Answer:
[tex]P(X<10.1)=P(\frac{X-\mu}{\sigma}<\frac{10.1-\mu}{\sigma})=P(Z<\frac{10.1-12.5}{2.4})=P(z<-1)[/tex]
And we can find this probability with the normal standard distirbution or excel:
[tex]P(z<-1)=0.1587[/tex]
And then the probability that a lion lives less than 10.1 years is 0.1587
Step-by-step explanation:
Let X the random variable that represent the lifespans of lions population, and for this case we know the distribution for X is given by:
[tex]X \sim N(12.5,2.4)[/tex]
Where [tex]\mu=12.5[/tex] and [tex]\sigma=2.4[/tex]
We want to find this probability:
[tex]P(X<10.1)[/tex]
And we can use the z score formula givrn by:
[tex]z=\frac{x-\mu}{\sigma}[/tex]
If we apply this formula to our probability we got this:
[tex]P(X<10.1)=P(\frac{X-\mu}{\sigma}<\frac{10.1-\mu}{\sigma})=P(Z<\frac{10.1-12.5}{2.4})=P(z<-1)[/tex]
And we can find this probability with the normal standard distirbution or excel:
[tex]P(z<-1)=0.1587[/tex]
And then the probability that a lion lives less than 10.1 years is 0.1587
