Question

a. Consider the following distribution of a random variable X

X	– 1	0	1	2
P(X)	1/8	2/8	3/8	2/8

                Find the mean, variance and standard deviation

b. There are 6 Republican, 5 Democrat, and 4 Independent candidates. Find the probability the committee will be made of 3 Republicans, 2 Democrats, and 2 Independent be selected?

c. At a large university, the probability that a student takes calculus and is on the dean’s list is 0.042. The probability that a student is on the dean’s list is 0.21. Find the probability that the student is taking calculus given that the student is on the dean’s list.

d. In a statistics class there are 16 juniors and 12 seniors; 8 of the senior students are females, and 12 of the juniors are males. If a student is selected at random, find the probability that the student is neither a junior nor a male.

Answer 1

a)

X	P(X)	X*P(X)	X² * P(X)
-1	1/8	-0.125	0.1250
0	2/8	0.000	0.0000
1	3/8	0.375	0.3750
2	2/8	0.500	1.0000

	P(X)	X*P(X)	X² * P(X)
total sum =	1	0.75	1.50

mean = E[X] = Σx*P(X) =            0.7500
          
E [ X² ] = ΣX² * P(X) =            1.5000
          
variance = E[ X² ] - (E[ X ])² =            0.9375
          
std dev = √(variance) =            0.9682

b)

6 Republican, 5 Democrat, and 4 Independent candidates

total candidates=6+5+4 = 15

P( 3 Republicans, 2 Democrats, and 2 Independent) = 6C3 *5C2 * 4C2 / 15C7 = 0.1865

c)

P(calculaus | dean's list) = P(calculus and dean's list)/P(deans list) = 0.042/0.21 = 0.2

d)

	male	female	total
junior	12	4	16
senior	4	8	12
total	16	12	28

P(neither a junior nor a male) = P(senior and female) = 8/28 = 0.2857