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 $390 on back-to-college electronics?
(c) They spend between $110 and $190 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 > 390) = enter the probability that they spend more than $390 on back-to-college electronics
(c) P(110 < x < 190) = enter the probability that they spend between $110 and $190 on back-to-college electronics

Answer 1

(A) use normalcdf

set lower = -1E99, upper = 145, mean = 237 and standard deviation = 54

we get

P(X<145) = normalcdf(-1E99,145,237,54)

P(X<145) = 0.0446

(B) use normalcdf

set lower = 390, upper = 1E99, mean = 237 and standard deviation = 54

we get

P(X>390) = normalcdf(390,1E99,237,54)

P(X>390) = 0.0023

(C) use normalcdf

set lower = 110, upper = 390, mean = 237 and standard deviation = 54

we get

P(100<X<390) = normalcdf(110,390,237,54)

P(100<X<390) = 0.9883