Answer:
Step-by-step explanation:
given that in a recent survey of 600 working Americans ages 25-34, the average weekly amount spent on lunch as S45.32 with standard deviation S2.87.
The weekly amounts are approximately bell-shaped
so for practical purposes we assume X = ages of working Americans is N(45.32, 2.87)
Estimate the percentage of amounts that were less than S42.45
First let us find probability
P(X<42.45) = F(42.45)
= [tex]P(Z<\frac{42.45-45.32}{2.87} \\=P(Z<-1)[/tex]
= 0.5-0.3413
=0.1687
So 16.87% of the amounts were less than 42.45 dollars