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|
|
Passed |
Did not pass |
|
Female |
430 |
60 |
|
Male |
410 |
100 |
A class consists of 490 female and 510 male students. The students are divided according to their marks If one person is selected randomly, the probability that it did not pass given that it is female is: *0.12 (it is male.*0.196)
|
X |
0 |
2 |
4 |
6 |
|
P(x) |
0.05 |
0.17 |
0.43 |
0.35 |
Find the variance for the given probability distribution. *2.8544
Probability mass function for discrete random variable X is represented by the graph. Find Var(X).*1
|
Х |
0 |
2 |
4 |
|
Р |
0,1 |
0,5 |
0,4 |
PMF of discrete random variable is given by
Find the value of CDF of X in the interval (2, 4].*0.6
|
Х |
0 |
2 |
4 |
|
Р |
0,3 |
0,1 |
0,6 |
PMF of discrete random variable is given by
Find F(2).*0.4
|
Х |
-1 |
5 |
|
Р |
0,4 |
0,6 |
PMF of discrete random variable X is given by
Find standard deviation of X.*2.9393
|
Х |
-1 |
5 |
|
Р |
0,4 |
0,6 |
PMF of discrete random variable X is given by
Find variance of X.*8.64
|
Х |
0 |
5 |
|
|
Р |
0,6 |
0,1 |
0,3 |
PMF of discrete random variable X is given by
If E[X]=3.5 then find the value of x3.*10
|
X |
1 |
2 |
3 |
4 |
|
F(X) |
1/8 |
3/8 |
3/4 |
1 |
Table shows the cumulative distribution function of a random variable X. Determine *7/8
|
X |
1 |
2 |
3 |
4 |
|
F(X) |
1/8 |
3/8 |
3/4 |
1 |
Table shows the cumulative distribution function of a random variable X. Determine .*0
|
xi |
0 |
x2 |
5 |
|
pi |
0.1 |
0.2 |
0.7 |
The table below shows the probability mass function of a random variable X. If E[X]=5,5 find the value of x2. *10
|
X |
-2 |
1 |
2 |
|
Р |
0,1 |
0,6 |
0,3 |
The table below shows the pmf of a random variable X. What is the Var(X)?*1.2
|
X |
-1 |
0 |
1 |
|
P |
0.2 |
0.3 |
0.5 |
The table below shows the pmf of a random variable X. Find E[X] and Var(X).*E[X]= 0,3; Var(X) =0.61
We are given the pmf of two random variables X and Y shown in the tables below.
|
Х |
1 |
3 |
У |
2 |
4 |
||
|
px |
0,4 |
0,6 |
py |
0,2 |
0,8 |
Find E[X+Y].*5,8
A random variable Y has the following distribution: Y | -1 0 1 2
P(Y)| 3C 2C 0.4 0.1 The value of the constant C is:*0.1
A random variable X has a probability distribution as follows: X | 0 1 2 3
P(X) | 2k 3k 13k 2k Then the probability that P(X < 2.0) is equal to*0.25
A random variable X has the cumulative distribution function
Suppose that the random variable T has the following probability distribution:
t | 0 1 2
P(T = t) | .5 .3 .2 Find . *0.5
Suppose that the random variable T has the following probability distribution:
t | 0 1 2 P(T = t) | .5 .3 .2 Find . *0.8
Suppose that the random variable T has the following probability distribution:
t | 0 1 2
P(T = t) | .5 .3 .2 Compute the mean of the random variable T.*0.7