Question

Question 1: Motherboard can be ordered in lots of 1, 2, or 3 units. The random variable is the number of units ordered by a customer with the following probability mass function:

x	1	2	3	4	5	6
P(x)	k	2k	3k	3k	2k	k

1. Find the value of k.
2. Calculate the probability of between 3 and 5 (inclusive) units are ordered.
3. Calculate the probability of more than 3 units are ordered.
4. Calculate the mean and standard deviation of the distribution

Answer 1

Solution:

a) To find the value of k

We know that Total probability is always 1

Sum of P_i = 1

k + 2k + 3k + 3k + 2k + k = 1

12k = 1

k = 1/12 =  0.08333333333

b)

First we rewrite the PMF of the distribution of the X

x	1	2	3	4	5	6	Sum
P(x)	1/12	2/12 = 1/6	3/12 = 1/4	3/12 = 1/4	2/12 = 1/6	1/12	1

P(between 3 and 5)

= P(3 X 5 )

= P(X = 3) + P(X = 4) + P(X = 5)

= (3/12) + (3/12) + 2/12)

= 8/12

= 2/3

P(between 3 and 5) = 2/3

c)

P(More than 3)

= P(X = 4) + P(X = 5) + P(X = 6)

= (3/12) + (2/12) + (1/12)

= 6/12

= 1/2

P(More than 3) = 1/2

d)

We prepare a table.

x	1	2	3	4	5	6	Sum
P(x)	1/12	2/12	3/12	3/12	2/12	1/12	1
x2	1	4	9	16	25	36	-
x* P(x)	1/12	4/12	9/12	12/12	10/12	6/12	42/12= 3.5
x2 * P(x)	1/12	8/12	27/12	48/12	50/12	36/12	170/12 = 14.16666667

Mean = E(X) = Summation of x * P(x) = 42/12 = 3.5

E(X² ) = Summation of x² * P(x) = 170/12

Variance = E(X² ) - {E(X)² } = (170/12) - (3.5)² = 1.9167

Standard deviation = variance = 1.3844

Answer : Mean = 3.5

Standard deviation = 1.3844