Question

During the past six months, 73.2 percent of US households purchased sugar. Assume that these expenditures are approximately normally distributed with a mean of $8.22 and a standard deviation of $1.10. What proportion of the households spent between $5.00 and $9.00 on sugar

Answer 1

Answer:

75.94% of the households spent between $5.00 and $9.00 on sugar.

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:

$\mu = 8.22, \sigma = 1.10$

What proportion of the households spent between $5.00 and $9.00 on sugar?

This is the pvalue of Z when X = 9 subtracted by the pvalue of Z when X = 5. So

X = 9

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

$Z = \frac{9 - 8.22}{1.10}$

$Z = 0.71$

$Z = 0.71$ has a pvalue of 0.7611

X = 5

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

$Z = \frac{5 - 8.22}{1.10}$

$Z = -2.93$

$Z = -2.93$ has a pvalue of 0.0017.

So 0.7611 - 0.0017 = 0.7594 = 75.94% of the households spent between $5.00 and $9.00 on sugar.