Question

A popular soft drink is sold in 2-liter (2000 milliliter) bottles. Because of variation in the filling process, bottles have a mean of 2000 milliliters and a standard deviation of 18, normally distributed. A) If the manufacturer samples 100 bottles, what is the probability that the mean is less than 1995 milliliters? B) What mean overfill or more will occur only 10% of the time for the sample of 100 bottles?

Answer 1

Answer:

a) 0.27% probability that the mean is less than 1995 milliliters.

b) 2002.3 milliliters or more will occur only 10% of the time for the sample of 100 bottles.

Step-by-step explanation:

To solve this problem, it is important to understand the normal probability distribution and the central limit theorem

Normal probability distribution

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.

Central Limit theorem

The Central Limit Theorem estabilishes that, for a random variable X, with mean $\mu$ and standard deviation $\sigma$, a large sample size can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$

In this problem, we have that:

$\mu = 2000, \sigma = 18$

A) If the manufacturer samples 100 bottles, what is the probability that the mean is less than 1995 milliliters?

So $n = 100, s = \frac{18}{\sqrt{100}} = 1.8$

This probability is the pvalue of Z when X = 1995. So

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

By the Central Limit Theorem

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

$Z = \frac{1995 - 2000}{1.8}$

$Z = -2.78$

$Z = -2.78$ has a pvalue of 0.0027

So 0.27% probability that the mean is less than 1995 milliliters.

B) What mean overfill or more will occur only 10% of the time for the sample of 100 bottles?

This is the value of X when Z has a pvalue of 1-0.1 = 0.9.

So it is X when $Z = 1.28$

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

$1.28 = \frac{X - 2000}{1.8}$

$X - 2000 = 1.8*1.28$

$X = 2002.3$

2002.3 milliliters or more will occur only 10% of the time for the sample of 100 bottles.