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%?

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?

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?


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


There's no null hypothesis in this problem, because you aren't doing significance testing. In fact, this isn't even a problem of statistical inference. It's more of a probability problem.

  1. the lack of a specified distribution is problematic

My guess is that the assumption "38% favourability among all the voters" means that votes are IID Bernoulli random variables.

  1. 38% thing is the null hypothesis

No, but it does tell you the parameter of the Bernoulli distribution, namely .38.

We are allowed to assume normality of $X$

There's no need to.

As for the three questions stated in the problem, they can be answered directly by using the cumulative distribution function or the probability mass function of the binomial distribution. Since this is a self-study problem, I'll leave the details to you.

  • $\begingroup$ Suppose we don't use bernoulli $\endgroup$ – BCLC Jul 9 '16 at 14:54
  • 1
    $\begingroup$ What alternative do you have in mind? Individual votes for candidate 1 can't be normally distributed, since they're either 0 or 1. $\endgroup$ – Kodiologist Jul 9 '16 at 15:31
  • $\begingroup$ Kodiologist, well what's wrong with my solution please? $\endgroup$ – BCLC Jul 9 '16 at 15:57
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    $\begingroup$ Besides what I mentioned above, you got a z of about -1,750, implying a probability around log(-1.5e6), much smaller than the correct answer of .0089. $\endgroup$ – Kodiologist Jul 9 '16 at 17:34
  • $\begingroup$ (+1) But the calculation for the normal quantile shown in the q. doesn't come to -1750 $\endgroup$ – Scortchi Jul 11 '16 at 16:19

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