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.
(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
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