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.

Answer 1

33%

Explanation:

33%

Answer 2

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%