Answer :
Answer:
(a) N (16, 2.5²)
(b) 0.241
(c) Low: 15.4 years
High: 16.6 years
Step-by-step explanation:
The random variable X is defined as the age at death of a randomly selected indoor cat.
(a)
The distribution of X is:
[tex]X\sim N(\mu = 16, \sigma^{2}=2.5^{2})[/tex]
(b)
Compute the probability that an indoor cat dies when it is between 17.2 and 19.6 years old as follows:
[tex]P(17.2<X<19.6)=P(\frac{17.2-16}{2.5}<\frac{X-\mu}{\sigma}<\frac{19.6-16}{2.5})[/tex]
[tex]=P(0.48<Z<1.44)\\=P(Z<1.44)-P(Z<0.48)\\=0.92507-0.68439\\=0.24068\\\approx 0.241[/tex]
Thus, the probability that an indoor cat dies when it is between 17.2 and 19.6 years old is 0.241.
(c)
Compute the two numbers within which 20% of indoor cats' age of death lies as follows:
[tex]P(x_{1}<X<x_{2})=0.20\\\\\Rightarrow P(-z<Z<z)=0.20\\\\P(Z<z)-P(Z<-z)=0.20\\\\P(Z<z)-[1-P(Z<z)]=0.20\\\\2P(Z<z)-1=0.20\\\\2P(Z<z)=1.20\\\\P(Z<z)=0.60[/tex]
The corresponding value of z is, 0.25.
Compute the value of x₁ and x₂ as follows:
[tex]z=\frac{x_{1}-\mu}{\sigma}\\\\0.25=\frac{x_{1}-16}{2.5}\\\\x_{1}=16+(0.25\times 2.5}\\\\x_{1}=16.625\\\\x_{1}\approx 16.6[/tex] [tex]z=\frac{x_{1}-\mu}{\sigma}\\\\-0.25=\frac{x_{2}-16}{2.5}\\\\x_{2}=16-(0.25\times 2.5}\\\\x_{2}=15.375\\\\x_{2}\approx 15.4[/tex]
Low: 15.4 years
High: 16.6 years