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

Answer:

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

Step-by-step 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  =  $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$  ~ 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( $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ < $\frac{16.1-16.18}{\frac{0.4}{\sqrt{65} } }$ ) = 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.