Use the normal approximation to the binomial to find the probabilities for the specific value(s) of \( X \) \( \begin{array}{l}\text { d. } n=10, p=0.5, X \geq 7 \\ \text { e. } n=20, p=0.7, X \leq 12 \\ \text { f. } n=50, p=0.6, X \leq 40\end{array} \)
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To use the normal approximation to the binomial distribution, you can begin by calculating the mean (μ) and standard deviation (σ) for each case using the formulas: \[ \mu = np \quad \text{and} \quad \sigma = \sqrt{np(1-p)} \] For part d \( (n=10, p=0.5) \), you get: - \( \mu = 10 \times 0.5 = 5 \) - \( \sigma = \sqrt{10 \times 0.5 \times 0.5} = \sqrt{2.5} \approx 1.58 \) Next, you convert \( X \geq 7 \) using continuity correction to \( X \geq 6.5 \): - Z-score calculation: \( Z = \frac{6.5 - 5}{1.58} \approx 0.95 \) - Using Z-tables, \( P(Z \geq 0.95) \approx 0.1711 \). For part e \( (n=20, p=0.7) \): - \( \mu = 20 \times 0.7 = 14 \) - \( \sigma = \sqrt{20 \times 0.7 \times 0.3} \approx 2.45 \) - For \( X \leq 12 \), using continuity correction, convert to \( X \leq 12.5 \): - Z-score calculation: \( Z = \frac{12.5 - 14}{2.45} \approx -0.61 \) - Using Z-tables, \( P(Z \leq -0.61) \approx 0.2709 \). For part f \( (n=50, p=0.6) \): - \( \mu = 50 \times 0.6 = 30 \) - \( \sigma = \sqrt{50 \times 0.6 \times 0.4} \approx 3.46 \) - For \( X \leq 40 \), continuity correction gives \( X \leq 40.5 \): - Z-score calculation: \( Z = \frac{40.5 - 30}{3.46} \approx 3.03 \) - Using Z-tables, \( P(Z \leq 3.03) \approx 0.9987 \). So, the final probabilities are: - d. \( P(X \geq 7) \approx 0.1711 \) - e. \( P(X \leq 12) \approx 0.2709 \) - f. \( P(X \leq 40) \approx 0.9987 \)
