- The mean is \(\mu = 38\) and standard deviation is \(\sigma \approx 3.02\).
- Using the continuity correction, we evaluate the normal curve between \(z_1 = 0.83\) and \(z_2 = 1.16\).
- The cumulative probability difference is \(\Phi(1.16) - \Phi(0.83) \approx 0.8770 - 0.7967\).
- The final probability is \(\boxed{0.0803}\).
Explanation
- Analyze the data and model selection
To find the probability that exactly \(41\) out of \(50\) small business owners do not have a college degree, we use the binomial distribution model. Here, the number of trials is \(n = 50\), and the probability of "success" (not having a college degree) is \(p = 0.76\). Since the conditions \(np = 50 \times 0.76 = 38\) and \(nq = 50 \times 0.24 = 12\) are both at least \(10\), we can use the normal approximation to the binomial distribution. We apply the continuity correction to find the probability of exactly \(41\) by calculating the area between \(40.5\) and \(41.5\).
- Calculate parameters and z-values
First, we calculate the mean \(\mu\) and the standard deviation \(\sigma\) for the distribution:
- Mean: \(\mu = np = 50 \times 0.76 = 38\)
- Standard Deviation: \(\sigma = \sqrt{npq} = \sqrt{50 \times 0.76 \times 0.24} = \sqrt{9.12} \approx 3.0199\)
Next, we calculate the \(z\)-values for our boundaries \(40.5\) and \(41.5\), rounding them to two decimal places:
- \(z_1 = \frac{40.5 - 38}{3.0199} \approx 0.8278 \approx 0.83\)
- \(z_2 = \frac{41.5 - 38}{3.0199} \approx 1.1590 \approx 1.16\)
- Find the final probability
Using the standard normal distribution table or a cumulative distribution function, we find the probability:
- \(\Phi(1.16) \approx 0.8770\)
- \(\Phi(0.83) \approx 0.7967\)
The probability that \(X = 41\) is the difference between these two values:
- \(P(40.5 < X < 41.5) = \Phi(1.16) - \Phi(0.83) = 0.8770 - 0.7967 = 0.0803\)
(Note: Using the values from the standard normal distribution function, \(0.8770005 - 0.796731 = 0.0802695\), which rounds to \(0.0803\).)
Examples
This method is useful in quality control or market research. For example, if a firm needs to estimate the likelihood of a specific number of customers having a certain characteristic (like education level) in a random sample, they use these binomial approximations to make informed business decisions without needing to check every single customer. If the sample size is large enough, normal approximation allows for quick estimations of such outcomes. For a variable with success probability \(p\) and sample size \(n\), the probability of observing exactly \(k\) successes is approximately calculated using the normal curve between \(k-0.5\) and \(k+0.5\): \[P(X=k) \approx \Phi\left(\frac{k+0.5 - np}{\sqrt{npq}}\right) - \Phi\left(\frac{k-0.5 - np}{\sqrt{npq}}\right)\]