Questions: According to an article, 75% of high school seniors have a driver's license. Suppose we take a random sample of 300 high school seniors and find the proportion who have a driver's license. Find the probability that more than 78% of the sample have a driver's license. Begin by verifying that the conditions for the Central Limit Theorem for Sample Proportions have been met.

Solution Steps

To apply the Central Limit Theorem for sample proportions, we need to verify the following conditions:

1. \( np \geq 10 \)
2. \( n(1-p) \geq 10 \)

Given:

• \( n = 300 \)
• \( p = 0.75 \)

Calculating: \[ np = 300 \times 0.75 = 225.0 \] \[ n(1-p) = 300 \times (1 - 0.75) = 300 \times 0.25 = 75.0 \]

Both conditions are satisfied since \( 225.0 \geq 10 \) and \( 75.0 \geq 10 \).

We want to find the probability that more than \( 78\% \) of the sample has a driver's license. This can be expressed as: \[ P(\hat{p} > 0.78) \]

Using the normal approximation, we first calculate the Z-score for \( \hat{p} = 0.78 \): \[ Z = \frac{\hat{p} - p}{\sigma} \] where \( \sigma = \sqrt{\frac{p(1-p)}{n}} \).

Calculating \( \sigma \): \[ \sigma = \sqrt{\frac{0.75 \times (1 - 0.75)}{300}} = \sqrt{\frac{0.75 \times 0.25}{300}} = \sqrt{\frac{0.1875}{300}} \approx 0.0250 \]

Now, substituting into the Z-score formula: \[ Z = \frac{0.78 - 0.75}{0.0250} = \frac{0.03}{0.0250} = 1.2 \]

Next, we find the probability: \[ P(\hat{p} > 0.78) = P(Z > 1.2) = 1 - P(Z \leq 1.2) \]

Using the standard normal distribution, we find: \[ P(Z \leq 1.2) \approx 0.8849 \] Thus, \[ P(Z > 1.2) = 1 - 0.8849 = 0.1151 \]

Final Answer