Hommork6-4 Question 2 of 7 (3 points) I Question Altempt: 1 of 5 \( \checkmark 1 \) \( =2 \) 3 4 5 6 7 Use the normal approximation to the binomial to find the probability for \( n=10, p=0.5 \), and \( X \geq 7 \). Use \( \because \) The Standard Normal Distribution Table. Round \( z \) value calculations to 2 decimal places and final answer to 4 decimal places. The probability is \( \square \) 01711

To find the probability \( P(X \geq 7) \) using the normal approximation to the binomial distribution, we first need to calculate the mean (\( \mu \)) and standard deviation (\( \sigma \)) of the binomial distribution with \( n=10 \) and \( p=0.5 \). The mean is given by \( \mu = n \cdot p = 10 \cdot 0.5 = 5 \). The standard deviation is calculated as \( \sigma = \sqrt{n \cdot p \cdot (1-p)} = \sqrt{10 \cdot 0.5 \cdot 0.5} = \sqrt{2.5} \approx 1.58 \). Next, for the normal approximation, we use continuity correction. Thus, we want \( P(X \geq 7) \) which is approximated as \( P(X > 6.5) \). We convert this to a z-score: \[ z = \frac{x - \mu}{\sigma} = \frac{6.5 - 5}{1.58} \approx 0.95 \] Now looking up \( z = 0.95 \) in the Standard Normal Distribution Table, we find \( P(Z < 0.95) \approx 0.8264 \). Thus, \[ P(X \geq 7) = 1 - P(Z < 0.95) = 1 - 0.8264 = 0.1736 \] Final answer rounded to four decimal places is \( 0.1736 \). So the calculated probability is \( 0.1736 \).