Problem 9 [25 marks]. Suppose the life expectancy of house cats can be modeled w
ID: 3315216 • Letter: P
Question
Problem 9 [25 marks]. Suppose the life expectancy of house cats can be modeled with a a) (5 points) Find the central range of life expectancies of 95% of house cats. Interpret b) (4 points) To what age do the top 2% of cats live? What percentile does this age normal distribution with an average of 14.5 years and a standard deviation of 3 years what this interval means in this context (try not to restate the question) represent? Explain what your results mean c) (3 points) What is the probability that a randomly selected house cat will live more than 18 years? d) (4 points) What is the probability that a randomly selected house cat will live more than 18 years if we know it is already 15 years old? Justify in plain English why the result is different from (c) e) (3 points) Consider more generally a random variable X where X ~ N(, *). Find the value of k in terms of number of standard deviations such that P(-ko-X + ko) = 0.95. Comment on any similarities you notice to part (a) f) (2 points) Referring to part (d), what does the random variable Z, or your Z-scores in the normal probability table, represent in terms of and for a normal random variable X ~ N(, *)? g) (4 points) Using X ~ N(, 2), show that PIX-Explanation / Answer
a)
mean = 14.5 sd = 3
95 % confidence interval
(14.5 + 1.96*3 , 14.5 + 3*1.96)
= (8.62,20.38)
this means we are 95 % confident that the actual life expectancy is between this confidence interval
b)
top 2 %
P(Z> z*) = 0.02
z*= 2.054
X = mean + z* sd
= 14.5 +2.054*3
= 20.662
percentile = 98
this means 98 % of house cats have life expectancy less than 20.662 years
c)
P(X> 18)
=P(Z>(18 -14.5)/3)
=P(Z> 1.16666)
= 0.1217
d)
P(X> 18|X>15)= P(X>18)/P(X>15)
P(X>15) = P(Z>(15 -14.5)/3)
= P(Z> 0.166666)
= 0.4338
hence required probability = 0.1217/0.4338 = 0.280544
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