Advertisements
Advertisements
Question
The cumulative distribution function of a discrete random variable is given by
F(x) = `{{:(0, - oo < x < - 1),(0.15, - 1 ≤ x < 0),(0.35, 0 ≤ x < 1),(0.60, 1 ≤ x < 2),(0.85, 2 ≤ x < 3),(1, 3 ≤ x < oo):}`
Find the probability mass function
Advertisements
Solution
Given F(x) = `{{:(0, - oo < x < - 1),(0.15, - 1 ≤ x < 0),(0.35, 0 ≤ x < 1),(0.60, 1 ≤ x < 2),(0.85, 2 ≤ x < 3),(1, 3 ≤ x < oo):}`
The value of 'x' are –1, 0, 1, 2, 3
F(–1) = P(X = –1)
= F(–1) – F(–1)
= 0.15 – 0
= 0.15
F(0) = P(X = 0)
= F(0) – F(–1)
= 0.35 – 0.15
= 0.20
F(1) = P(X = 1)
= F(1) – F(0)
= 0.60 – 0.35 =
0.25
F(2) = P(X = 2)
= F(2) – F(1)
= 0.85 – 0.60
= 0.25
F(3) = P(X = 3)
= F(3) – F(2)
= 1 – 0.85
= 0.15
Probability mass function table
| x | –1 | 0 | 1 | 2 | 3 |
| F(x) | 0.15 | 0.20 | 0.25 | 0.25 | 0.15 |

APPEARS IN
RELATED QUESTIONS
Suppose error involved in making a certain measurement is continuous r.v. X with p.d.f.
f (x) = k `(4 – x^2 )`, for –2 ≤ x ≤ 2 and = 0 otherwise.
P(x > 0)
Given the p.d.f. of a continuous r.v. X , f (x) = `x^2/3` ,for –1 < x < 2 and = 0 otherwise
Determine c.d.f. of X hence find
P( x < 1)
Given the p.d.f. of a continuous r.v. X ,
f (x) = `x^2/3` , for –1 < x < 2 and = 0 otherwise
Determine c.d.f. of X hence find P(1 < x < 2)
The p.m.f. of a r.v. X is given by P (X = x) =`("" ^5 C_x ) /2^5` , for x = 0, 1, 2, 3, 4, 5 and = 0, otherwise.
Then show that P (X ≤ 2) = P (X ≥ 3).
F(x) is c.d.f. of discrete r.v. X whose p.m.f. is given by P(x) = `"k"^4C_x` , for x = 0, 1, 2, 3, 4 and P(x) = 0 otherwise then F(5) = _______
Fill in the blank :
The value of continuous r.v. are generally obtained by _______
Solve the following problem :
Identify the random variable as discrete or continuous in each of the following. Identify its range if it is discrete.
A person on high protein diet is interested in the weight gained in a week.
A six sided die is marked ‘1’ on one face, ‘3’ on two of its faces, and ‘5’ on remaining three faces. The die is thrown twice. If X denotes the total score in two throws, find P(4 ≤ X < 10)
The cumulative distribution function of a discrete random variable is given by
F(x) = `{{:(0, - oo < x < - 1),(0.15, - 1 ≤ x < 0),(0.35, 0 ≤ x < 1),(0.60, 1 ≤ x < 2),(0.85, 2 ≤ x < 3),(1, 3 ≤ x < oo):}`
Find P(X < 1)
The cumulative distribution function of a discrete random variable is given by
F(x) = `{{:(0, - oo < x < - 1),(0.15, - 1 ≤ x < 0),(0.35, 0 ≤ x < 1),(0.60, 1 ≤ x < 2),(0.85, 2 ≤ x < 3),(1, 3 ≤ x < oo):}`
Find P(X ≥ 2)
A random variable X has the following probability mass function.
| x | 1 | 2 | 3 | 4 | 5 |
| F(x) | k2 | 2k2 | 3k2 | 2k | 3k |
Find P(2 ≤ X < 5)
Choose the correct alternative:
A pair of dice numbered 1, 2, 3, 4, 5, 6 of a six-sided die and 1, 2, 3, 4 of a four-sided die is rolled and the sum is determined. Let the random variable X denote this sum. Then the number of elements in the inverse image of 7 is
The p.m.f. of a random variable X is
P(x) = `(5 - x)/10`, x = 1, 2, 3, 4
= 0, otherwise
The value of E(X) is ______
If the probability function of a random variable X is defined by P(X = k) = a`((k + 1)/2^k)` for k - 0, 1, 2, 3, 4, 5, then the probability that X takes a prime value is ______
The c.d.f. of a discrete r.v. x is
| x | 0 | 1 | 2 | 3 | 4 | 5 |
| F(x) | 0.16 | 0.41 | 0.56 | 0.70 | 0.91 | 1.00 |
Then P(1 < x ≤ 4) = ______
The p.d.f. of a continuous random variable X is
f(x) = 0.1 x, 0 < x < 5
= 0, otherwise
Then the value of P(X > 3) is ______
At random variable X – B(n, p), if values of mean and variance of X are 18 and 12 respectively, then total number of possible values of X are ______.
For the following distribution function F(x) of a rv.x.
| x | 1 | 2 | 3 | 4 | 5 | 6 |
| F(x) | 0.2 | 0.37 | 0.48 | 0.62 | 0.85 | 1 |
P(3 < x < 5) =
If f(x) = `k/2^x` is a probability distribution of a random variable X that can take on the values x = 0, 1, 2, 3, 4. Then, k is equal to ______.
