Question

According to the Statistical Abstract of the US, 1995 , the mean per capita consumption of fresh fruits in the US in 1993 was 99.9 pounds. If the distribution of amounts consumed was normal with standard deviation of 30 pounds, how much fruit did a person have to consume to be in the top 5%

Answer 1

Answer:

To be in the top 5%, a person must have consumed at least 149.25 pounds of fruit.

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:

$\mu = 99.9, \sigma = 30$

How much fruit did a person have to consume to be in the top 5%

At least X pounds, in which X is found when Z has a pvalue of 1-0.05 = 0.95. So it is X when Z = 1.645.

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

$1.645 = \frac{X - 99.9}{30}$

$X - 99.9 = 30*1.645$

$X = 149.25$

To be in the top 5%, a person must have consumed at least 149.25 pounds of fruit.