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.

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AmbarG261601 AmbarG261601 · Answer 1 · 2022-10-30T12:36:11+01:00

Given:

[tex]\mu=42.3\text{ }h[/tex][tex]\sigma=1.5\text{ }h[/tex]

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

[tex]P=P(X<40)[/tex]

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:

[tex]Z=\frac{X-\mu}{\sigma}[/tex]

The value of "X" you must use is:

[tex]X=40[/tex]

Then, substituting values and evaluating, you get:

[tex]Z=\frac{40-42.3}{1.5}=\frac{-2.3}{1.5}\approx-1.53[/tex]

Therefore, now you need to find:

[tex]P=P(Z<-1.53)[/tex]

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

[tex]P=0.0630[/tex]

Hence, the answer is:

[tex]P=0.0630[/tex]