Question

Solve the problems. Use the graph for the standard deviation points.

In a fishing contest, the length of the fish caught ranged from 11.5 inches to 23 inches with a mean length of 16.40 inches and standard deviation of 4.71 inches. Assume there was a normal distribution. What percent of the fish caught would have a length greater than 21.11 inches, one standard deviation above the mean?
a. 84.1 %
b. 15.9%
c. 2.1%
d. 13.6%
e. 2%

Answer 1

84.1% of the fish caught would have a length greater than 21.11 inches. Hence, a) is correct.

Step-by-step explanation:

To find the percentage of fish caught that would have a length greater than 21.11 inches, we need to find the area under the normal distribution curve to the right of 21.11 inches. We can use the z-score formula to convert the given length to a z-score, which measures the number of standard deviations a value is from the mean. The z-score formula is z = (x - mean) / standard deviation.

Substituting the given values, we have z = (21.11 - 16.40) / 4.71 = 0.997. Using a standard normal distribution table, we can find that the area to the right of 0.997 is approximately 0.1587.

Since we want the area to the left of 21.11 inches, we subtract 0.1587 from 1 to get the area to the right: 1 - 0.1587 = 0.8413.

Therefore, approximately 84.1% of the fish caught would have a length greater than 21.11 inches.