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.

Question

Justice Mathews

Mathematics

9 February, 08:48

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.

+2

Answers (1)

Know the Answer?

Romeo Weaver · Answer 1

The z-score is given by:

z = (x-μ) / σ

where:

μ is the mean

σ is the standard deviation

from the question:

x=276

μ=300

σ=53

substituting the value in our formula we get:

z = (276-300) / 53

z=-0.453

the probability associated with this z-score is 0.3264~32.64%

thus the percentage pickers that picked less than Jill is 32.64%~33%