Normal continuous distribution: The U.S. Bureau of Labor Statistics reports that

Question

Normal continuous distribution:

The U.S. Bureau of Labor Statistics reports that the average annual expenditure on food and beverages for a family is $ 5700 (Money, December 2003). Suppose that annual expenditures on food and beverages are distributed in the normal way and that the standard deviation is $ 1500.
a. What is the range of the expenses of 10% of the families that have the lowest annual expenses in food and beverages?

B. What percentage (families' probability spends more than $ 7,000 annually on food and beverages?)
c. What is the range of average expenses for 5% of families with the highest food and drink costs?

Explanation / Answer

let X denotes the annual expenditure on food and beverages for a family in $

by question X~N(5700,15002)

a) let the desired range be $k

so we must have P[X<k]=0.10

or, P[(X-5700)/1500<(k-5700)/1500]=0.10

or, P[Z<(k-5700)/1500]=0.10 where Z~N(0,1)

or, P[Z<(k-5700)/1500]=0.10=P[Z<-1.281552]

so (k-5700)/1500=-1.281552

or, k=5700-1.281552*1500=$3777.672 [answer]

so the range is less than $3777.672

b) % of families spending more than $7000 annually on food and beverages is

100*P[X>7000]=100*P[(X-5700)/1500>(7000-5700)/1500]=100*P[Z>0.8666667]=100*0.1930623=19.30623% [answer]

c) let the rang be s

so P[X>s]=0.05 or, P[(X-5700)/1500>(s-5700)/1500]=0.05

or, P[Z>(s-5700)/1500]=0.05

or, P[Z<(s-5700)/1500]=0.95=P[Z<1.644854]

or, (s-5700)/1500=1.644854

or, s=5700+1500*1.644854=8167.281

hence the range is greater than $8167.281 [answer]

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