Question

According to the U.S. Department of Agriculture, the average American consumed 55.2 pounds of salad and cooking oils in 2015. Suppose that the current distribution of salad and cooking oil consumption is approximately normally distributed with a mean of 55.2 pounds and a standard deviation of 14.7 pounds. What percentage of Americans' annual salad and cooking oil consumption is between 42 and 67 pounds? Round your answers to two decimal places.

Answer 1

We know that if X~N($\mu,\sigma^{2}$) then $\frac{X-\mu}{\sigma}$ ~ N(0,1) .

And we know that standard normal curve is symmetric about zero it means area on both sides of zero is equally divided that is 0.5. hence we can write P(Z<z) =0.5+P(0<Z<z)

And P(Z<-z) = P(Z>z)

And P(Z>z) = 0.5 - P(0<Z<z)Hence the percentage of American's annual salad and cooking oil consumption is between 42 and 67 pounds is 60.4%.

Soß tet Х. Annual salad and oil consumption In pounds. NN (55.2, 14.72) ~ N(55.2, 216.09) Given x Now Percentage of American's annual salad and Cooking or consumptoom is bestween 42 and 67 pounds loox PC42 < X < 67) P( 42-55-2<x-4 < 67-55-2) < 1,8 joox Z OOX p(-13-2 L 147 14.7 = 100x P fo.897 < Z < 0.802 = toon pl 0.90 <ztorlo) = 100w {PCZ <0.80) – P(Z<-0,90)} = 100x 30.5 + PCO< Z<0.30) P(270.90) = 100x20.5+ Pozz<0.80)– (0.5-PCOZZ<0.90) toor {pfo<z.40.80) + pfoezco.9093 yoor{0.2881 + 0.3159} 100*0.604 = 60.4 %