Answer :
Answer:
[tex]z=-0.253<\frac{a-100}{15}[/tex]
And if we solve for a we got
[tex]a=100 -0.253*15=96.205[/tex]
And for this case the value would be 96.2 for the P40
Step-by-step explanation:
Let X the random variable that represent the IQ scores of a population, and for this case we know the distribution for X is given by:
[tex]X \sim N(100,15)[/tex]
Where [tex]\mu=100[/tex] and [tex]\sigma=15[/tex]
We want to find the P40 ( a) so we need to satisfy the following condition:
[tex]P(X>a)=0.60[/tex] (a)
[tex]P(X<a)=0.40[/tex] (b)
For this case we can look for the critical value in the normal standar ddistirbution who accumulate 0.4 of the area in the left and 0.6 in the right and we got z=-0.253.
If we use condition (b) from previous we have this:
[tex]P(X<a)=P(\frac{X-\mu}{\sigma}<\frac{a-\mu}{\sigma})[/tex]
[tex]P(z<\frac{a-\mu}{\sigma})=0.40[/tex]
And we can set up the following equationL
[tex]z=-0.253<\frac{a-100}{15}[/tex]
And if we solve for a we got
[tex]a=100 -0.253*15=96.205[/tex]
And for this case the value would be 96.2 for the P40