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.)

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Let the standard score (z-score) be computed using the formula

$$
z = \frac{\bar{x} - \mu}{\sigma_{\bar{x}}}
$$

**(a)** Given $$\bar{x}=16.28$$ ounces, $$\mu=16.00$$ ounces, and $$\sigma_{\bar{x}}=0.04$$ ounce, we have

$$
z = \frac{16.28 - 16.00}{0.04} = \frac{0.28}{0.04} = 7.00
$$

So, the sample mean is $$7.00$$ standard deviations away from the population mean.

**(b)** We need to find the probability

$$
P(\bar{x} \geq 16.28) = P(z \geq 7.00)
$$

For a standard normal distribution, the probability of obtaining a z-score of 7 or more is extremely small. In fact,

$$
P(z \geq 7.00) \approx 0.00
$$

when rounded to two decimal places.
**(a)** The sample mean is 7.00 standard deviations away from the population mean. **(b)** The probability that a random sample of size 36 has a mean of at least 16.28 ounces is approximately 0.00.

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