Question

Based on historical data, your manager believes that 42% of the company's orders come from first-time customers. A random sample of 124 orders will be used to estimate the proportion of first-time-customers. What is the probability that the sample proportion is between 0.23 and 0.37?

Note: You should carefully round any z-values you calculate to 4 decimal places to match wamap's approach and calculations.

Answer =
(Enter your answer as a number accurate to 4 decimal places.)

Answer 1

To find the probability that the sample proportion \(\hat{p}\) is between 0.23 and 0.37, we can use the normal approximation to the binomial distribution.

1. **Calculate the mean and standard deviation of the sample proportion**:
- The population proportion \(p = 0.42\).
- The sample size \(n = 124\).

The mean of the sample proportion is given by:
\[
\mu_{\hat{p}} = p = 0.42
\]

The standard deviation of the sample proportion is given by:
\[
\sigma_{\hat{p}} = \sqrt{\frac{p(1-p)}{n}} = \sqrt{\frac{0.42 \times (1 - 0.42)}{124}}
\]

We first calculate \(1 - p\):
\[
1 - p = 1 - 0.42 = 0.58
\]

Now calculate \(\sigma_{\hat{p}}\):
\[
\sigma_{\hat{p}} = \sqrt{\frac{0.42 \times 0.58}{124}} = \sqrt{\frac{0.2436}{124}} \approx \sqrt{0.00196452} \approx 0.0443
\]

2. **Convert sample proportions to z-scores**:
We want to find the probability that \(\hat{p}\) is between 0.23 and 0.37. We will calculate the z-scores for both values.

For \(\hat{p} = 0.23\):
\[
z_1 = \frac{0.23 - \mu_{\hat{p}}}{\sigma_{\hat{p}}} = \frac{0.23 - 0.42}{0.0443} \approx \frac{-0.19}{0.0443} \approx -4.2875
\]

For \(\hat{p} = 0.37\):
\[
z_2 = \frac{0.37 - \mu_{\hat{p}}}{\sigma_{\hat{p}}} = \frac{0.37 - 0.42}{0.0443} \approx \frac{-0.05}{0.0443} \approx -1.1282
\]

3. **Use the z-scores to find probabilities**:
Now, we will look up these z-scores in the standard normal distribution table (or use a calculator).

- For \(z_1 \approx -4.2875\), the cumulative probability \(P(Z < -4.2875) \approx 0\) (very small).
- For \(z_2 \approx -1.1282\), using standard normal distribution tables:
- \(P(Z < -1.1282) \approx 0.1290\).

4. **Find the probability that \(\hat{p}\) is between 0.23 and 0.37**:
\[
P(0.23 < \hat{p} < 0.37) = P(Z < -1.1282) - P(Z < -4.2875) \approx 0.1290 - 0 \approx 0.1290
\]

Thus, the probability that the sample proportion is between 0.23 and 0.37 is approximately:

\[
\boxed{0.1290}
\]