Question

According to a government study, among adults in the 25- to 34-year age group, the mean amount spent per year on reading and entertainment is $2,060. Assume that the distribution of the amounts spent follows the normal distribution with a standard deviation of $495. Use Appendix B.3. (Round your z-score computation to 2 decimal places and final answers to 2 decimal places.) What percent of the adults spend more than $2,575 per year on reading and entertainment

Answer 1

The answer is
`15.00` %

Explanation:

Let's start defining the random variable. We have the following variable :

`X:` '' The amount spent per year on reading and entertainment among adults in the 25- to 34- year age group ''

We assume that
`X` follows the normal distribution. We can write :

`X` ~ N ( μ , σ )

Where μ is the mean of the distribution and σ is the standard deviation (both are parameters from the normal distribution). Using the data from the question :

`X` ~
`N(2060;495)`

In order to answer the question, we first must calculate the probability :

`P(X>2575)` (I)

We are going to calculate this probability by making a substitution. If we substract the mean to the variable
`X` and then divide by the standard deviation, we obtain a new variable
`Z` which can be modeled as a
`N(0;1)`. This is convenient because the cumulative distribution from
`Z` is tabulated and can be found on any book or either in Internet.This process is called standardizing the variable :

[ (
`X`-μ) / σ ] =
`Z` ~
`N(0;1)` ⇒ If we apply this to the equation (I) ⇒

`P(X>2575)=P(Z>(2575-2060)/(495))=P(Z>1.04)`

Then,

`P(Z>1.04)=1-P(Z\leq 1.04)` (II)

Looking in any cumulative distribution table of
`Z` ⇒
`P(Z\leq 1.04)=0.85`

If we replace this value in (II) ⇒

`P(Z>1.04)=1-P(Z\leq 1.04)=1-0.85=0.15`

Using percent we obtain
`15.00` %