91Ó°ÊÓ

To decide if the normal approximation is appropriate, we will use the rule of thumb: if \(np \geq 10\) and \(n(1-p) \geq 10\), then the normal approximation is said to be appropriate. Given \(n=15\) and \(p=0.5\): \(np = 15 \times 0.5 = 7.5\) and \(n(1-p) = 15 \times 0.5 = 7.5\) In this case, since both \(np\) and \(n(1-p)\) are less than 10, the normal approximation is not considered appropriate according to the rule of thumb (part a).

Even though the normal approximation is not appropriate, we will use it to calculate the probabilities for parts b and c (as per the exercise instructions). For a binomial random variable with parameters \(n\) and \(p\), its corresponding normal distribution will have a mean of \(\mu=np\) and a variance of \(\sigma^2=np(1-p)\). This gives us a standard deviation of \(\sigma=\sqrt{np(1-p)}\). Using our values, we get \(\mu=7.5\) and \(\sigma=\sqrt{7.5 \times 0.5}=1.9365\). To find \(P(x \geq 6)\) using normal approximation, we will use the standard normal distribution. We will standardize the given value \(x = 6\) by using the formula \(z=\frac{x - \mu}{\sigma}\): \(z = \frac{6 - 7.5}{1.9365}=-0.7746\) Now, we need to find \(P(x \geq 6) = P(Z \geq -0.7746)\). Using a standard normal table or calculator, we get: \(P(Z \geq -0.7746) = 0.7794\) So, \(P(x \geq 6) \approx 0.7794\) (part b).

To find \(P(x > 6)\) using normal approximation, we will apply the continuity correction and consider \(x = 6.5\). Standardizing \(x = 6.5\), we get: \(z = \frac{6.5 - 7.5}{1.9365}=-0.5164\) Now, we need to find \(P(x > 6) = P(Z > -0.5164)\). Using a standard normal table or calculator, we obtain: \(P(Z > -0.5164) = 0.6979\) So, \(P(x > 6) \approx 0.6979\) (part c).

To find the exact probabilities for parts b and c using the binomial distribution, we will use the formula: \(P(X = k) = \binom{n}{k}p^k(1-p)^{(n-k)}\) For part b, we need \(P(x \geq 6) = P(x=6) + P(x=7) + \cdots + P(x=15)\): Using the formula, we get: \(P(x \geq 6) = \sum_{k=6}^{15} \binom{15}{k}0.5^k(0.5)^{(15-k)} = 0.8062\) For part c, we require \(P(x > 6) = P(x=7) + P(x=8) + \cdots + P(x=15)\): Using the formula, we obtain: \(P(x > 6) = \sum_{k=7}^{15} \binom{15}{k}0.5^k(0.5)^{(15-k)} = 0.7246\) Comparing the exact probabilities with the approximations: For part b, the exact probability is \(0.8062\) while the approximation is \(0.7794\). For part c, the exact probability is \(0.7246\) while the approximation is \(0.6979\). As we can see, the approximation does not yield very accurate results when the conditions for normal approximation are not met. Therefore, it is essential to first check if the normal approximation is appropriate (as we did in Step 1) before using it to estimate probabilities.