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] %
