English

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 and P[X < 2] - Mathematics and Statistics

Advertisements
Advertisements

Question

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 and P[X < 2]

Sum

Solution

As probability distribution of r. v. X is given, `sum"P"("X" = x)` = 1

i.e., P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) = 1

∴ 0.1 + k + 2k + 2k + k = 1

∴ 6k = 0.9

∴ k = 0.15   .....(i)

∴ P[X < 2] = P(X = 0 or 1)

= P(X = 0) + P(X = 1)

= 0.1 + k

= 0.1 + 0.15     ......[From (i)]

= 0.25

shaalaa.com
Probability Distribution of Discrete Random Variables
  Is there an error in this question or solution?
Chapter 2.7: Probability Distributions - Short Answers I

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
P(X) 0.1 0.6 0.3

Find expected value and variance of X for the following p.m.f.

x -2 -1 0 1 2
P(X) 0.2 0.3 0.1 0.15 0.25

Let X denote the sum of the numbers obtained when two fair dice are rolled. Find the standard deviation of X.


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 )


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.


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

Find probability that X is negative


Find k, if the following function represents p.d.f. of r.v. X.

f(x) = kx(1 – x), for 0 < x < 1 and = 0, otherwise.

Also, find `P(1/4 < x < 1/2) and P(x < 1/2)`.


Suppose that X is waiting time in minutes for a bus and its p.d.f. is given by f(x) = `1/5`, for 0 ≤ x ≤ 5 and = 0 otherwise.

Find the probability that waiting time is between 1 and 3.


Suppose that X is waiting time in minutes for a bus and its p.d.f. is given by f(x) = `1/5`, for 0 ≤ x ≤ 5 and = 0 otherwise.

Find the probability that the waiting time is more than 4 minutes.


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 = x) = `x^2 /(n (n + 1))`, for x = 1, 2, 3, . . ., n and = 0, otherwise 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) =


Solve the following :

Identify the random variable as either discrete or continuous in each of the following. Write down the range of it.

The person on the high protein diet is interested gain of weight in a week.


Solve the following :

The following probability distribution of r.v. X

X=x -3 -2 -1 0 1 2 3
P(X=x) 0.05 0.1 0.15 0.20 0.25 0.15 0.1

Find the probability that

X is positive


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

Find p.m.f. of X.
i. P(–1 ≤ X ≤ 2)
ii. P(X ≤ 3 / X > 0).


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 (X ≤ 3/ X > 0)


Let X be amount of time for which a book is taken out of library by randomly selected student and suppose X has p.d.f

f (x) = 0.5x, for 0 ≤ x ≤ 2 and = 0 otherwise.

Calculate: P(0.5 ≤ x ≤ 1.5)


Find the probability distribution of number of number of tails in three tosses of a coin


Find the probability distribution of number of heads in four tosses of a coin


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).


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]`


Given that X ~ B(n,p), if n = 10, E(X) = 8, find Var(X).


Fill in the blank :

If X is discrete random variable takes the value x1, x2, x3,…, xn then \[\sum\limits_{i=1}^{n}\text{P}(x_i)\] = _______


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.


State whether the following is True or False :

If p.m.f. of discrete r.v. X is

x 0 1 2
P(X = x) q2 2pq p2 

then E(x) = 2p.


If r.v. X assumes values 1, 2, 3, ……. n with equal probabilities then E(X) = `("n" + 1)/(2)`


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 following is the c.d.f of a 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

Find the probability distribution of X and P(–1 ≤ X ≤ 2).


Solve the following problem :

Find the expected value and variance of the r. v. X if its probability distribution is as follows.

x – 1 0 1
P(X = x) `(1)/(5)` `(2)/(5)` `(2)/(5)`

Solve the following problem :

Find the expected value and variance of the r. v. X if its probability distribution is as follows.

X 0 1 2 3 4 5
P(X = x) `(1)/(32)` `(5)/(32)` `(10)/(32)` `(10)/(32)` `(5)/(32)` `(1)/(32)`

Solve the following problem :

Let the p. m. f. of the r. v. X be

`"P"(x) = {((3 - x)/(10)", ","for"  x = -1", "0", "1", "2.),(0,"otherwise".):}`
Calculate E(X) and Var(X).


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 ______


Find the probability distribution of the number of successes in two tosses of a die, where a success is defined as number greater than 4 appears on at least one die.


Choose the correct alternative:

f(x) is c.d.f. of discete r.v. X whose distribution is

xi – 2 – 1 0 1 2
pi 0.2 0.3 0.15 0.25 0.1

then F(– 3) = ______


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

k = `square`


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 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 ≥ 3) = `square - square - square  = square`


Using the following activity, 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`

Solution: µ = E(X) = `sum_("i" = 1)^3 x_"i""p"_"i"`

E(X) = `square + square + square = square`

Var(X) = `"E"("X"^2) - {"E"("X")}^2`

= `sum"X"_"i"^2"P"_"i" - [sum"X"_"i""P"_"i"]^2`

= `square - square`

= `square`


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
  1. Determine the value of k.
  2. Find P(X ≤ 4)
  3. P(2 < X < 4)
  4. P(X ≥ 3)

The probability distribution of X is as follows:

x 0 1 2 3 4
P[X = x] 0.1 k 2k 2k k

Find

  1. k
  2. P[X < 2]
  3. P[X ≥ 3]
  4. P[1 ≤ X < 4]
  5. P(2)

Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×