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

The probability that the mean contents of the 625 sample cans is less than 9.995 ounces is about 10.56%, calculated using the Central Limit Theorem and the standard normal distribution.

Step-by-step explanation:

Finding the Probability of a Sample Mean:

To find the probability that the mean contents of the 625 sample cans is less than 9.995 ounces, we will use the Central Limit Theorem. Since we have a large sample size (n=625), the distribution of the sample mean will be approximately normally distributed with a mean (μ) of 10 ounces and a standard deviation (σ) of 0.10 ounce. First, we must calculate the standard error of the mean (SEM), which is σ divided by the square root of n.

SEM = 0.10 / √625 = 0.10 / 25 = 0.004.

Next, we calculate the z-score for 9.995 ounces, which is the difference between the sample mean and the population mean divided by the SEM.

Z = (9.995 - 10.00) / 0.004 = -0.005 / 0.004 = -1.25.

To find the probability associated with a z-score of -1.25, we look up the value in the standard normal distribution table or use a statistical software. The probability associated with a z-score of -1.25 is approximately 0.1056.

Therefore, the probability that the mean contents of the 625 sample cans is less than 9.995 ounces is about 10.56%.

Answer 2

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
`(\sigma)/(√(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 = (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 = (0.1)/(√(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 = (X - \mu)/(s)`

`Z = (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.