Question

The amount of soft drink in a bottle is a Normal random variable. Suppose that in 7% of the bottles containing this soft drink there are less than 15.5 ounces, and in 10% of them there are more than 16.3 ounces. What are the mean and standard deviation of the amount of soft drink in a randomly selected bottle

Answer 1

Answer:

The mean is 15.93 ounces and the standard deviation is 0.29 ounces.

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.

7% of the bottles containing this soft drink there are less than 15.5 ounces

This means that when X = 15.5, Z has a pvalue of 0.07. So when X = 15.5, Z = -1.475.

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

$-1.475 = \frac{15.5 - \mu}{\sigma}$

$15.5 - \mu = -1.475\sigma$

$\mu = 15.5 + 1.475\sigma$

10% of them there are more than 16.3 ounces.

This means that when X = 16.3, Z has a pvalue of 1-0.1 = 0.9. So when X = 16.3, Z = 1.28.

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

$1.28 = \frac{16.3 - \mu}{\sigma}$

$16.3 - \mu = 1.28\sigma$

$\mu = 16.3 – 1.28\sigma$

From above

$\mu = 15.5 + 1.475\sigma$

So

$15.5 + 1.475\sigma = 16.3 – 1.28\sigma$

$2.755\sigma = 0.8$

$\sigma = \frac{0.8}{2.755}$

$\sigma = 0.29$

The mean is

$\mu = 15.5 + 1.475\sigma = 15.5 + 1.475*0.29 = 15.93$

The mean is 15.93 ounces and the standard deviation is 0.29 ounces.