Question

the mean time taken to learn the basics of a software program by all students is 200 minutes with a standard deviation of 20 minutes, the shape of the distribution is unknown. Find the interval that contains the times taken by at least 60% of all students to learn this software program

Answer 1

By the Chebyshev's inequality: $P(|X-\mu |<k\sigma)>1-\frac{1}{k^2}$.

In our case: $1-\frac{1}{k^2}=0.6.$

Thus, $k=\sqrt{\frac{1}{0.4}}=1.58.$

So, the interval $(\mu-k\sigma, \mu+k\sigma)=(200-1.58*20,200+1.58*20)=($168.4, 231.6)

contains the times taken by at least 60% of all students.