Answer:
[tex]z=1.64<\frac{a-14.4}{1.1}[/tex]
And if we solve for a we got
[tex]a=14.4 +1.64*1.1=16.204[/tex]
The 95th percentile of the hip breadth of adult men is 16.2 inches.
Step-by-step explanation:
Let X the random variable that represent the hips breadths of a population, and for this case we know the distribution for X is given by:
[tex]X \sim N(14.4,1.1)[/tex]
Where [tex]\mu=14.4[/tex] and [tex]\sigma=1.1[/tex]
For this part we want to find a value a, such that we satisfy this condition:
[tex]P(X>a)=0.05[/tex] (a)
[tex]P(X<a)=0.95[/tex] (b)
We can find a quantile in the normal standard distribution who accumulates 0.95 of the area on the left and 0.05 of the area on the right it's z=1.64
Using this value we can set up the following equation:
[tex]P(X<a)=P(\frac{X-\mu}{\sigma}<\frac{a-\mu}{\sigma})=0.95[/tex]
[tex]P(z<\frac{a-\mu}{\sigma})=0.95[/tex]
And we have:
[tex]z=1.64<\frac{a-14.4}{1.1}[/tex]
And if we solve for a we got
[tex]a=14.4 +1.64*1.1=16.204[/tex]
The 95th percentile of the hip breadth of adult men is 16.2 inches.