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.
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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 \(\sigma_{\hat{p}}\): \[ \sigma_{\hat{p}} = \sqrt{\frac{0.42 \times 0.58}{124}} = \sqrt{\frac{0.2436}{124}} \approx \sqrt{0.00196452} \approx 0.0443 \]
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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 \(\hat{p} = 0.37\): \[ z_2 = \frac{0.37 - \mu_{\hat{p}}}{\sigma_{\hat{p}}} = \frac{0.37 - 0.42}{0.0443} \approx \frac{-0.05}{0.0443} \approx -1.1282 \]
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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:
- \(P(Z < -1.1282) \approx 0.1290\).
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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:
\[ \boxed{0.1290} \]