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 $165 on back-to-college electronics?

(b) They spend more than $380 on back-to-college electronics?

(c) They spend between $130 and $180 on back-to-college electronics?

Answer 1

Part a)

X ~ N ( µ = 237 , σ = 54 )
P ( X < 165 )
Standardizing the value
Z = ( X - µ ) / σ
Z = ( 165 - 237 ) / 54
Z = -1.3333
P ( ( X - µ ) / σ ) < ( 165 - 237 ) / 54 )
P ( X < 165 ) = P ( Z < -1.3333 )
P ( X < 165 ) = 0.0912

Part b)

X ~ N ( µ = 237 , σ = 54 )
P ( X > 380 ) = 1 - P ( X < 380 )
Standardizing the value
Z = ( X - µ ) / σ
Z = ( 380 - 237 ) / 54
Z = 2.6481
P ( ( X - µ ) / σ ) > ( 380 - 237 ) / 54 )
P ( Z > 2.6481 )
P ( X > 380 ) = 1 - P ( Z < 2.6481 )
P ( X > 380 ) = 1 - 0.996
P ( X > 380 ) = 0.004

Part c)

X ~ N ( µ = 237 , σ = 54 )
P ( 130 < X < 180 )
Standardizing the value
Z = ( X - µ ) / σ
Z = ( 130 - 237 ) / 54
Z = -1.9815
Z = ( 180 - 237 ) / 54
Z = -1.0556
P ( -1.98 < Z < -1.06 )
P ( 130 < X < 180 ) = P ( Z < -1.06 ) - P ( Z < -1.98 )
P ( 130 < X < 180 ) = 0.1456 - 0.0238
P ( 130 < X < 180 ) = 0.1218