Question

The lengths of a professor's classes has a continuous uniform distribution between 50.0 min and 52.0 min. If one such class is randomly selected, find the probability that the class length is more than 50.2 min. PX> 50.2)( (Report answer accurate to 2 decimal places.) The lengths of a professor's classes has a continuous uniform distribution between 50.0 min and 52.0 min. If one such class is randomly selected, find the probability that the class length is between 50.3 and 50.8 min. P(50.3 X50.8)-| Let X have a uniform distribution on the interval (10, 16). Find the probability that the sum of 2 independent observations of X is greater than 29.

Answer 1

Answer:

a)  P(X > 50.2) = 0.90

b) P(50.3 < X < 50.8) = 0.25

c) P(X+Y >29) = 0.9167

Step-by-step explanation:

From the given information:

$f(x) = \dfrac{1}{b-a} \ \ \ \ a<x<b$

$f(x) = \dfrac{1}{52.0-50.0} \ \ \ \ 50.0<x<52.0$

$f(x) = \dfrac{1}{2} \ \ \ \ 50.0<x<52.0$

a)

Let X be the random variable that follows a uniform distribution.

Such that $X \sim U (50.0, 52.0)$

where;

a = 50

b = 52

$P(X> 50.2) = \dfrac{b - x}{b-a}$

$P(X> 50.2) = \dfrac{52 - 50.2}{52-50}$

$P(X> 50.2) = \dfrac{1.8}{2.0}$

P(X > 50.2) = 0.90

b.)

To find the probability that the class length is between 50.3 and 50.8 min.

i.e.

$P(50.3 <X<50.8) = \dfrac{(X_2-X_1)}{(b-a)}$

$P(50.3 <X<50.8) = \dfrac{(50.8-50.3)}{(52-50)}$

$P(50.3 <X<50.8) = \dfrac{(0.5)}{(2)}$

P(50.3 < X < 50.8) = 0.25

c)

Given that:

the interval = (101,16)

The probability that the sum of 2 independent observations of X is greater than 29 can be computed as follows.

here;

$c = \dfrac{a+b}{2}$

where;

a = 10*2 = 20

b = 16*2 = 32

$c = \dfrac{20+32}{2}$

c = 52/2

c =26

The required probability:

$P(X+Y >29) = 1 - \dfrac{(b-x)\times 2}{(b-a) \times (b - c)}$

$P(X+Y >29) = 1 - \dfrac{(32-29)\times 2}{(32-20) \times (32 - 26)}$

$P(X+Y >29) = 1 - \dfrac{(3)\times 2}{(12) \times (6)}$

$P(X+Y >29) = 1 - \dfrac{6}{72}$

P(X+Y >29) = 1 - 0.0833

P(X+Y >29) = 0.9167