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A worker in the automobile industry works an average of 43.7 hours per week. If the distribution is approximately normal with a standard deviation of 1.6 hours what is the probability that a randomly selected automobile worker works less than 40 hours per week

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gutierrezmpx

Answer:

1.04%

Step-by-step explanation:

Using the z-score formula, we have:

z = (X - P) / σ

z = score

x = given value

P = average

σ = standard deviation

Z = (40 - 43.7) / 1.6 = 2.31 > 0.9896 (tabulated in the z-table)

Now, with that number, we find the probability that a randomly selected worker works less than 40 hours per week.

1 - 0.9896 = 0.0104 * 100 = 1.04%

Using the normal distribution, it is found that there is a 0.0104 = 1.04% probability that a randomly selected automobile worker works less than 40 hours per week.

In a normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the z-score of a measure X is given by:

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

  • It measures how many standard deviations the measure is from the mean.  
  • After finding the z-score, we look at the z-score table and find the p-value associated with this z-score, which is the percentile of X.

In this problem:

  • Mean of 43.7 hours, thus [tex]\mu = 43.7[/tex]
  • Standard deviation of 1.6 hours, thus [tex]\sigma = 1.6[/tex].

The probability is the p-value of Z when X = 40, thus:

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

[tex]Z = \frac{40 - 43.7}{1.6}[/tex]

[tex]Z = -2.31[/tex]

[tex]Z = -2.31[/tex] has a p-value of 0.0104.

0.0104 = 1.04% probability that a randomly selected automobile worker works less than 40 hours per week.

A similar problem is given at https://brainly.com/question/24663213

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