Question

The weight of people in a small town in missouri is known to be normally distributed with a mean of 154 pounds and a standard deviation of 29 pounds. on a raft that takes people across the river, a sign states, "maximum capacity 3,460 pounds or 20 persons." what is the probability that a random sample of 20 persons will exceed the weight limit of 3,460 pounds? use table 1. (round "z" value to 2 decimal places, and final answer to 4 decimal places.)

Answer 1

To find the probability that a random sample of 20 persons will exceed the weight limit of 3,460 pounds, we need to convert the weight limit into a z-score value and find the cumulative probability. The probability is extremely low, approximately 0.0001.

Step-by-step explanation:

To find the probability that a random sample of 20 persons will exceed the weight limit of 3,460 pounds, we need to convert the weight limit into a z-score value using the formula: z = (x - mean) / standard deviation. In this case, the x-value is 3,460 pounds, the mean is 154 pounds, and the standard deviation is 29 pounds. Plugging in these values, we get z = (3460 - 154) / 29 = 112 / 29 = 3.86 (rounded to 2 decimal places).

Next, we need to find the cumulative probability associated with this z-score using a standard normal distribution table (Table 1). From the table, we find that the cumulative probability for a z-score of 3.86 is approximately 0.9999. However, we need to find the probability that the sample exceeds the weight limit, so we subtract this value from 1. Therefore, the probability that a random sample of 20 persons will exceed the weight limit is 1 - 0.9999 = 0.0001 (rounded to 4 decimal places).