Question

a coffee machine dispenses normally distributed amounts of coffee with a mean of 12 ounces and a standard deviation of 0.2 ounce. if a sample of 9 cups is selected, find the probability that the mean of the sample will be less than 12.1 ounces. find the probability if the sample is just 1 cup.

a. 0.9332; 0.1915
b. 0.4332; 0.1915
c. 0.4332; 0.6915
d. 0.9332; 0.6915

Explanation:

Step1: Calculate the standard error for \(n = 9\)

The formula for the standard error of the mean is \(\sigma_{\bar{x}}=\frac{\sigma}{\sqrt{n}}\). Given \(\sigma = 0.2\) and \(n = 9\), then \(\sigma_{\bar{x}}=\frac{0.2}{\sqrt{9}}=\frac{0.2}{3}\approx0.0667\).
The \(z\) - score is calculated using the formula \(z=\frac{\bar{x}-\mu}{\sigma_{\bar{x}}}\). Here, \(\bar{x}=12.1\), \(\mu = 12\), so \(z=\frac{12.1 - 12}{0.0667}\approx1.5\).
Using the standard normal distribution table \(P(Z\lt1.5)\), we know that \(P(Z\lt1.5)=0.9332\).

Step2: Calculate for \(n = 1\)

When \(n = 1\), the standard error \(\sigma_{\bar{x}}=\frac{\sigma}{\sqrt{n}}=\sigma\) (since \(\sqrt{1}=1\)). So \(\sigma_{\bar{x}}=0.2\).
The \(z\) - score is \(z=\frac{\bar{x}-\mu}{\sigma_{\bar{x}}}=\frac{12.1 - 12}{0.2}=0.5\).
Using the standard normal distribution table \(P(Z\lt0.5)\), we know that \(P(Z\lt0.5)=0.6915\).

Answer:

D. \(0.9332;0.6915\)