91Ó°ÊÓ

Interpret the problem and identify the given parameters. The mean interpupillary distance is 65mm, the standard deviation is 5mm. The sample size for part a) is 25, and for part b) is 100.

Since the number of trials \( n \) is 25, which is less than 30, we assume that the interpupillary distances are normally distributed. The standard error \( \sigma_{\bar{x}} \) is given by the formula \( \frac{\sigma}{\sqrt{n}} \), where \( \sigma \) is the standard deviation and \( n \) is the number of trials. This gives \( \sigma_{\bar{x}} = \frac{5}{\sqrt{25}} = 1 \). Now, we convert the sample means 64 and 67 to z-values using the formula \( z = \frac{x - \mu}{\sigma} \). This gives us \( Z_1 = \frac{64 - 65}{1} = -1 \) and \( Z_2 = \frac{67 - 65}{1} = 2 \). Finally, we find these probabilities from the standard normal table: \( P(Z_1 < z < Z_2) = P(-1 < z < 2) = 0.9772 - 0.1587 = 0.8185 \). Similarly, finding the probability for the mean being at least 68 gives \( z = \frac{68 - 65}{1} = 3 \), so \( P(\bar{x} \geq 68) = P(z \geq 3) = 1 - P(z < 3) = 1 - 0.9987 = 0.0013 \).

The sample size is now 100 which is much larger than 30. The central limit theorem states that the sample means will be normally distributed for large values of n, regardless of the original distribution. The standard error is now \( \sigma_{\bar{x}} = \frac{5}{\sqrt{100}} = 0.5 \). Converting the desired values to z-scores, and looking up those values on a standard normal table results in \( P(64 < \bar{x} < 67) = P(-2 < z < 4) = 1 - 0.0228 = 0.9772 \). For the mean being at least 68, we find \( z = 6 \) so, \( P(\bar{x} \geq 68) = P(z \geq 6) \approx 0.00 \). This means that the probability of the mean being at least 68mm is essentially zero.