Question

On average, indoor cats live to 15 years old with a standard deviation of 2.5 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 10.9 and 14.8 years old.  

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

Please answer entirely and accuratley

Answer 1

Solution :

Given that ,

mean = = 15

standard deviation = = 2.5

(a)

X ~ N(15 , 2.5)

(b)

P(10.9 < x < 14.8) = P((10.9 - 15 / 2.5) < (x - ) / < (14.8 - 15) / 2.5) )

= P(-1.64 < z < -0.08)

= P(z < -0.08) - P(z < -1.64)

= 0.4681 - 0.0505 = 0.4176

Probability = 0.4176

(c)

P(Z < z) = 0.30

P(Z < -0.5244) = 0.30

z = -0.5244

Using z-score formula,

x = z * +

x = -0.5244 * 2.5 + 15 = 13.689

Low = 13.689ears

P(Z > z) = 30%

1 - P(Z < z) = 0.30

P(Z < z) = 1 - 0.30 = 0.70

P(Z < 0.5244) = 0.70

z = 0.5244

Using z-score formula,

x = z * +

x = 0.5244 * 2.5 + 15 = 16.311

High = 16.311 years