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:

The answer is [tex]15.00[/tex] %

Step-by-step explanation:

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

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

We assume that [tex]X[/tex] follows the normal distribution. We can write :

[tex]X[/tex] ~ 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 :

[tex]X[/tex] ~ [tex]N(2060;495)[/tex]

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

[tex]P(X>2575)[/tex] (I)

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

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

[tex]P(X>2575)=P(Z>\frac{2575-2060}{495})=P(Z>1.04)[/tex]

Then,

[tex]P(Z>1.04)=1-P(Z\leq 1.04)[/tex] (II)

Looking in any cumulative distribution table of [tex]Z[/tex] ⇒ [tex]P(Z\leq 1.04)=0.85[/tex]

If we replace this value in (II) ⇒

[tex]P(Z>1.04)=1-P(Z\leq 1.04)=1-0.85=0.15[/tex]

Using percent we obtain [tex]15.00[/tex] %

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