Question

g A particular fruit's weights are normally distributed, with a mean of 738 grams and a standard deviation of 14 grams. The heaviest 2% of fruits weigh more than how many grams? Give your answer to the nearest gram.

Answer 1

Answer:

The heaviest 2% of fruits weigh more than 767 grams.

Step-by-step explanation:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$, the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

$\mu = 738, \sigma = 14$

The heaviest 2% of fruits weigh more than how many grams?

More than X when Z has a pvalue of 0.98. So X when Z = 2.054.

$Z = \frac{X - \mu}{\sigma}$

$2.054 = \frac{X - 738}{14}$

$X - 738 = 2.054*14$

$X = 766.76$

Rounded to the nearest gram

The heaviest 2% of fruits weigh more than 767 grams.