Finding probability sample proportion is less than 33% assuming null hypothesis is true

Question

Candidates 1,2 and 3 are running for a position in a company. Candidate 1 claims 38% favourability among all the voters. Assuming this is true, what is the probability that in a random sample of 500 laborers,

1. less than 33% will vote for candidate 1?
2. more than 40%?
3. Between 35% and 45%?

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I haven't done statistics in awhile, but I'm guessing:

1. the lack of a specified distribution is problematic.

2. 38% thing is the null hypothesis

If I assume normality, is this how I shall proceed?

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Let $X$ be proportion of voters in favour of candidate 1.

Then under $H_0$, we have

$\mu = E[X] = 0.38$ and $\sigma^2 = Var[X] = 0.38(1-0.38)$

We want to compute

$$P(X < 0.33 | H_0)$$

$$ = P(X < 0.33 | \mu = 0.38, \sigma^2 = 0.38(1-0.38))$$

$$ = P(\frac{X - 0.38}{\frac{\sigma}{\sqrt{500}}} < \frac{0.33 - 0.38}{\frac{\sigma}{\sqrt{500}}})$$

$$ = P(Z < \frac{0.33 - 0.38}{\frac{\sqrt{0.38(1-0.38)}}{\sqrt{500}}}) \tag{*}$$

Then I just look up $$\frac{0.33 - 0.38}{\frac{\sqrt{0.38(1-0.38)}}{\sqrt{500}}}$$ in the z-table, and that's it?

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$(*)$

We are allowed to assume normality of $X$ because $np = 500(0.38) \ge 5$ right?

Answer 1

Some confusion here: $X$ is a count, and .33 is a proportion. You are trying to use the normal approximation to the binomial distribution. The binomial random variable $X$ is a count. Specifically, you have the distribution Binom(n = 500, p = .38), You're right about n being large enough to use a normal approximation.

The average number of voters out of 500 voters (workers?) supporting the candidate is .38(500) = 190. And the standard deviation is about 5.68. So it seems unlikely fewer than 33% (or 165 people) would vote for the candidate.

An exact computation of $P(X \le 164)$ in R software is:

 pbinom(164, 500, .38)
 ## 0.008932127

See how close you get to that with the normal approximation. (Don't expect more than 2 decimal places of accuracy.) Then on to 2 and 3.