Normal Approximation of Sums

Question

Let $X$ be a random variable with

$P(X=-1)=\frac{1}{4}$

$P(X=0)=\frac{1}{4}$

$P(X=1)=\frac{1}{2}$

Let $S$ be the sum of 25 independent random variables with the same distribution as $X$. Calculate approximately $P(S<0)$.

$X_i$, $1 \le i \le 25$, has the same distribution. Thus, over the long run, the distribution of $S$ is approximately normal so we can use normal approximation.

$P(S<0)$

$= P(\frac{S-E(S)}{σ} < \frac{0-E(S)}{σ})$

$= P(z < \frac{0-E(S)}{σ})$ where $z = \frac{S-E(S)}{σ}$

$= P(z < \frac{0-\frac{25}{4}}{\sqrt{Var(S)}})$

$= P(z < \frac{0-\frac{25}{4}}{\sqrt{\frac{275}{16}}})$

$= P(z < \frac{0-\frac{25}{4}}{\sqrt{\frac{275}{16}}})$

$= P(z < -1.507556723)$

$=\Phi(-1.507556723)$

$=1 - \Phi(1.507556723)$

$=1 - 0.9345$

$=0.0655$

Textbook Answer:

0.05

Answer 1

The reason why your answer differs from the textbook is because you did not employ continuity correction for the normal approximation. Specifically, $$\Pr[S < 0] \approx \Pr\left[\frac{S - \operatorname{E}[S]}{\sqrt{\operatorname{Var}[S]}} < \frac{\color{red}{-\frac{1}{2}} - \frac{25}{4}}{\sqrt{\frac{275}{16}}}\right].$$ The use of $-1/2$ instead of $0$ arises because the exact distribution is discrete, hence $\Pr[S = 0]$ has a nonzero probability; under the normal approximation, using $0$ would in a sense include half of this probability mass in your answer.