Question

pizza mass: slices of pizza for a certain brand of pizza have a mass that is approximately normally distributed with a mean of 67.5 grams and a standard deviation of 2.13 grams. round your answers to four decimal places.
a) for samples of size 20 pizza slices, what is the mean for the sampling distribution of the sample mean?
b) for samples of size 20 pizza slices, what is the standard deviation for the sampling distribution of the sample mean?
c) what is the probability of finding a random single slice of pizza with a mass of less than 67.1 grams?
d) what is the probability of finding a sample of 20 slices of pizza with a mean mass of less than 67.1 grams?
e) what sample mean (for a sample of size 20) would represent the bottom 15% (the 15th percentile)?
grams

Explanation:

Step1: Identify sampling mean

For the sampling distribution of the sample mean, the mean of the sampling distribution equals the population mean.
$\mu_{\bar{x}} = \mu = 67.5$

Step2: Calculate standard error

The standard deviation of the sampling distribution (standard error) is $\frac{\sigma}{\sqrt{n}}$, where $\sigma=2.13$, $n=20$.
$\sigma_{\bar{x}} = \frac{2.13}{\sqrt{20}} \approx \frac{2.13}{4.4721} \approx 0.4763$

Step3: Find z-score for single slice

For a single slice, calculate z-score: $z = \frac{x-\mu}{\sigma}$, where $x=67.1$.
$z = \frac{67.1 - 67.5}{2.13} \approx -0.1878$
Find $P(Z < -0.1878) \approx 0.4255$

Step4: Find z-score for sample mean

For sample mean $\bar{x}=67.1$, z-score is $z = \frac{\bar{x}-\mu}{\sigma/\sqrt{n}}$.
$z = \frac{67.1 - 67.5}{0.4763} \approx -0.8398$
Find $P(Z < -0.8398) \approx 0.2007$

Step5: Find 15th percentile z-score

The z-score for the 15th percentile is $z \approx -1.0364$. Use $\bar{x} = \mu + z \cdot \frac{\sigma}{\sqrt{n}}$.
$\bar{x} = 67.5 + (-1.0364)(0.4763) \approx 67.5 - 0.4936 = 67.0064$

Answer:

a) 67.5000
b) 0.4763
c) 0.4255
d) 0.2007
e) 67.0064