Question

homework: section 7.1
score: 16.38/25 answered: 17/25
question 18
pizza mass: slices of pizza for a certain brand of pizza have a mass that is approximately normally
distributed with a mean of 68.3 grams and a standard deviation of 2.25 grams. round your answers to four
decimal places.
a) for samples of size 22 pizza slices, what is the standard deviation for the sampling distribution of the
sample mean?
b) what is the probability of finding a random slice of pizza with a mass of less than 68 grams?
c) what is the probability of finding a sample of 22 slices of pizza with a mean mass of less than 68 grams?
d) what sample mean (for a sample of size 22) would represent the bottom 15% (the 15th percentile)?
grams

Explanation:

Step1: Calculate the standard deviation for the sampling distribution of the sample mean (part a)

The formula for the standard deviation of the sampling distribution of the sample mean (also known as the standard error) is \(\sigma_{\bar{x}}=\frac{\sigma}{\sqrt{n}}\), where \(\sigma\) is the population standard deviation and \(n\) is the sample size.
Given \(\sigma = 2.25\) grams and \(n = 22\), we have \(\sigma_{\bar{x}}=\frac{2.25}{\sqrt{22}}\approx\frac{2.25}{4.6904}\approx0.4797\)

Step2: Calculate the probability of finding a random slice of pizza with a mass of less than 68 grams (part b)

First, calculate the z - score using the formula \(z=\frac{x-\mu}{\sigma}\), where \(x = 68\), \(\mu=68.3\), and \(\sigma = 2.25\)
\(z=\frac{68 - 68.3}{2.25}=\frac{- 0.3}{2.25}\approx - 0.1333\)
Using the standard normal table (or a calculator with a normal - distribution function), \(P(X\lt68)=\Phi(-0.1333)\)
\(P(X\lt68)=0.4470\)

Step3: Calculate the probability of finding a sample of 22 slices of pizza with a mean mass of less than 68 grams (part c)

Calculate the z - score for the sample mean using the formula \(z=\frac{\bar{x}-\mu}{\sigma_{\bar{x}}}\)
We know \(\mu = 68.3\), \(\bar{x}=68\), and \(\sigma_{\bar{x}}\approx0.4797\) (from part a)
\(z=\frac{68 - 68.3}{0.4797}=\frac{-0.3}{0.4797}\approx - 0.6254\)
Using the standard normal table (or a calculator with a normal - distribution function), \(P(\bar{X}\lt68)=\Phi(-0.6254)\)
\(P(\bar{X}\lt68)=0.2659\)

Step4: Find the sample mean that represents the bottom 15% (part d)

First, find the z - score corresponding to the 15th percentile. Using the standard normal table (or a calculator), the z - score \(z\) such that \(P(Z\lt z)=0.15\) is approximately \(z=-1.0364\)
We use the formula \(z=\frac{\bar{x}-\mu}{\sigma_{\bar{x}}}\) and solve for \(\bar{x}\)
\(\bar{x}=\mu+z\sigma_{\bar{x}}\)
Substitute \(\mu = 68.3\), \(z=-1.0364\), and \(\sigma_{\bar{x}}\approx0.4797\)
\(\bar{x}=68.3+(-1.0364)\times0.4797\)
\(\bar{x}=68.3-0.4976\)
\(\bar{x}=67.8024\)

Answer:

a) \(0.4797\)
b) \(0.4470\)
c) \(0.2659\)
d) \(67.8024\)