Question

How can i calculate the following statistics of engine life-span expectancy? (statistics, would really appreciate help with this)

the life expectancy of an engine from specific kind is a continuous random variable x, with the following probability density function `f_x(x)=cx^2(10-x)`.

a)what is the variance of the life expectancy of the engine?

b)knowing that a certain engine from the kind given in the question, works for the last 7 years, what are the odds he'll continue working for at least 2 additional years?

c)a man bought 20 engines of this kind. assuming that there's no dependency between the life span of different engines, calculate the probability of the life span of exactly 14 of the engines will be bigger than 5.5 years.

please help me with this as i am stuck and don't know how to solve it. would really appreciate learning from your answers and understanding how to solve this kind of questions.

thank you very much

Answer 1

`"a) "4`
`"b) 0.150158"`
`"c) 0.133705"`

Explanation:

`"Note that a probability cannot be negative, therefore i guess"`
`"we have to assume that x goes from 0 to 10."`
`"First of all we need to determine c so that the sum of all"`
`"probabilities is 1 :"`
`int_0^10 cx^2(10 - x) "" dx= c int_0^10 x^2(10 - x)""dx`
`= 10 c int_0^10 x^2 dx - c int_0^10 x^3 dx`
`= 10 c [ x^3 / 3]_0^10 - c [ x^4 / 4 ]_0^10`
`= 10000 c / 3 - 10000 c / 4`
`= 10000 c (1/3 - 1/4)`
`= 10000 c (4 - 3)/12`
`= 10000 c / 12`
`= 1`
`=> c = 12/10000 = 0.0012`

`"a) variance = "E(X^2) - (E(X))^2`
`E(X) = int_0^10 0.0012 x^3(10 - x)dx`
`= 0.0012 int_0^10 x^3(10-x)dx`
`= 0.012 int_0^10 x^3 dx - 0.0012 int_0^10 x^4 dx`
`= 0.012 * 10^4/4 - 0.0012 * 10^5 / 5`
`= 30 - 24`
`= 6`
`E(X^2) = int_0^10 0.0012 x^4(10 - x)dx`
`= 0.012 int_0^10 x^4 dx - 0.0012 int_0^10 x^5 dx`
`= 0.012 * 10^5/5 - 0.0012 * 10^6/6`
`= 240 - 200`
`= 40`
`=> "variance = "40 - 6^2 = 4`

`"b) "P("Engine works > 9 years | It works at least 7 years") =`
`(P("It works at least 7 years AND it works > 9 years")) / (P("It works at least 7 years"))`
`= (P("It works > 9 years")) / (P("It works > 7 years"))`
`= (int_9^10 0.0012 x^2(10 - x) dx) / (int_7^10 0.0012 x^2(10-x) dx)`
`= [10 x^3/3 - x^4/4]_9^10 / [10 x^3/3 - x^4/4]_7^10`
`= (10000/3 - 10000/4 - 10*9^3/3 + 9^4/4)/(10000/3 - 10000/4 - 10*7^3/3 + 7^4/4)`
`= (10000/12 - 789.75)/(10000/12 - 543.0833)`
`= 0.150158`

`"c) "P("Engine works >= 5.5 years") = int_5.5^10 0.0012 x^2(10-x) dx`
`= 0.0012 [10 x^3/3 - x^4/4]_5.5^10`
`= 0.0012 [ 10000/12 - 10*5.5^3/3 + 5.5^4/4 ]`
`= 0.60901875`
`"Now we need to apply the binomial distribution with"`
`"n=20, k=14, p = 0.60901875"`
`P = C(20,14) 0.60901875^14 (1-0.60901875)^6`
`= 0.133705`