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.)

1 answer

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:

\[ \boxed{0.1290} \]