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 $150 on back-to-college electronics?
(b) They spend more than $350 on back-to-college electronics?
(c) They spend between $120 and $175 on back-to-college electronics?

Answer 1

Solution :

Given that ,

mean =   = 237

standard deviation = σ   = 54   

P(X< 150) = P[(X- ) / σ   < (150-237) /54 ]

= P(z < -1.61)

Using z table

= 0.0537   

probability =0.0537

b)

Solution :

Given ,

mean = = 237

standard deviation = = 54

P(x >530 ) = 1 - P(x<530 )

= 1 - P[ X - / / (530-237) / 54]

= 1 - P(z < 5.43)

Using z table

= 1 - 1

= 00

probability= 0

C)

Solution :

Given that ,

mean =   = 237

standard deviation = = 54   

P(120< x <175) = P[(120-237) / 54< (x - ) / < (175-237) / 54)]

= P(-2.17 < Z < -1.15)

= P(Z < -1.15) - P(Z < -2.17)

Using z table   

= 0.1251-0.015

probability= 0.1101