Question

the weights of ice cream cartons are normally distributed with a mean weight of 8 ounces and a standard deviation of 0.6 ounce.
(a) what is the probability that a randomly selected carton has a weight greater than 8.28 ounces?
(b) a sample of 25 cartons is randomly selected. what is the probability that their mean weight is greater than 8.28 ounces?
(a) the probability is
(round to four decimal places as needed.)
(b) the probability is
(round to four decimal places as needed.)

Explanation:

Step1: Calculate z - score for part (a)

The formula for the z - score is \(z=\frac{x-\mu}{\sigma}\), where \(x = 8.28\), \(\mu=8\), and \(\sigma = 0.6\).

$$z=\frac{8.28 - 8}{0.6}=\frac{0.28}{0.6}\approx0.47$$

Step2: Find probability for part (a)

We want \(P(X>8.28)\), which is equivalent to \(P(Z > 0.47)\). Using the property \(P(Z>z)=1 - P(Z\leq z)\). From the standard normal table, \(P(Z\leq0.47)=0.6808\). So \(P(Z > 0.47)=1 - 0.6808 = 0.3192\)

Step3: Calculate z - score for part (b)

The formula for the z - score of the sample mean is \(z=\frac{\bar{x}-\mu}{\frac{\sigma}{\sqrt{n}}}\), where \(\bar{x}=8.28\), \(\mu = 8\), \(\sigma=0.6\), and \(n = 25\).

$$z=\frac{8.28 - 8}{\frac{0.6}{\sqrt{25}}}=\frac{0.28}{\frac{0.6}{5}}=\frac{0.28}{0.12}\approx2.33$$

Step4: Find probability for part (b)

We want \(P(\bar{X}>8.28)\), which is equivalent to \(P(Z>2.33)\). Using the property \(P(Z > z)=1 - P(Z\leq z)\). From the standard normal table, \(P(Z\leq2.33)=0.9901\). So \(P(Z>2.33)=1 - 0.9901=0.0099\)

Answer:

(a) \(0.3192\)
(b) \(0.0099\)