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

Explanation:

Step1: Calculate \( z \)-score for \( n = 9 \)

The formula for the \( z \)-score of the sample mean is \( z=\frac{\bar{x}-\mu}{\frac{\sigma}{\sqrt{n}}} \).
Given \( \mu = 12 \), \( \sigma=0.2 \), \( n = 9 \), \( \bar{x}=12.1 \).
Substitute into the formula: \( z=\frac{12.1 - 12}{\frac{0.2}{\sqrt{9}}}=\frac{0.1}{\frac{0.2}{3}}=\frac{0.1\times3}{0.2}=1.5 \).
Using the standard normal distribution table, \( P(Z < 1.5)\).

Step2: Calculate \( z \)-score for \( n = 1 \)

The formula for the \( z \)-score of a single observation (\( n = 1 \)) is \( z=\frac{x-\mu}{\sigma} \).
Given \( x = 12.1 \), \( \mu = 12 \), \( \sigma=0.2 \).
Substitute into the formula: \( z=\frac{12.1 - 12}{0.2}=\frac{0.1}{0.2}=0.5 \).
Using the standard normal distribution table, \( P(Z < 0.5)\).

From the standard - normal table:

• For \( z = 1.5 \), \( P(Z<1.5)=0.9332 \) (using the property that \( P(Z < z)\) where \( z = 1.5\), and looking up the value in the standard - normal table: \( 0.5+0.4332\)).
• For \( z = 0.5 \), \( P(Z < 0.5)=0.6915 \) (since \( P(Z<0.5)=0.5 + 0.1915\)).

Answer:

C. \( 0.9332;0.6915 \)