Answer to Question #127567 in Statistics and Probability for Jessica Sean

Question

The talk time (in hours) on a cell phone in a month is approximated by the probability density function.

f(x) = x-5 / 5h for 5 < x < 10, 1/h for 10 less than or equal to x less than or equal to 15, - x-20 / 5h for 15 less than or equal to x less than or equal 20.

Determine the following:

a. h
b. P(X<30) Round the answers to two decimal places.
c. P(X<16.0) Round the answers to two decimal places.
d. x such that P(X<x)=0.95

Answer 1

Solution:

a. Let's use the normalization condition

$\int^{+\infty}_{-\infty}P(x)dx=1$.

Then

$\int ^{10} _{5}(x-5)/(5h)dx+\int ^{15} _{10}(1/h)dx+\int ^{20} _{15}(-(x-20)/(5h))dx$

$=(x-5)^2/(10h)|^{10}_{5}+x/h|^{15}_{10}-(x-20)^2/(10h)|^{20}_{15}$

$=5/(2h)+5/h+5/(2h)=10/h$ .

$10/h=1$ =>

$h=10$.

b. $P(X<30)=\int^{30}_{-\infty}P(x)dx=1$.

c. $P(X<16)=\int^{16}_{-\infty}P(x)dx=$

$=\int ^{10} _{5}(x-5)/(5h)dx+\int ^{15} _{10}(1/h)dx+\int ^{16} _{15}(-(x-20)/(5h))dx=$

$=5/(20)+5/10-(x-20)^2/(10h)|^{16}_{15}=1/4+1/2-(16/100-25/100)=0.84$

So, $P(X<16)=0.84$

d. If $x<15$ then $P(X<x)<P(X<15)=\int^{15}_{-\infty}P(x)dx=\int ^{10} _{5}(x-5)/(5h)dx+\int ^{15} _{10}(1/h)dx=1/4+1/2=0.75$

Therefore $x>15$.

If $x>20$ then $P(X<x)>P(X<20)=1$

Therefore

$x<20$ (1)

So, $P(X<x)=\int^{x}_{-\infty}P(t)dt=$

$=\int ^{10} _{5}(t-5)/(5h)dt+\int ^{15} _{10}(1/h)dt+\int ^{x} _{15}(-(t-20)/(5h))dt=$

$=1/4+1/2-(t-20)^2/(10h)|^{x}_{15}=3/4-((x-20)^2/100-25/100)=1-(x-20)^2/100$

From the condition $P(X<x)=0.95$ we obtain the equation

$1-(x-20)^2/100=0.95$

$(x-20)^2/100=0.05$

$(x-20)^2=5$

$x_1=20-\sqrt5$

$x_2=20+\sqrt5$ - does not satisfy the condition (1).

So,

$x=20-\sqrt5$

$x\approx17.76$