Question

Use technology or a z-score table to answer the question.

The number of baby carrots in a bag is normally distributed with a mean of 94 carrots and a standard deviation of 8.2 carrots.

Approximately what percent of the bags of baby carrots have between 90 and 100 carrots?


23.3%

31.2%

45.5%

76.73%

Answer 1

46% is the percent of the bags of baby carrots that have between 90 and 100 carrots

Explanation:

The baby carrot is normally distributed.

z = (x - µ)/σ

x is equal to number of baby carrots

µ = mean

σ = standard deviation

Substituting the given values, we get -

90 ≤ x ≤ 100

z = (90 - 94)/8.2 = - 0.49

For z value of -0.49, the probability is 0.31

For x = 100

z = (100 - 94)/8.2 = 0.73

For z value of 0.73, the probability is 0.77

P(90 ≤ x ≤ 100) = 0.77 - 0.31 = 0.46

The percent of the bags of baby carrots having carrots between 90 and 100 carrots is 0.46 × 100 = 46%