The average diameter of sand dollars on a certain island is 3.00 centimeters with a standard deviation of 1.00 centimeters. If 9 sand dollars are chosen at random for a
Collection, find the probability that the average diameter of those sand dollars is more
Than 2.60 centimeters. Assume that the variable is normally distributed.
A) 0.885
B) 0.385
C) 0.615
D) 0.820

Question

Quizplus · Accepted Answer

To find the probability that the average diameter of the 9 randomly chosen sand dollars is more than 2.60 cm, we use the formula for the standard error of the mean: $ \sigma_{\bar{x}} = \frac{\sigma}{\sqrt{n}} $, where $ \sigma $ is the standard deviation of the population and $ n $ is the sample size. Here, $ \sigma = 1.00 $ cm and $ n = 9 $, so $ \sigma_{\bar{x}} = \frac{1.00}{\sqrt{9}} = \frac{1.00}{3} = 0.333 $ cm. To find the Z-score for 2.60 cm, we use $ Z = \frac{\bar{x} - \mu}{\sigma_{\bar{x}}} = \frac{2.60 - 3.00}{0.333} = -1.20 $. Looking up -1.20 on the Z-table gives a left-tail probability of approximately 0.115. Since we want the probability of being more than 2.60 cm, we subtract this from 1, giving $ 1 - 0.115 = 0.885 $.

The Average Diameter of Sand Dollars on a Certain Island