According to an NRF survey conducted by BIGreserch, the average family spends ab

Question

According to an NRF survey conducted by BIGreserch, the average family spends about $237 on electronics (computers, cell phones, etc.) in back-to-college spending per student. Suppose back-to-college family spending on electronics is normally distributed with a standard deviation of $54. If a family of a returning college student is randomly selected, what is the probability that:

A) They spend less than $160 on back-to-college electronics? P(x < 160) = 0.0764

B) They spend more than $350 on back-to-college electronics? P(x > 350) = ???

C) They spend between $115 and $175 on back-to-college electronics? P(115 < x < 175) = ???

(Round the valurs of z to 2 decimal places. Round your answers to 4 decimal places.)

Explanation / Answer

A. z=160-237/54=-1.43 so P(x<160)=> P(z<-1,43)=0.0764

B. z=350-237/54=2.09 now P(x>350)=> P(z>2.09)=1-P(x<=2.09)=1-0.9817=0.0183

C. z1=115-237/54=-2.26 and z2=117-237/54=-2.22

Now P(115<x<117)=> P(-2.26<z<-2.22)=P(-2.22)-P(z<=-2.26)=0.0132-0.0119=0.0013

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