Question

A professor sees students during regular office hours. Time spent with students follow an
exponential distribution with mean of 20 minutes.
a. Write the p.d.f of X, E(X) and Var(X). ( 2 marks)
b. Find the probability that a given student spends less than 0.4 hours with the professor.
(1mark)
c. Find the probability that a given student spends more than 0.25 hours with the professor.
(1mark)
d. Find the probability that a given student spends between 0.20 and 0.5 hours with the
professor. (1mark)
e. Find the probability that a given student spends at least 0.75 hours with the professor.
(1mark)
B) If X is N (9,0.25?
2
), find:
(a) P(X> 9), (b) P( 9−? < ? < 9 + ?), (c) P(9−2? < ? < 9 + 3?), (d) P( X< 9 + 4?)

Answer 1

Ques A)

Let X be the time spent with students.

Given X is exponential with mean 20 => X ~ exp(1/20) = exp(0.05)

For an exponential distribution, pdf, mean and variance should be known because it is one of the most common type of distribution.

a. pdf = =

E[X] = = 20

Var(X) = = 400

CDF = F(X<x) = =

b. The probability that a given student spends less than 0.4 hours with the professor.

0.4 hours = 24 minutes

So, P(X<24) = F(X < 24) = = 0.698 ~ 0.7

c. The probability that a given student spends more than 0.25 hours with the professor.

0.25 hours = 15 minutes

P(X > 15) = 1 - P(X<15) = = = 0.47

d. The probability that a given student spends between 0.20 and 0.5 hours with the professor.

P(12 < X < 30) = F(30) - F(12) = = 0.55-0.22 = 0.33

e. The probability that a given student spends at least 0.75 hours with the professor.

P(X > 45) = 1 - P(X<45) = = 0.105

Ques B)

Given X ~  N (9,0.25)

Standard deviation = =

To solve this question, you must be familiar with standardizing the normal distribution.

In standardizing, we use the transformation on X as Z = . Now this value of Z corresponds to standard normal distribution, probability values of which can be found in standard normal distribution table or z-score table.

(a) P(X>9)

Since 9 is the mean of the distribution, and normal distribution is symmetric about mean or mean and median coincide, so it divides the curve into two equal halves. Hence P(X>9) = 0.5

(b)P( 9−? < ? < 9 + ?)

On standardizing this normal distribution

P( 9−? < ? < 9 + ?) = P( (9−? -9)/ < Z < (9+? -9)/ ) = P(-2 < Z < 2) = 0.95 [from standard normal table]

(c) P(9−2? < ? < 9 + 3?)

On standardizing this normal distribution

P( 9−2? < ? < 9+2?) = P( (9−2? -9)/ < Z < (9+2? -9)/​​​​​​​ ) = P(-4 < Z < 4) = 1 [from standard normal table]

(d) P( X< 9 + 4?)

On standardizing this normal distribution

P( X< 9 + 4?) = P( Z < (9+4? -9)/​​​​​​​ ) = P(Z<8) = 1    [from standard normal table]