Question

slices of pizza for a certain brand of pizza have a mass that is approximately normally distributed with a mean of 67.3 grams and a standard deviation of 2.24 grams.
a) for samples of size 18 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 66.9 grams?
c) what is the probability of finding a 18 random slices of pizza with a mean mass of less than 66.9 grams?
d) what sample mean (for a sample of size 18) would represent the bottom 15% (the 15th percentile)? grams

Explanation:

Step1: Standard deviation of sampling distribution (part a)

The formula for the standard deviation of the sampling distribution of the sample mean (\(\sigma_{\bar{x}}\)) is \(\sigma_{\bar{x}}=\frac{\sigma}{\sqrt{n}}\), where \(\sigma = 2.24\) (population standard deviation) and \(n = 18\) (sample size).

$$ \sigma_{\bar{x}}=\frac{2.24}{\sqrt{18}}\approx\frac{2.24}{4.2426}\approx0.528 $$

Step2: Probability for a single slice (part b)

First, calculate the \(z\) - score using the formula \(z=\frac{x-\mu}{\sigma}\), where \(x = 66.9\), \(\mu=67.3\), and \(\sigma = 2.24\).

$$ z=\frac{66.9 - 67.3}{2.24}=\frac{- 0.4}{2.24}\approx - 0.179 $$

Then, use the standard normal distribution table (or a calculator with a normal - distribution function). \(P(X\lt66.9)=P(Z\lt - 0.179)\). Using a calculator or table, \(P(Z\lt - 0.179)\approx0.4286\)

Step3: Probability for the sample mean (part c)

Calculate the \(z\) - score for the sample mean using the formula \(z=\frac{\bar{x}-\mu}{\sigma_{\bar{x}}}\). We know \(\bar{x}=66.9\), \(\mu = 67.3\), and \(\sigma_{\bar{x}}\approx0.528\) (from part a)

$$ z=\frac{66.9 - 67.3}{0.528}=\frac{-0.4}{0.528}\approx - 0.758 $$

Then, \(P(\bar{X}\lt66.9)=P(Z\lt - 0.758)\). Using a calculator or table, \(P(Z\lt - 0.758)\approx0.224\)

Step4: 15th percentile (part d)

First, find the \(z\) - score corresponding to the 15th percentile. Using a standard normal distribution table or calculator, the \(z\) - score \(z\) such that \(P(Z\lt z)=0.15\) is approximately \(z=-1.036\)
Then, use the formula \(\bar{x}=\mu+z\sigma_{\bar{x}}\). Substitute \(\mu = 67.3\), \(z=-1.036\), and \(\sigma_{\bar{x}}\approx0.528\)

$$ \bar{x}=67.3+(-1.036)\times0.528=67.3 - 0.547=66.753 $$

Answer:

a) \(0.528\)
b) \(0.4286\)
c) \(0.224\)
d) \(66.753\)