Question

We can choose the average number of ounces (µ) a bottling machine puts into each bottle. The actual amount put into each bottle has a normal distribution with µ and a standard deviation of 0.3. If the bottles are 8 ounces, what value of µ would give us a 1% probability of putting more than 8 ounces in a bottle (overflowing) normal distribution

Answer 1

Answer:

$\mu = 7.301$

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:

$\sigma = 0.3$

If the bottles are 8 ounces, what value of µ would give us a 1% probability of putting more than 8 ounces in a bottle (overflowing) normal distribution

This means that Z when X = 8 has a pvalue of 1-0.01 = 0.99. So when X = 8, Z = 2.33.

We want to find $\mu$

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

$2.33 = \frac{8 - \mu}{0.3}$

$8 - \mu = 2.33*0.3$

$\mu = 8 - 2.33*0.3$

$\mu = 7.301$