Answer to Question #98389 in Statistics and Probability for wyatt

Question

On average, indoor cats live to 15 years old with a standard deviation of 2.3 years. Suppose that the distribution is normal. Let X = the age at death of a randomly selected indoor cat. Round answers to 4 decimal places where possible.

a. What is the distribution of X? X ~ N(
,
)

b. Find the probability that an indoor cat dies when it is between 13.9 and 16.1 years old.


c. The middle 20% of indoor cats' age of death lies between what two numbers?
Low:
years
High:
years

Answer 1

a) Given that $\mu=15 \ years, \sigma=2.3\ years$

$X\sim N(\mu, \sigma^2)$

$X\sim(15, 2.3^2)$

b) $X\sim N(\mu, \sigma^2).$ Then

$$Z={X-\mu \over \sigma}\sim N(0, 1)$$

$\mu=15 \ years, \sigma=2.3\ years$

$$P(13.9<X<16.1)=P(X<16.1)-P(X<13.9)=$$

$$=P(Z<{16.1-15 \over 2.3})-P(Z<{13.9-15 \over 2.3})\approx$$

$$\approx0.683768-0.316232\approx0.3675$$

$P(13.9<X<16.1)=0.3675$

c)

$$P(\mu-\delta<X<\mu+\delta)=0.2$$

$$P(Z<{\mu+\delta-\mu \over \sigma})-P(Z<{\mu-\delta-\mu \over \sigma})=0.2$$

$$P(Z<{\delta \over \sigma})-P(Z<-{\delta \over \sigma})=0.2$$

$$P(Z<-{\delta \over \sigma})={1-0.2 \over 2}=0.4$$

$$-{\delta \over \sigma}\approx-0.253348$$

$$\delta\approx2.3\cdot0.253348=0.5827004$$

$$\mu-\delta\approx15-0.5827004\approx14.4173$$

$$\mu+\delta\approx15+0.5827004\approx15.5827$$

$Low:14.4173 \ years$

$High:15.5827\ years$