Small Business Owners Seventy-six percent of small business owners do not have a college degree. If a random sample of 60 small business owners is selected, find the probability that exactly 49 will not have a college degree. Round the final answer to at least 4 decimal places and intermediate \( z \)-value calculations to 2 decimal places. \( P(X=49)= \)

To find the probability \( P(X=49) \) for the given scenario, we can use the binomial distribution formula. The parameters here are \( n = 60 \) (the sample size) and \( p = 0.76 \) (the probability of not having a college degree). First, let's calculate the mean \( \mu \) and standard deviation \( \sigma \) of the distribution: \[ \mu = n \times p = 60 \times 0.76 = 45.6 \] \[ \sigma = \sqrt{n \times p \times (1-p)} = \sqrt{60 \times 0.76 \times 0.24} \approx \sqrt{10.944} \approx 3.30 \] Next, we can use the normal approximation to the binomial distribution since \( n \) is large. To use this approximation, we first find the z-score for \( X = 49 \): \[ z = \frac{X - \mu}{\sigma} = \frac{49 - 45.6}{3.30} \approx \frac{3.4}{3.30} \approx 1.03 \] Using the z-score, we can find probabilities from the standard normal distribution. However, since we're interested in the probability \( P(X=49) \) and with the continuity correction, we need to calculate: \[ P(48.5 < X < 49.5) \] Calculating the z-scores for 48.5 and 49.5: \[ z_{48.5} = \frac{48.5 - 45.6}{3.30} \approx \frac{2.9}{3.30} \approx 0.88 \] \[ z_{49.5} = \frac{49.5 - 45.6}{3.30} \approx \frac{3.9}{3.30} \approx 1.18 \] Using a z-table or calculator, we find: \[ P(Z < 0.88) \approx 0.8106 \] \[ P(Z < 1.18) \approx 0.8810 \] Now we can find \( P(48.5 < X < 49.5) \): \[ P(48.5 < X < 49.5) \approx P(Z < 1.18) - P(Z < 0.88) \approx 0.8810 - 0.8106 = 0.0704 \] Finally, rounding to four decimal places, we have: \[ P(X=49) \approx 0.0704 \]