Find out the expected value and the standard deviation of the number of aces obtained in 60 rolls of a fair 6-face die. Ditto for 600 rolls. Use them to explain why the observed count of aces obtained is more likely to be within 2 from the expected value with 60 rolls than with 600 rolls.

Note that both X and Y are obtained from counting the amount of successful tries from repetitions of independent experiments that have 1/6 of probability of success. Thus, both X and Y are random variables with binomial distribution, with n = 60 and 600 respectively and probability 1/6.

Remember that if Z is a random variable, Z ≈ Bi(n,p), then

E(Z) = np, where E(Z) denotes the expected value of Z
V(Z) = np(1-p), where V(Z) denotes the variance of Z. Hence, the standard deviation is the square root of V(Z), √(np(1-p)).

As a result

E(X) = 60*1/6 = 10
V(X) = 10*(1-1/6) = 50/6 ≅ 8.666
sd(X) = √8.666 = 2.886
E(Y) = 600*1/6 = 100
V(Y) = 100*(1-1/6) = 500/6 ≅ 86.666

sd(Y) = √86.666 = 9.1287

The observed amount of aces is more likely to be closer from the expected value with 60 rolls because, since we have less rolls, it is more difficult to obtain spread results.

You can also notice that X and Y can be obtained by summing independent variables with distribution BI(1,p) (also called Bernoulli(p) ). When you sum independent variables with the same distribution you have this property:

E(r1+r2+...+rn) = n*E(r1)
V(r1+r2+...+rn) = n*V(r1)

sd(r1+r2+...+rn) = √n*sd(r1)

X can be obtained by summing 60 independent variables r1, ...., r60 with mean 1/6 and variance 1/6*(5/6) = 5/36. So we obtain that V(X) = 60*5/36, and sd(X) = √60 * √(5/36). While for the same argument sd(Y) = √600*√(5/36). The higher the number of rolls, the more spread the results are.

I hope this helped you!