standard deviation 2

Question

In a large class of 500 students, the mean and standard deviation of a final exam are 65 and 11, respectively
If the grades distribution is not symmetric at least how many students have grade within 9/5 standard deviation of the mean?

Answer 1

n=500, x bar = 65 sd=11

not symmetric, how many within 9/5 of 11 of the mean=65

use chebyshev's theorem at least 1-1/k^2 k=9/5

1-1/k^2 = 1-1/(9/5)^2 = 1-25/81 = 56/81 minimum

= at least 56/81 x 500 = at least 346

9/5 x 11 = 99/5 = 19.8

65+/-19.8 = 45.2 to 84.8

at least 346 between 45.2 and 84.8