Math Problem Statement

Below, n is the sample size, p is the population proportion and \( \hat{p} \) is the sample proportion. Use the Central Limit Theorem and the TI-84 calculator to find the probability. Round the answer to at least four decimal places. n = 111 p = 0.54 \( P(0.50 < \hat{p} < 0.60) \) = ?

Solution

To solve this problem, we need to calculate the probability that the sample proportion p^\hat{p} is between 0.50 and 0.60, given the population proportion p=0.54p = 0.54 and sample size n=111n = 111.

Step-by-Step Solution:

  1. Determine the Standard Deviation of the Sampling Distribution of p^\hat{p}: The standard deviation of p^\hat{p}, also known as the standard error, is calculated using the formula: σp^=p(1p)n\sigma_{\hat{p}} = \sqrt{\frac{p(1 - p)}{n}} Substituting p=0.54p = 0.54 and n=111n = 111: σp^=0.54×(10.54)111\sigma_{\hat{p}} = \sqrt{\frac{0.54 \times (1 - 0.54)}{111}}

  2. Calculate σp^\sigma_{\hat{p}}: σp^=0.54×0.46111=0.24841110.0473\sigma_{\hat{p}} = \sqrt{\frac{0.54 \times 0.46}{111}} = \sqrt{\frac{0.2484}{111}} \approx 0.0473

  3. Convert the Bounds to Z-Scores: To find P(0.50<p^<0.60)P(0.50 < \hat{p} < 0.60), convert 0.50 and 0.60 to z-scores using the formula: z=p^pσp^z = \frac{\hat{p} - p}{\sigma_{\hat{p}}}

    • For p^=0.50\hat{p} = 0.50: z=0.500.540.04730.8452z = \frac{0.50 - 0.54}{0.0473} \approx -0.8452

    • For p^=0.60\hat{p} = 0.60: z=0.600.540.04731.2676z = \frac{0.60 - 0.54}{0.0473} \approx 1.2676

  4. Find the Probability Using the Z-Scores: Now, find the probability between these z-scores, P(0.8452<Z<1.2676)P(-0.8452 < Z < 1.2676).

    Using the z-table or a calculator:

    • The probability corresponding to z=0.8452z = -0.8452 is approximately 0.1992.
    • The probability corresponding to z=1.2676z = 1.2676 is approximately 0.8975.

    So, P(0.50<p^<0.60)=0.89750.1992=0.6983P(0.50 < \hat{p} < 0.60) = 0.8975 - 0.1992 = 0.6983

Answer:

P(0.50<p^<0.60)0.6983P(0.50 < \hat{p} < 0.60) \approx 0.6983

Would you like more details or have any questions on this solution?

Related Questions:

  1. How do we interpret the z-scores in this context?
  2. Why is the standard error of p^\hat{p} used in the calculation?
  3. How does sample size impact the standard error of the sample proportion?
  4. Can this method be applied to different confidence intervals?
  5. What assumptions are required for the Central Limit Theorem to hold in this context?

Tip:

When using the Central Limit Theorem for sample proportions, ensure the sample size is large enough to approximate a normal distribution, generally np10np \geq 10 and n(1p)10n(1 - p) \geq 10.

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Math Problem Analysis

Mathematical Concepts

Statistics
Sampling Distribution
Normal Distribution
Z-scores

Formulas

Standard Error of \( \hat{p} \): \( \sigma_{\hat{p}} = \sqrt{\frac{p(1 - p)}{n}} \)
Z-score formula: \( z = \frac{\hat{p} - p}{\sigma_{\hat{p}}} \)

Theorems

Central Limit Theorem

Suitable Grade Level

College Level Statistics