Respuesta :
Answer:
The mean of this distribution is 36 years of age.
Step-by-step explanation:
Problems of normally distributed samples are 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:
There is a 0.02275 probability that the age of any randomly chosen person in the department is less than 22.
This means that when X = 22, Z has a pvalue of 0.02275. So when X = 22, [tex]Z = -2[/tex]
So
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]-2 = \frac{22 - \mu}{\sigma}[/tex]
[tex]22 - \mu = -2\sigma[/tex]
[tex]\mu = 22 + 2\sigma[/tex]
0.15866 probability that the age of any randomly chosen person is greater than 43.
Tis means that when X = 43, Z has a pvalue of 1-0.15866 = 0.84134. So when X = 43, [tex]Z = 1[/tex]
So
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]1 = \frac{43 - \mu}{\sigma}[/tex]
[tex]43 - \mu = \sigma[/tex]
[tex]\sigma = 43 - \mu[/tex]
We have that:
[tex]\mu = 22 + 2\sigma[/tex]
So
[tex]\mu = 22 + 2(43 - \mu)[/tex]
[tex]\mu = 22 + 86 - 2\mu[/tex]
[tex]3\mu = 108[/tex]
[tex]\mu = \frac{108}{3}[/tex]
[tex]\mu = 36[/tex]
What is the mean of this distribution?
The mean of this distribution is 36 years of age.
