Question

The average amount of a beverage in randomly selected 16-ounce beverage can is 16.18 ounces with a standard deviation of 0.4 ounces. If a random sample of sixty-five 16-ounce beverage cans are selected, what is the probability that the mean of this sample is less than 16.1 ounces of beverage? Answer: (round to 4 decimal places)

Answer 1

The probability that the mean of this sample is less than 16.1 ounces of beverage is 0.0537.

Explanation:

We are given that the average amount of a beverage in randomly selected 16-ounce beverage can is 16.18 ounces with a standard deviation of 0.4 ounces.

A random sample of sixty-five 16-ounce beverage cans are selected

Let
`\bar X` = sample mean amount of a beverage

The z-score probability distribution for the sample mean is given by;

Z =
`(\bar X-\mu)/((\sigma)/(√(n) ) )` ~ N(0,1)

where,
`\mu` = population mean amount of a beverage = 16.18 ounces

`\sigma` = standard deviation = 0.4 ounces

n = sample of 16-ounce beverage cans = 65

Now, the probability that the mean of this sample is less than 16.1 ounces of beverage is given by = P(
`\bar X` < 16.1 ounces)

P(
`\bar X` < 16.1 ounces) = P(
`(\bar X-\mu)/((\sigma)/(√(n) ) )` <
`(16.1-16.18)/((0.4)/(√(65) ) )` ) = P(Z < -1.61) = 1 - P(Z
`\leq` 1.61)

= 1 - 0.9463 = 0.0537

The above probability is calculated by looking at the value of x = 1.61 in the z table which has an area of 0.9591.