Given $X_1, \dots , X_{81}$ iid random variables with mean $54$ and variance $225$, find the possible difference between the sample mean and the population mean with a probability of at least $0.75$. Then $E(\bar X) = 54$ and $Var(\bar X) = 25/9$
Chebyshev: $$P(|\bar X -54| \geq 5k/3) \leq 1/k^2$$ $$ P(|\bar X -54| \leq 5k/3) \geq 1- 1/k^2 = 0.75$$ Then $k=2$ and thus $5k/3$ = $3.\bar 3$. And so the difference between the sample mean and population mean with a probability of at least $0.75$ is $3.33$.
CLT: $$P(|\bar X -54| \leq K) \geq 0.75$$ $$P(|Z| \leq 9K/(5/3)) \geq 0.75$$ $$\Phi(9K/(5/3)) \geq 0.875$$ $$9K/(5/3) = 1.15$$ $$K=0.213$$
The provided answers are $0.0926$ and $1.92$ respectively. Could someone point out where my mistakes are?
As a commenter noted, your only problem with the second one is you have an extra $9$: you already took care of the $9$ when you divided by $81$ in your computation of the variance so don't need to put it in again.
Your first answer is correct: Chebyshev says there's $75\%$ chance of being less than two standard deviations away and two standard deviations is $10/3.$ $0.0926$ doesn't remotely make sense and I suspect either you are mistakenly reading an answer to a different question or they have mistakenly written down the answer to another question where the answer to this one should be.