Question

Assuming the life expectancy of horses follows the normal distribution with a mean of 27 years and a standard deviation of 1.2 years.
a) If we randomly select 13 horses, what is the probability that the average lifetime of the horses is greater than or equal to 28 years?
b) In a group of 35 horses, approximately how many of them will live between 25 to 30 years?
c) Knowing that a horse died earlier than 63% of all horses, approximately how many years does this horse live?

Answer 1

X = life expectancy of horses

X ~ N(μ=27, σ=1.2)

a) n = 13

$P(\bar{X}>28) = 1 - P(\bar{X}<28) \\ = 1 -P(\frac{\bar{X}-μ}{σ/\sqrt{n}}<\frac{28-27}{1.2/\sqrt{13}}) \\ = 1 -P(Z<3.0046) \\ = 1 -0.9987 \\ = 0.0013$

b) n = 35

$P(25< \bar{X}<30) = P(\frac{25-27}{1.2}< \bar{X} < \frac{30-27}{1.2}) \\ = P(-1.667<Z<2.5) \\ = P(Z<2.5)-P(Z<-1.667) \\ = 0.99379 -0.04779 \\ = 0.946 \\ 35 \times 0.946 = 33.11$

In a group of 35 horses, approximately 33 will live between 25 to 30 years.

c) Let this horse live will be Y.

$P(X<Y) = 1-0.63 \\ P(\frac{X-μ}{σ}<\frac{Y-27}{1.2})=0.37 \\ P(Z<\frac{Y-27}{1.2})=0.37 \\ P(Z<-0.33)=0.37 \\ \frac{Y-27}{1.2}=-0.33, \, Y = 26.6$

Approximately this horse live for 26.6 years.