Question

Consider a population consisting of the number of teachers per college at small 2-year colleges. Suppose that the number of teachers per college has an average μ = 160 and a standard deviation σ = 15.

A) Use Chebyshev's Rule to make a statement about the minimum percentage of colleges that have between 115 and 205 teachers. (Round your answer to two decimal places if necessary.)

(I got 88.89%)

B) Assume that the population is mound-shaped symmetrical. What proportion of colleges have more than 130 teachers?

how do I solve B?

Thanks.

Answer 1

(a)

= 160

= 15

To find P(115 < X < 205):

Case 1: For X from 115 to mid mid value:

Transforming to Standard Normal Variate:
Z = (X - )/

= (115 - 160)/15 = - 3

Case 2: For X from mid value to 205:
Z = (205 - 160)/15 = 3

By Chebyshev's theorem, 88.89% is the minimum percentage of colleges that have between 115 and 205 teachers.

(b)

To find P(X>130):

Z = (130 - 160)/15 = - 2

By Chebyshev's theorem, % within 2 standard deviation of mean = 0.75

% within 2 standard deviation of mean corresponds to area from Z = - 0.025 on the LHS of mid value to Z = + 0.025 on RHS of mid value.

But, P(X>130) is required.

This corresponds to from Z = - 0.25 to extreme right of Standard Normal curve.

So, P(X>130) + 0.75 + 0.025 = 0.975

Thus, the answer is:

Proportion of colleges have more than 130 teachers:

0.975