A recent study done by the National Retail Federation found that 2019 back-to-school spending for all US households who have school-aged children follows a Normal distribution with mean $697 and a standard deviation $120.
Use this information to answer the following questions.
(a) What is the probability that 2019 back-to-school spending for a US household with school-aged children is greater than $893?
(b) Provide the z-score corresponding to the 2019 back-to-school spending of $893.
(c) Based on your answer in (b), what is the probability of 2019 back-to-school spending for a US household with school-aged children is greater than $893?

a) P(x > 893) = 0.051

b) $893 has a z-score of 1.633

c) P(z > 1.633) = 0.051

Step-by-step explanation:

This is a normal distribution problem

μ = mean = $697

σ = standard deviation = $120

We first normalize the $893

The standardized score for any value is the value minus the mean then divided by the standard deviation.

z = (x - μ)/σ = (893 - 697)/120 = 1.633

To determine the probability of 2019 back-to-school spending for a US household with school-aged children is greater than $893 = P(x > 893) = P(z > 1.633)

We'll use data from the normal probability table for these probabilities

P(x > 893) = P(z > 1.633) = 1 - P(z ≤ 1.633) = 1 - 0.949 = 0.051

step-by-step explanation:

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