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 15.6 and 18.8 years old.

c. The middle 30% of indoor cats' age of death lies between what two numbers?
     Low:  years
     High:  years

Answer 1

a)

X ~ N(15,2.3)

b)

Here, μ = 15, σ = 2.3, x1 = 15.6 and x2 = 18.8. We need to compute P(15.6<= X <= 18.8). The corresponding z-value is calculated using Central Limit Theorem

z = (x - μ)/σ
z1 = (15.6 - 15)/2.3 = 0.26
z2 = (18.8 - 15)/2.3 = 1.65

Therefore, we get
P(15.6 <= X <= 18.8) = P((18.8 - 15)/2.3) <= z <= (18.8 - 15)/2.3)
= P(0.26 <= z <= 1.65) = P(z <= 1.65) - P(z <= 0.26)
= 0.9505 - 0.6026
= 0.3479

c)

z value at 30% = +/- 0.39

LOw value = -0.39 = (x - 15)/2.3
x = 2.3 * -0.39 + 15
x = 14.1030

High value = 0.39 = (x - 15)/2.3
x = 2.3 * 0.39 + 15
x = 15.8970