Advertisements
Advertisements
प्रश्न
Suppose a discrete random variable can only take the values 0, 1, and 2. The probability mass function is defined by
`f(x) = {{:((x^2 + 1)/k"," "for" x = 0"," 1"," 2),(0"," "otherwise"):}`
Find the value of k
Advertisements
उत्तर
Given f is a probability mass function
`sum_x "f"(x)` = 1
Probability mass function is
| x | 0 | 1 | 2 |
| F(x) | `1/k` | `2/k` | `5/k` |
`1/k + 2/k + 5/k` = 1
`8/k` = 1
k = 8
APPEARS IN
संबंधित प्रश्न
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 (–0·5 < x or x > 0·5)
The following is the p.d.f. of continuous r.v.
f (x) = `x/8` , for 0 < x < 4 and = 0 otherwise.
Find F(x) at x = 0·5 , 1.7 and 5
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).
It is felt that error in measurement of reaction temperature (in celsius) in an experiment is a continuous r.v. with p.d.f.
f(x) = `{(x^3/(64), "for" 0 ≤ x ≤ 4),(0, "otherwise."):}`
Find P(0 < X ≤ 1).
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.
The probability distribution of a r.v. X is
| X = x | -3 | -2 | -1 | 0 | 1 |
| P(X = x) | 0.3 | 0.2 | 0.25 | 0.1 | 0.15 |
Then F (-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 the probability mass function
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)
The cumulative distribution function of a discrete random variable is given by
F(x) = `{{:(0, "for" - oo < x < 0),(1/2, "for" 0 ≤ x < 1),(3/5, "for" 1 ≤ x < 2),(4/5, "for" 2 ≤ x < 4),(9/5, "for" 3 ≤ x < 4),(1, "for" ≤ x < oo):}`
Find P(X ≥ 2)
Choose the correct alternative:
The probability mass function of a random variable is defined as:
| x | – 2 | – 1 | 0 | 1 | 2 |
| f(x) | k | 2k | 3k | 4k | 5k |
Then E(X ) is equal to:
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 probability distribution of a random variable X is given below. If its mean is 4.2, then the values of a and bar respectively
| X = x | 1 | 2 | 3 | 4 | 5 | 6 |
| P(X = x) | a | a | a | b | b | 0.3 |
Two coins are tossed. Then the probability distribution of number of tails is.
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 ______
