Question

Design specifications for filling a bottled soda claim that bottles should contain 350-360 milliliters of liquid. Sample data indicate that the bottles contain an average of 355 milliliters of liquid, with a standard deviation of 2 milliliters. Is the filling operation capable of meeting the design specifications?

Answer 1

Answer:

It is high likely that the filling operation is capale of meeting design specifications.

Step-by-step explanation:

Problems of normally distributed samples can be 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:

$\mu = 355, \sigma = 2$

Is the filling operation capable of meeting the design specifications?

It will be capable if it is highly likely that the specifications will be met. A probability is said to be high likely when it is of at least 95%.

In this case, the probability of containing between 350 and 360 ml of liquid is the pvalue of Z when X = 360 subtracted by the pvalue of Z when X = 350.

X = 360

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

$Z = \frac{360 - 355}{2}$

$Z = 2.5$

$Z = 2.5$ has a pvalue of 0.9938.

X = 350

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

$Z = \frac{350 - 355}{2}$

$Z = -2.5$

$Z = -2.5$ has a pvalue of 0.0062.

This means that there is a 0.9938 - 0.0062 = 0.9876 = 98.76% probability that the filling operation is capable of meeting the design specifications. It is high likely that the filling operation is capale of meeting design specifications.