Math Snap

STEP 1

What is this asking?
What's the chance that our sample proportion $\hat{p}$ is less than 0.22, given a population proportion $p$ of 0.24?
Watch out!
Don't forget to check if the Central Limit Theorem conditions are met!

STEP 2

1. Check Central Limit Theorem Applicability
2. Calculate the z-score
3. Find the Probability

STEP 3

Before we dive in, let's make sure we can actually use the Central Limit Theorem!
We need to check if $n \cdot p$ and $n \cdot (1-p)$ are both greater than or equal to 10.
Since the problem doesn't give us $n$, we'll assume it's large enough for the Central Limit Theorem to apply.
In a real-world scenario, you'd always want to make sure you have that $n$ value!

STEP 4

Alright, now for the z-score!
This tells us how far our sample proportion is from the population proportion, in terms of standard deviations.
The formula is:
$$z = \frac{\hat{p} - p}{\sqrt{\frac{p \cdot (1-p)}{n}}}$$

STEP 5

We know $\hat{p} = \textbf{0.22}$ and $p = \textbf{0.24}$.
Since we are assuming $n$ is large, let's plug those values into our formula, still keeping $n$ as a variable:
$$z = \frac{\textbf{0.22} - \textbf{0.24}}{\sqrt{\frac{\textbf{0.24} \cdot (1-\textbf{0.24})}{n}}}$$

STEP 6

Let's simplify the numerator: $\textbf{0.22} - \textbf{0.24} = \textbf{-0.02}$.
And the denominator: $1 - \textbf{0.24} = \textbf{0.76}$, so we have $\textbf{0.24} \cdot \textbf{0.76} = \textbf{0.1824}$.
Now our equation looks like this:
$$z = \frac{\textbf{-0.02}}{\sqrt{\frac{\textbf{0.1824}}{n}}}$$

STEP 7

We can rewrite the fraction in the denominator as a product with $n$:
$$z = \frac{\textbf{-0.02}}{\sqrt{\textbf{0.1824} \cdot \frac{1}{n}}}$$ $$z = \frac{\textbf{-0.02}}{\sqrt{\textbf{0.1824}} \cdot \sqrt{\frac{1}{n}}}$$$$z = \frac{\textbf{-0.02}}{\sqrt{\textbf{0.1824}} \cdot \frac{1}{\sqrt{n}}}$$$$z = \frac{\textbf{-0.02} \cdot \sqrt{n}}{\sqrt{\textbf{0.1824}}}$$$$z = \frac{\textbf{-0.02} \cdot \sqrt{n}}{\textbf{0.42708}}$$$$z \approx \textbf{-0.0468} \cdot \sqrt{n}$$

STEP 8

Now, we need to find $P(\hat{p} < 0.22)$, which is the same as finding $P(z < \textbf{-0.0468} \cdot \sqrt{n})$.
Since we're assuming a large $n$, this probability will be very close to the population proportion $p = 0.24$.
We're asked to round to four decimal places, so let's use a "very large" $n$ like $n = 10000$.

STEP 9

With $n = 10000$, we have $z \approx \textbf{-0.0468} \cdot \sqrt{10000} = \textbf{-0.0468} \cdot 100 = \textbf{-4.68}$.

STEP 10

Using a calculator or a z-table, we find that $P(z < \textbf{-4.68})$ is a very small number, approximately 0.0000.

$P(\hat{p} < 0.22) \approx \textbf{0.0000}$