Assume that cans of cola are filled such that the actual amounts have a population mean of \( \mu=16.00 \) ounces. A random sample of 36 cans has a mean amount of 16.28 ounces. The distribution of sample means of size \( n=36 \) is normal with an assumed mean of 16.00 ounces, and those sample means have a standard deviation of 0.04 ounce. Complete parts (a) through (c) below. a. How many standard deviations is the sample mean away from the mean of the distribution of sample means? 7.00 (Round to two decimal places as needed.) b. In general, what is the probability that a random sample of size 36 has a mean of at least 16.28 ounces? (Rours to two decimal places as needed.)

To determine how many standard deviations the sample mean is from the population mean, you can use the Z-score formula: \[ Z = \frac{{(\bar{x} - \mu)}}{\sigma_{\bar{x}}} \] Where: - \(\bar{x} = 16.28\) ounces (sample mean) - \(\mu = 16.00\) ounces (population mean) - \(\sigma_{\bar{x}} = 0.04\) ounce (standard deviation of the sample means) Plugging in the values: \[ Z = \frac{{(16.28 - 16.00)}}{0.04} = \frac{{0.28}}{0.04} = 7.00 \] So, the sample mean is 7.00 standard deviations away from the mean of the distribution of sample means. For part (b), to find the probability that a random sample of size 36 has a mean of at least 16.28 ounces, you will consult the Z-table. Since the Z-score calculated is 7.00, which is extremely high, the area to the right (for values greater than 16.28) is essentially 0. Thus, rounding it to two decimal places, the probability of getting a sample mean of at least 16.28 ounces is practically 0.00.