Questions: Question number 8. A random sample of 1444 16-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 16 ounces and a standard deviation of 0.16 ounce. What is the probability that the mean contents of the 1444 sample cans is less than 15.992 ounces? 0.039 0.029 0.069 0.059 0.079 None of the above

Solution Steps

To solve this problem, we need to use the Central Limit Theorem (CLT) to find the probability that the sample mean is less than 15.992 ounces. The CLT states that the sampling distribution of the sample mean will be normally distributed with the same mean as the population mean and a standard error equal to the population standard deviation divided by the square root of the sample size. We can then use the Z-score formula to find the probability.

The standard error (SE) is calculated using the formula: \[ SE = \frac{\sigma}{\sqrt{n}} \] where \(\sigma = 0.16\) and \(n = 1444\). Thus, \[ SE = \frac{0.16}{\sqrt{1444}} = 0.004210526315789474 \approx 0.004211 \]

The Z-score is calculated using the formula: \[ Z = \frac{\bar{x} - \mu}{SE} \] where \(\bar{x} = 15.992\), \(\mu = 16\), and \(SE = 0.004211\). Thus, \[ Z = \frac{15.992 - 16}{0.004211} = -1.8999999999997907 \approx -1.900 \]

The probability that the sample mean is less than 15.992 ounces is found using the cumulative distribution function (CDF) of the standard normal distribution: \[ P(\bar{x} < 15.992) = P(Z < -1.900) \approx 0.0287 \]

Final Answer