Question

On average, indoor cats live to 16 years old with a standard deviation of 2.6 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 16.2 and 18.4 years old.

c. The middle 40% of indoor cats' age of death lies between what two numbers?
     Low:  years
     High:  years

Answer 1

Solution :

Given that ,

mean = = 16

standard deviation = = 2.6

a)

The distribution of X ~ N( , )

b)

P(16.2 < x < 18.4) = P((16.2 - 16)/ 2.6) < (x - ) /  < (18.4 - 16) / 2.6) )

= P(0.08 < z < 0.92)

= P(z < 0.92) - P(z < 0.08)

= 0.8212 - 0.5319 Using standard normal table,  

Probability = 0.2893

c)

P( -z < Z < z ) = 0.40

P( Z < z ) - P( Z < -z ) = 0.40

2*P(Z < z ) - 1 = 0.40

2*P(Z < z ) = 1 + 0.40

P( Z < z ) = 1.40 / 2

P( Z < z ) = 0.7

P( Z < 0.524 ) = 0.7

z = 0.524

and

z = -0.524

Using z - score formula,

X = z * +

= -0.524 * 2.6 + 16  

= 14.6

and

X = z * +

= 0.524 * 2.6 + 16  

= 17.4

Low: 14.6 year

High: 17.4 years