Question

The distribution of weights of 16-ounce bags of Lay’s barbeque potato chips is approximately normal with a standard deviation of 0.28 ounce. How does the weight of a bag at the 40th percentile compare with the mean weight?

Answer 1

$P(Z<z)=0.4\to\;z=-0.2534.$

$\frac{\bar x-\mu}{\sigma}=z\to\;\bar x=\mu+z\sigma=\mu-0.2534*0.28=\mu-0.0710.$

The weight of a bag at the 40th percentile is less than the mean weight by 0.071 ounces.