Let Y be a random variable with mean 11 and variance 9

Step 1 of 2

a) We have to lower bound for \(P(6<Y<16)\)

Given mean=11

And Variance \(\left(\sigma^{2}\right)=9\)

                      \(\sigma=3\)

By Tchebysheff’s theorem \(P(|Y-\mu|<k \sigma) \geq 1-\frac{1}{k^{2}}\)

\(\begin{array}{l} P(-k \sigma<(Y-\mu)<k \sigma) \geq 1-\frac{1}{k^{2}} \\ P(\mu-k \sigma<Y<\mu+k \sigma) \geq 1-\frac{1}{k^{2}} \end{array}\)

Here given that \(P(6<Y<16)\)

Now \(\mu-k \sigma=6\) and \(\mu+k \sigma=16\)

11 - 3k = 6     and    11 + 3k = 16

By solving the above 2 equations we can get the values of  ‘k’

Then k = 5/3

Now

\(\begin{aligned}P(6<Y<16) \geq & 1-\frac{1}{(5 / 3)^{2}} \\ & =1-(9 / 25) \\ & =16 / 25 \\ & =0.64 \end{aligned}\)

Hence the lower bound is 0.64