Answer to Question #119849 in Statistics and Probability for MalaikaTahir

Question

Suppose a special type of small data processing firm is so specialized that some have difficulty making a profit in their first year operation. The pdf that characterizes proportion Y that makes the profit is given by:

f(y)= Ky(1-y)2 0< y < 1
(a): What is the value of “K” that renders the above a valid density function?
(b): Find the probability that at most 40% of the firms make a profit in the first year.

Answer 1

We have the pdf that characterizes proportion Y that makes the profit is given by:

f(y)= Ky(1-y)² , 0< y < 1

(a) Now, f(y) will be a valid density function if,

$\int_0^1f(y)dy=1$

$\implies\int_0^1Ky(1-y)^2dy=1$

$\implies K\int_0^1(y-2y^2+y^3)dy=1$

$\implies K[\int_0^1ydy-2\int_0^1y^2dy+\int_0^1y^3dy]=1$

$\implies K[\frac{1}{2}y^2-\frac{2}{3}y^3+\frac{1}{4}y^4]_0^1=1$

$\implies K[\frac{1}{2}-\frac{2}{3}+\frac{1}{4}]=1$

$\implies K.\frac{1}{12}=1$

$\implies K = 12$

Answer: The value of K = 12 will render the given function a valid density function.

(b) The probability that at most 40% of the firms make a profit in the first year

P(Y $\leq$ 40%)

= P(Y $\leq$ 0.4)

= $\int_0^{0.4}f(y)dy$

= $12\int_0^{0.4}y(1-y)^2dy$

= $12[\int_0^{0.4}ydy-2\int_0^{0.4}y^2dy+\int_0^{0.4}y^3dy]$

= $12[\frac{1}{2}y^2-\frac{2}{3}y^3+\frac{1}{4}y^4]_0^{0.4}$

= $12[\frac{1}{2}\times{0.4}^2-\frac{2}{3}\times{0.4}^3+\frac{1}{4}\times{0.4}^4]$

= 0.5248

Answer: The probability that at most 40% of the firms make a profit in the first year is 0.5248.