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 \(\mu=175\) and a standard deviation \(\sigma=15 .\) a. Use Tchebysheff's Theorem to make a statement about the percentage of colleges that have between 145 and 205 teachers. b. Assume that the population is normally distributed. What fraction of colleges have more than 190 teachers?

Short answer

Answer: At least 75% of small 2-year colleges have between 145 and 205 teachers.

Step-by-step solution

a. Use Tchebysheff's Theorem

According to Tchebysheff's theorem, for any k > 0, at least \(1 - \frac{1}{k^2}\) of the data values lie within k standard deviations of the mean. First, we need to determine k, the number of standard deviations away from the mean for both 145 and 205. Lower limit: \(k_l = \frac{(145 - \mu)}{\sigma} = \frac{(145 - 175)}{15} = -2\) Upper limit: \(k_u = \frac{(205 - \mu)}{\sigma} = \frac{(205 - 175)}{15} = 2\) Since both lower and upper limits are within 2 standard deviations, we can use Tchebysheff's theorem with k = 2: Percentange of colleges within 145 and 205 teachers: \(1 - \frac{1}{k^2} = 1 - \frac{1}{4} = \frac{3}{4} = 75 \%\) So, at least 75% of colleges will have between 145 and 205 teachers.

b. Find the fraction of colleges with more than 190 teachers

First, we need to find the z-score for 190 teachers: \(z = \frac{(190 - \mu)}{\sigma} = \frac{(190 - 175)}{15} = 1\) Using the standard normal distribution table or calculator, we can find the area to the left (cumulative probability) of z = 1, which is approximately 0.8413. Now, to find the fraction of colleges with more than 190 teachers, we subtract the cumulative probability from 1: \(1 - 0.8413 = 0.1587\) Therefore, approximately 15.87% of colleges have more than 190 teachers.