Question

According to an NRF survey conducted by BIGresearch, 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 $145 on back-to-college electronics? (b) They spend more than $350 on back-to-college electronics? (c) They spend between $140 and $180 on back-to-college electronics? (Round the values of z to 2 decimal places. Round your answers to 4 decimal places.) (a) P(x < 145) = enter the probability that they spend less than $145 on back-to-college electronics (b) P(x > 350) = enter the probability that they spend more than $350 on back-to-college electronics (c) P(140 < x < 180) = enter the probability that they spend between $140 and $180 on back-to-college electronics.

Answer 1

Answer)

As the data is normally distributed we can use standard normal z table to estimate the answers

Z = (x-mean)/s.d

Given mean = 237

S.d = 54

A)

P(x<145)

Z = (145 - 237)/54 = -1.7

From z table, P(z<-1.7) = 0.0446

B)

P(x>350)

Z = (350 - 237)/54 = 2.09

From z table, P(z>2.09) = 0.0183

C)

P(140<x<180) = P(x<180) - P(x<140)

P(x<180)

Z = (180 - 237)/54 = -1.06

From z table, P(z<-1.06) = 0.1446

P(x<140)

Z = -1.8

From.z table, P(z<-1.8) = 0.0359

Required probability is 0.1446 - 0.0359 = 0.1087