Advertisements
Advertisements
Question
Solve the following problem :
Let X∼B(n,p) If n = 10 and E(X)= 5, find p and Var(X).
Solution
Let X ~ B(n, p)
n = 10, E(X) = 5 ...[Given]
But E(X) = np
∴ 5 = 10(p)
∴ p = `(1)/(2)`
∴ q = 1 – p = `1 - (1)/(2) = (1)/(2)`
Var (X) = npq
= `10 xx (1)/(2) xx (1)/(2)`
= `1 xx (1)/(4)`
= 2.5
APPEARS IN
RELATED QUESTIONS
State if the following is not the probability mass function of a random variable. Give reasons for your answer.
X | 0 | 1 | 2 | 3 | 4 |
P(X) | 0.1 | 0.5 | 0.2 | − 0.1 | 0.2 |
The following is the p.d.f. of r.v. X :
f(x) = `x/8`, for 0 < x < 4 and = 0 otherwise
P ( 1 < x < 2 )
If a r.v. X has p.d.f.,
f (x) = `c /x` , for 1 < x < 3, c > 0, Find c, E(X) and Var (X).
Choose the correct option from the given alternative:
If the p.d.f of a.c.r.v. X is f (x) = 3 (1 − 2x2 ), for 0 < x < 1 and = 0, otherwise (elsewhere) then the c.d.f of X is F(x) =
Choose the correct option from the given alternative:
If p.m.f. of a d.r.v. X is P (X = x) = `x^2 /(n (n + 1))`, for x = 1, 2, 3, . . ., n and = 0, otherwise then E (X ) =
The following is the c.d.f. of r.v. X
x | -3 | -2 | -1 | 0 | 1 | 2 | 3 | 4 |
F(X) | 0.1 | 0.3 | 0.5 | 0.65 | 0.75 | 0.85 | 0.9 |
*1 |
P (–1 ≤ X ≤ 2)
70% of the members favour and 30% oppose a proposal in a meeting. The random variable X takes the value 0 if a member opposes the proposal and the value 1 if a member is in favour. Find E(X) and Var(X).
If F(x) is distribution function of discrete r.v.x with p.m.f. P(x) = `(x - 1)/(3)` for x = 0, 1 2, 3, and P(x) = 0 otherwise then F(4) = _______.
State whether the following is True or False :
x | – 2 | – 1 | 0 | 1 | 2 |
P(X = x) | 0.2 | 0.3 | 0.15 | 0.25 | 0.1 |
If F(x) is c.d.f. of discrete r.v. X then F(–3) = 0
If r.v. X assumes values 1, 2, 3, ……. n with equal probabilities then E(X) = `("n" + 1)/(2)`
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 the p.m.f. of a d.r.v. X is P(X = x) = `{{:(x/("n"("n" + 1))",", "for" x = 1"," 2"," 3"," .... "," "n"),(0",", "otherwise"):}`, then E(X) = ______
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 ______
Find mean for the following probability distribution.
X | 0 | 1 | 2 | 3 |
P(X = x) | `1/6` | `1/3` | `1/3` | `1/6` |
Find the expected value and variance of r.v. X whose p.m.f. is given below.
X | 1 | 2 | 3 |
P(X = x) | `1/5` | `2/5` | `2/5` |
The probability distribution of a discrete r.v.X is as follows.
x | 1 | 2 | 3 | 4 | 5 | 6 |
P(X = x) | k | 2k | 3k | 4k | 5k | 6k |
Complete the following activity.
Solution: Since `sum"p"_"i"` = 1
P(X ≤ 4) = `square + square + square + square = square`
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)