normal approximation to the binomial distribution in fair die

Question

Suppose a fair die is independently tossed 1000 times.

(a) Use the normal approximation to approximate the probability that the number 1 appears at least 180 times.

(b) Use the normal approximation to approximate the probability that odd outcomes for the die (1,3,5) appear between 480 and 510 times (both 480 and 510 are included).

(a) Since, $$p = {1\over 6} , n= 1000$$ I calculated that: $$μ = {500\over 3}, σ = {50\sqrt{2}\over 6} $$

Since I am asked to use normal approximation:

$$P(X\geq180) = P(X> 179.5) = P(Z>{179.5 - μ\over σ}) = 1 - Φ(1.088944) = 0.1381$$

(b) Since, $$p = {1\over 2} , n= 1000$$ I calculated that: $$μ = 500, σ = {5\sqrt{10}} $$

$$P(480 \leq X \leq 510) = P(479.5 < X < 510.5) = P({479.5 - μ\over σ} < Z < {510.5 - μ\over σ})$$

Thus, we get:

$$Φ(0.664078309) - Φ(-1.29653384) = 0.7467 - 0.0974 = 0.6493 $$

Are all my steps and answers correct?

Answer 1

Your answer to (a) looks reasonable - the exact binomial probability is closer to $0.18343$, but you were asked for an approximation. The continuity correction is sensible. In R:

1 - pbinom(179, 1000, 1/6)
# 0.1384309
1 - pnorm(179.5, 1000/6, sqrt(1000*1/6*5/6))
# 0.1380892

Your approach to (b) is also reasonable including the continuity correction, and the exact binomial probability is about $0.649287$ compared to your calculation of $0.649284$. In R

sum(dbinom(480:510, 1000, 3/6))
# 0.6492868
pnorm(510+0.5, 1000/2, sqrt(1000/4)) - pnorm(480-0.5, 1000/2, sqrt(1000/4))
# 0.6492841

(I had originally misread the question, thinking you only wanted odd numbers between $480$ and $510$ so $481,483, \ldots, 509$. If that had been the case, there are $15$ odd numbers between $480$ and $510$, while $510-480 = 30$, so perhaps you want to approximate about half of the probability in this interval, without any further continuity correction. So you might have used something like $\frac12\left(\Phi\left(\frac{510-500}{5 \sqrt{10}}\right) - \Phi\left(\frac{480-500}{5 \sqrt{10}}\right)\right) \approx 0.31675$ while the exact binomial probability would have been closer to $0.31686$.)