Question

Use technology or a z-score table to answer the question. The number of huckleberries picked during a huckleberry contest are normally distributed with a mean of 300 and a standard deviation of 53. Jill picked 276 huckleberries in the contest. What percent of huckleberry pickers picked less than Jill? Round your answer to the nearest whole number. O 24% O 33% O 45% O 52%​

Answer 1

To solve the problem, we need to calculate the z-score for Jill's huckleberry picking and then use the z-score table to find the percentage of huckleberry pickers who picked less than Jill.

The z-score for Jill's huckleberry picking is:

z = (276 - 300) / 53 = -0.453

Using the z-score table, we can find that the percentage of huckleberry pickers who picked less than Jill is 32.8%, which rounds to 33%.

Therefore, the answer is 33%.