# Finding lower proportion limit of a sample distribution

Problem description

If a population proportion is 0.28, and if the sample size is 140, 30% of the time the sample proportion will be less than what value if you are taking random samples?

This is a sample proportion problem for a binomial sample distribution. Sample distributions tends to become normally distributed when enough samples are taken.

$p = 0.28$

$n = 140$

$P \ (\hat{p} < x) = 30 \%$

My reasoning has been to find the $z$-score for the left part of the normal curve, up to the proportion value $P = 30 \%$. According to a $z$-score table, the $z$-value for $0.2995$ is $z = -0.84$.

Using the z-score formula for sample proportions:

$$z = \frac{\hat{p}-p}{\sqrt{\frac{p*q}{n}}} = -0.84 = \frac{\hat{p}-0.28}{\sqrt{\frac{0.28*(1-0.28)}{140}}}$$

We solve the equation for $\hat{p}$, which should give $\hat{p} = 0.24$.

However, my solution sheet says the correct answer should be $0.26$.

Have I solved the problem incorrectly in identifying $\hat{p} = 0.24$?

I realize that we are actually looking for the value of $x$, however I'm not sure if my logic is correct to assume that $x = \hat{p}$ for these limit values.

• Because this distribution is not Normal--it is Binomial--the answers you are getting could be misleading in some applications. The crucial idea is that the Binomial distribution is discrete. That matters, because there are only a small number of proportions you are likely to see. Thus, $24.54\%$ of the time the sample proportion will be $35/140=0.25$ or less and $30.93\%$ of the time the proportion will be $36/140=0.257$ or less. That's an appreciable gap in proportions--and neither is exactly $30\%$. – whuber Jul 24 '16 at 16:37

It looks like you made an error looking at the $z$-score table. The $z$-value for .30 is around -.525. Using that you will get the right answer.