I have the following question:
Suppose $40\%$ of the population possess a given characteristic. If a random sample of size $300$ is drawn from the population, then the probability that $44\%$ or fewer of the samples possess the characteristic is:
My steps
Given: $\hat{p} = 40\%$, $n = 300$ and $p = 44\%$.
Therefore, we can find the $\sigma$ using $p$ and $n$.
\begin{align} \sigma &= \sqrt{\frac{0.44\times 0.56}{300}} \\ \sigma &\approx 0.02865891 \end{align}
We are asked to find probability of $p \lt 44\%$. Therefore, we can do \begin{align} &z = \frac{\hat{p} - p}{\sigma} \\ &=\frac{40\%-44\%}{0.02865891} \\ &\approx1.395726 \end{align}
Now, here's the tricky part. According to the answer sheet, the answer is $0.9207$; however, the $z$-score for $1.395726$ is $0.9177$ or if we round the answer to $z = 1.40$ then we have $0.9192$. The answer's $z$-score is of $1.41$. So my question is: am I doing something incorrectly i.e. rounding or steps or is the answer key wrong?
Thanks!
When finding the standard deviation, use the prior estimate of proportion, $\hat p$, which is to say the population's proportion, $0.4$. This gives
$$\sigma = \sqrt{\frac{0.4\times0.6}{300}}\approx0.0282$$ $$z = \frac{0.44 - 0.4}{\sigma}\approx1.414$$
Then the normal distribution gives $P\approx0.9214$.
The answer key's value comes from $z=1.41$, but usually you want to interpolate between $z=1.41$ and $z=1.42$ when using a table; I just used a calculator.