A recent study of the hourly wages of maintenance crew members for major airlines showed that the mean hourly salary was $20.50, with a standard deviation of $3.50. Assume the distribution of hourly wages follows the normal probability distribution. We select a crew member at random.



What is the probability the crew member earns:


(a) Between $20.50 and $24.00 per hour (Round your answer to 4 decimal places.)


Probability

Respuesta :

Answer:

There is a 34.15% probability the crew member earns between $20.50 and $24.00 per hour

Step-by-step explanation:

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the zscore of a measure X is given by:

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

The Z-score 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. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

A recent study of the hourly wages of maintenance crew members for major airlines showed that the mean hourly salary was $20.50, with a standard deviation of $3.50. This means that [tex]\mu = 20.50, \sigma = 3.50[/tex].

a) What is the probability the crew member earns between $20.50 and $24.00 per hour?

This is the pvalue of the zscore of X = 24.00 subtracted by the pvalue of the zscore of X = 20.50.

X = 24

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

[tex]Z = \frac{24 - 20.50}{3.50}[/tex]

[tex]Z = 1[/tex]

[tex]Z = 1[/tex] has a pvalue of 0.8413.

X = 20.50

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

[tex]Z = \frac{20.50 - 20.50}{3.50}[/tex]

[tex]Z = 0[/tex]

[tex]Z = 0[/tex] has a pvalue of 0.50.

This means that there is a 0.8413 - 0.50 = 0.3413 = 34.15% probability the crew member earns between $20.50 and $24.00 per hour