Question

the weights of ice cream cartons are normally distributed with a mean weight of 13 ounces and a standard deviation of 0.5 ounce. (a) what is the probability that a randomly selected carton has a weight greater than 13.22 ounces? (b) a sample of 25 cartons is randomly selected. what is the probability that their mean weight is greater than 13.22 ounces? (a) the probability is . (round to four decimal places as needed.)

Explanation:

Step1: Calculate z - score for single - carton case

Use the z - score formula $z=\frac{x-\mu}{\sigma}$, where $x = 13.22$, $\mu=13$, and $\sigma = 0.5$.
$z=\frac{13.22 - 13}{0.5}=\frac{0.22}{0.5}=0.44$

Step2: Find the probability for single - carton case

We want $P(X>13.22)$, which is equivalent to $P(Z > 0.44)$. Since $P(Z>z)=1 - P(Z\leq z)$, and from the standard normal table $P(Z\leq0.44)=0.6700$, then $P(Z > 0.44)=1 - 0.6700 = 0.3300$.

Step3: Calculate z - score for sample - mean case

For a sample of size $n = 25$, the standard deviation of the sample mean $\sigma_{\bar{x}}=\frac{\sigma}{\sqrt{n}}$, where $\sigma = 0.5$ and $n = 25$. So $\sigma_{\bar{x}}=\frac{0.5}{\sqrt{25}}=\frac{0.5}{5}=0.1$.
The z - score for the sample mean is $z=\frac{\bar{x}-\mu}{\sigma_{\bar{x}}}$, with $\bar{x}=13.22$, $\mu = 13$, and $\sigma_{\bar{x}}=0.1$. Then $z=\frac{13.22 - 13}{0.1}=\frac{0.22}{0.1}=2.2$.

Step4: Find the probability for sample - mean case

We want $P(\bar{X}>13.22)$, which is equivalent to $P(Z > 2.2)$. Since $P(Z>z)=1 - P(Z\leq z)$, and from the standard normal table $P(Z\leq2.2)=0.9861$, then $P(Z > 2.2)=1 - 0.9861=0.0139$.

Answer:

(a) 0.3300
(b) 0.0139