sold be x.

then 60% of x = 45

⇒

⇒ x(.6) = 45

by dividing with .6 on both the sides of the equation

⇒ x = 75

therefore 75 tickets have been sold so far.

the probability that a bird is green (.30) is true

step-by-step explanation: