There is a 0.02275 probability that the age of any randomly chosen person in the department is less than 22 and 0.15866 probability that the age of any randomly chosen person is greater than 43. What is the mean of this distribution?

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 and standard deviation , the zscore of a measure X is given by:

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,

So

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,

So

We have that:

So

What is the mean of this distribution?

The mean of this distribution is 36 years of age.

area of part 1:

a1=base*height/2=(8+2)*(8-2-2-1)/2=10*3/2=15 ft^2

area of part 2:

a2=length*width=(8+2)*1=10 ft2

area of part 3:

a3=length*width=8*2=16 ft2

total shaded area : a1+a2+a3=41 ft2

total area of the reactangle : a=16*8=128 ft2

total area of the nonshaded region : 128-41=87 ft2

. + λrh() e 1 1 2 4 4 4 4 4 3 4 4 4 4 4 + th−1(− s(x)2 dx ω + λrh−1() e 2 1 2 4 4

step-by-step explanation: