Question

A worker in the automobile industry works an average of 42.3 hours per week. If the distribution is approximately normal with a standard deviation of 1.5 hours, what is the probability that a randomly selected automobile worker works less than 40 hours per week? Round the final answer to at least four decimal places and intermediate z value calculations to two decimal places.

Answer 1

Given:

`\mu=42.3\text{ }h`
`\sigma=1.5\text{ }h`

Where the Mean (μ ) and Standard Deviation (σ) for a Normal Distribution, you need to find:

`P=P(X<40)`

Where "X" is the number of hours worked per worker.

In order to calculate that probability, you should approximate to a Standard Normal Distribution. Therefore, you need to find z-statistic:

`Z=(X-\mu)/(\sigma)`

The value of "X" you must use is:

`X=40`

Then, substituting values and evaluating, you get:

`Z=(40-42.3)/(1.5)=(-2.3)/(1.5)\approx-1.53`

Therefore, now you need to find:

`P=P(Z<-1.53)`

Now you need to use the Standard Normal Distribution Table in order to find the value of the probability. Then, you get:

`P=0.0630`

Hence, the answer is:

`P=0.0630`