Aworker 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

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 and standard deviation , the z-score of a measure X is given by:

In this problem:

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

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 link

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%

an advantage of the standard deviation is that it increases as the dispersion of the data increases. e. the interquartile range is preferred when the data are not skewed or no have outliers. an advantage of the standard deviation is that it uses all the observations in its computation.

answer: 0

step-by-step explanation: the positive, like 5, + its opposite, like -5 would be 0

5+(-5)=0, or 5-5=0