To solve these problems, we will use the Central Limit Theorem (CLT). The CLT states that, for a sufficiently large sample size, the sampling distribution of the sample mean will be normally distributed, regardless of the shape of the population distribution.
Problem 1:
Given:
- Mean (μ) = 70
- Standard deviation (σ) = 3.0
- Sample size (n) = 36
To find the probability that the average score exceeds 71:
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Calculate the standard error (SE): \[ SE = \frac{\sigma}{\sqrt{n}} = \frac{3.0}{\sqrt{36}} = \frac{3.0}{6} = 0.5 \]
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Convert the score of 71 to a z-score: \[ z = \frac{X - \mu}{SE} = \frac{71 - 70}{0.5} = \frac{1}{0.5} = 2 \]
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Use a z-table or calculator to find the probability: \[ P(Z > 2) = 1 - P(Z \leq 2) \approx 1 - 0.9772 = 0.0228 \]
Therefore, the probability that the average score of the 36 pro golfers exceeds 71 is approximately 0.0228.
Problem 2:
Given:
- Mean (μ) = R3.26 million
- Standard deviation (σ) = R1.2 million
- Sample size (n) = 100
2.1 Find the probability that the mean salary of the 100 players exceeds R3.5 million:
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Calculate the SE: \[ SE = \frac{\sigma}{\sqrt{n}} = \frac{1.2}{\sqrt{100}} = \frac{1.2}{10} = 0.12 \]
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Calculate z: \[ z = \frac{3.5 - 3.26}{0.12} = \frac{0.24}{0.12} = 2 \]
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Find the probability: \[ P(Z > 2) = 1 - P(Z \leq 2) \approx 0.0228 \]
2.2 Find the probability that the mean salary exceeds R4.0 million:
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Calculate z: \[ z = \frac{4.0 - 3.26}{0.12} = \frac{0.74}{0.12} \approx 6.17 \]
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Use the z-table: \[ P(Z > 6.17) \approx 0 \]
2.3 Find the probability that the mean salary is no more than R3.0 million:
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Calculate z: \[ z = \frac{3.0 - 3.26}{0.12} = \frac{-0.26}{0.12} \approx -2.17 \]
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Find the probability: \[ P(Z < -2.17) \approx 0.0150 \]
2.4 Find the probability that the mean salary is less than R2.5 million:
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Calculate z: \[ z = \frac{2.5 - 3.26}{0.12} = \frac{-0.76}{0.12} \approx -6.33 \]
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Use the z-table: \[ P(Z < -6.33) \approx 0 \]
2.5 Find \( P(\overline{X} < A) = 0.9616 \):
- From z-tables, find the corresponding z-score for 0.9616, approximately z = 1.75.
- Solve for A: \[ A = \mu + z \cdot SE = 3.26 + 1.75 \cdot 0.12 \approx 3.26 + 0.21 = 3.47 \]
Problem 3:
Given:
- Proportion (p) = 0.15
- Sample size (n) = 50
3.1 Mean of all sample proportions: \[ \mu_{\hat{p}} = p = 0.15 \]
3.2 Standard deviation of all sample proportions: \[ \sigma_{\hat{p}} = \sqrt{\frac{p(1-p)}{n}} = \sqrt{\frac{0.15 \cdot 0.85}{50}} \approx 0.0619 \]
3.3 Proportion of the samples with between 20% and 30% make a purchase:
- Calculate z-scores: \[ z_{20%} = \frac{0.20 - 0.15}{0.0619} \approx 0.81 \] \[ z_{30%} = \frac{0.30 - 0.15}{0.0619} \approx 2.42 \]
- Find probabilities: \[ P(Z < 0.81) \approx 0.7910 \] \[ P(Z < 2.42) \approx 0.9922 \]
- Thus, the proportion of samples: \[ P(0.20 < \hat{p} < 0.30) = P(Z < 2.42) - P(Z < 0.81) \approx 0.9922 - 0.7910 = 0.2012 \]
3.4 Proportion of the samples with less than 15%: Use \( z \) for 0.15: \[ P(Z < 0) = 0.5000 \Rightarrow 0.5 \]
3.5 Probability that at least 30% make a purchase: Use \( z \) for 0.30: \[ P(Z > 2.42) = 1 - 0.9922 \approx 0.0078 \]
3.6 What percentage will have less than what proportion? For ( P(\hat{p} < k) = 0.90:
- Find \( z \) corresponding to 0.90, which is approximately \( 1.28 \).
- Solve for \( k \): \[ k = \mu_{\hat{p}} + z \cdot \sigma_{\hat{p}} = 0.15 + 1.28 \cdot 0.0619 \approx 0.15 + 0.0795 \approx 0.2295 \]
Summary of Answers:
- \( P(\overline{X} > 71) \approx 0.0228 \) 2.1 \( P(\overline{X} > 3.5 \text{ million}) \approx 0.0228 \) 2.2 \( P(\overline{X} > 4.0 \text{ million}) \approx 0 \) 2.3 \( P(\overline{X} \leq 3.0 \text{ million}) \approx 0.0150 \) 2.4 \( P(\overline{X} < 2.5 \text{ million}) \approx 0 \) 2.5 \( A \approx 3.47 \text{ million} \) 3.1 Mean of sample proportions = 0.15 3.2 Standard deviation of sample proportions = 0.0619 3.3 \( P(0.20 < \hat{p} < 0.30) \approx 0.2012 \) 3.4 \( P(\hat{p} < 0.15) \approx 0.5 \) 3.5 \( P(\hat{p} \geq 0.30) \approx 0.0078 \) 3.6 \( k \approx 0.2295 \)