Question

A random sample of 625 10-ounce cans of fruit nectar is drawn from among all cans produced in a run. Prior experience has shown that the distribution of the contents has a mean of 10 ounces and a standard deviation of .10 ounce. What is the probability that the mean contents of the 625 sample cans is less than 9.995 ounces?

Answer 1

Answer:

10.57% probability that the mean contents of the 625 sample cans is less than 9.995 ounces.

Step-by-step explanation:

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 $\frac{\sigma}{\sqrt{n}}$.

Normal probability distribution

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:

$\mu = 10, \sigma = 0.1, n = 625, s = \frac{0.1}{\sqrt{625}} = 0.004$

What is the probability that the mean contents of the 625 sample cans is less than 9.995 ounces?

This is the pvalue of Z when X = 9.995. So

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

$Z = \frac{9.995 - 10}{0.004}$

$Z = -1.25$

$Z = -1.25$ has a pvalue of 0.1057

So there is a 10.57% probability that the mean contents of the 625 sample cans is less than 9.995 ounces.