Question

Suppose that the mean value of interpupillary distance (the distance between the pupils of the left and right eyes) for adult males is \(65 \mathrm{~mm}\) and that the population standard deviation is \(5 \mathrm{~mm}\). a. If the distribution of interpupillary distance is normal and a sample of \(n=25\) adult males is to be selected, what is the probability that the sample average distance \(\bar{x}\) for these 25 will be between 64 and \(67 \mathrm{~mm}\) ? at least \(68 \mathrm{~mm}\) ? b. Suppose that a sample of 100 adult males is to be obtained. Without assuming that interpupillary distance is normally distributed, what is the approximate probability that the sample average distance will be between 64 and \(67 \mathrm{~mm}\) ? at least \(68 \mathrm{~mm}\) ?

Short answer

The probabilities that the sample average will be between 64 and 67mm for sample sizes of 25 and 100 are approximately 0.8185 and 0.9772 respectively. The probabilities that the sample mean will be at least 68mm for these sample sizes are about 0.0013 and 0.00 respectively.

Step-by-step solution

Interpretation

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.

Calculate probabilities for a)

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 \( P(\bar{x} \geq 68) = P(z \geq 3) = 1 - P(z < 3) = 1 - 0.9987 = 0.0013 \).

Calculate probabilities for b)

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.