Advertisements
Advertisements
प्रश्न
Solve the following problem :
Let X∼B(n,p) If E(X) = 5 and Var(X) = 2.5, find n and p.
उत्तर
Let X ~ B(n, p)
E(X) = 5 and Var (X) = 2.5 …[Given]
But E(X) = np = 5 and
Var (X) = npq = 2.5
∴ 5(q) = 2.5
∴ q = `(2.5)/(5) = (1)/(2)`
∴ p = 1 – q = `1 - (1)/(2) = (1)/(2)`
Now, np = 5
∴ `"n"(1/2)` = 5
∴ n = 10.
APPEARS IN
संबंधित प्रश्न
State if the following is not the probability mass function of a random variable. Give reasons for your answer
Z | 3 | 2 | 1 | 0 | −1 |
P(Z) | 0.3 | 0.2 | 0.4 | 0 | 0.05 |
It is known that error in measurement of reaction temperature (in 0° c) in a certain experiment is continuous r.v. given by
f (x) = `x^2 /3` , for –1 < x < 2 and = 0 otherwise
Verify whether f (x) is p.d.f. of r.v. X.
Choose the correct option from the given alternative :
P.d.f. of a.c.r.v X is f (x) = 6x (1 − x), for 0 ≤ x ≤ 1 and = 0, otherwise (elsewhere)
If P (X < a) = P (X > a), then a =
If the p.d.f. of c.r.v. X is f(x) = `x^2/18`, for -3 < x < 3 and = 0, otherwise, then P(|X| < 1) = ______.
Choose the correct option from the given alternative :
If p.m.f. of a d.r.v. X is P (x) = `c/ x^3` , for x = 1, 2, 3 and = 0, otherwise (elsewhere) then E (X ) =
Choose the correct option from the given alternative:
If the a d.r.v. X has the following probability distribution :
x | -2 | -1 | 0 | 1 | 2 | 3 |
p(X=x) | 0.1 | k | 0.2 | 2k | 0.3 | k |
then P (X = −1) =
Choose the correct option from the given alternative:
Find expected value of and variance of X for the following p.m.f.
X | -2 | -1 | 0 | 1 | 2 |
P(x) | 0.3 | 0.3 | 0.1 | 0.05 | 0.25 |
Find the probability distribution of number of number of tails in three tosses of a coin
Find k if the following function represents the p. d. f. of a r. v. X.
f(x) = `{(kx, "for" 0 < x < 2),(0, "otherwise."):}`
Also find `"P"[1/4 < "X" < 1/2]`
State whether the following is True or False :
If P(X = x) = `"k"[(4),(x)]` for x = 0, 1, 2, 3, 4 , then F(5) = `(1)/(4)` when F(x) is c.d.f.
Solve the following problem :
The probability distribution of a discrete r.v. X is as follows.
X | 1 | 2 | 3 | 4 | 5 | 6 |
(X = x) | k | 2k | 3k | 4k | 5k | 6k |
Determine the value of k.
Solve the following problem :
The p.m.f. of a r.v.X is given by
`P(X = x) = {(((5),(x)) 1/2^5", ", x = 0", "1", "2", "3", "4", "5.),(0,"otherwise"):}`
Show that P(X ≤ 2) = P(X ≤ 3).
Solve the following problem :
Find the expected value and variance of the r. v. X if its probability distribution is as follows.
x | 1 | 2 | 3 |
P(X = x) | `(1)/(5)` | `(2)/(5)` | `(2)/(5)` |
If a d.r.v. X takes values 0, 1, 2, 3, … with probability P(X = x) = k(x + 1) × 5–x, where k is a constant, then P(X = 0) = ______
The p.m.f. of a d.r.v. X is P(X = x) = `{{:(((5),(x))/2^5",", "for" x = 0"," 1"," 2"," 3"," 4"," 5),(0",", "otherwise"):}` If a = P(X ≤ 2) and b = P(X ≥ 3), then
If a d.r.v. X has the following probability distribution:
X | –2 | –1 | 0 | 1 | 2 | 3 |
P(X = x) | 0.1 | k | 0.2 | 2k | 0.3 | k |
then P(X = –1) is ______
E(x) is considered to be ______ of the probability distribution of x.
The following function represents the p.d.f of a.r.v. X
f(x) = `{{:((kx;, "for" 0 < x < 2, "then the value of K is ")),((0;, "otherwise")):}` ______
The probability distribution of X is as follows:
x | 0 | 1 | 2 | 3 | 4 |
P[X = x] | 0.1 | k | 2k | 2k | k |
Find
- k
- P[X < 2]
- P[X ≥ 3]
- P[1 ≤ X < 4]
- P(2)
Given below is the probability distribution of a discrete random variable x.
X | 1 | 2 | 3 | 4 | 5 | 6 |
P(X = x) | K | 0 | 2K | 5K | K | 3K |
Find K and hence find P(2 ≤ x ≤ 3)