JEE Exam > JEE Questions > In a binomial distribution, mean is 18 and v...
Start Learning for Free
In a binomial distribution, mean is 18 and variance is 12 then p = . . . . . .
- a)2/3
- b)1/3
- c)3/4
- d)1/2
Correct answer is option 'B'. Can you explain this answer?
Most Upvoted Answer
In a binomial distribution, mean is 18 and variance is 12 then p = . ...
We know that, mean = np = 18 and variance = n p q = 12
Now, 
∴ p = 1 - q = 1 - (⅔) = (⅓)
Free Test
FREE
| Start Free Test |
Community Answer
In a binomial distribution, mean is 18 and variance is 12 then p = . ...
Binomial Distribution:
A binomial distribution represents the probability distribution of a random variable X that has a binomial distribution with parameters n and p, where n is the number of trials and p is the probability of success in each trial.
Mean and Variance of Binomial Distribution:
The mean of a binomial distribution is given by μ = np, where n is the number of trials and p is the probability of success in each trial.
The variance of a binomial distribution is given by σ^2 = np(1-p).
Given Mean and Variance:
In this problem, we are given the mean (μ) and variance (σ^2) of a binomial distribution.
μ = 18
σ^2 = 12
We can use these equations to solve for p:
μ = np
18 = n*p
σ^2 = np(1-p)
12 = n*p*(1-p)
Solving for n and p:
From the first equation, we can solve for n:
n = 18/p
Substituting this value of n into the second equation:
12 = (18/p)*p*(1-p)
12 = 18(1-p)
1-p = 2/3
p = 1/3
Therefore, the correct answer is option B: 1/3.
A binomial distribution represents the probability distribution of a random variable X that has a binomial distribution with parameters n and p, where n is the number of trials and p is the probability of success in each trial.
Mean and Variance of Binomial Distribution:
The mean of a binomial distribution is given by μ = np, where n is the number of trials and p is the probability of success in each trial.
The variance of a binomial distribution is given by σ^2 = np(1-p).
Given Mean and Variance:
In this problem, we are given the mean (μ) and variance (σ^2) of a binomial distribution.
μ = 18
σ^2 = 12
We can use these equations to solve for p:
μ = np
18 = n*p
σ^2 = np(1-p)
12 = n*p*(1-p)
Solving for n and p:
From the first equation, we can solve for n:
n = 18/p
Substituting this value of n into the second equation:
12 = (18/p)*p*(1-p)
12 = 18(1-p)
1-p = 2/3
p = 1/3
Therefore, the correct answer is option B: 1/3.
|
Explore Courses for JEE exam
|
|
Similar JEE Doubts
Top Courses for JEEView all
In a binomial distribution, mean is 18 and variance is 12 then p = . . . . . .a)2/3b)1/3c)3/4d)1/2Correct answer is option 'B'. Can you explain this answer?
Question Description
In a binomial distribution, mean is 18 and variance is 12 then p = . . . . . .a)2/3b)1/3c)3/4d)1/2Correct answer is option 'B'. Can you explain this answer? for JEE 2025 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about In a binomial distribution, mean is 18 and variance is 12 then p = . . . . . .a)2/3b)1/3c)3/4d)1/2Correct answer is option 'B'. Can you explain this answer? covers all topics & solutions for JEE 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for In a binomial distribution, mean is 18 and variance is 12 then p = . . . . . .a)2/3b)1/3c)3/4d)1/2Correct answer is option 'B'. Can you explain this answer?.
In a binomial distribution, mean is 18 and variance is 12 then p = . . . . . .a)2/3b)1/3c)3/4d)1/2Correct answer is option 'B'. Can you explain this answer? for JEE 2025 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about In a binomial distribution, mean is 18 and variance is 12 then p = . . . . . .a)2/3b)1/3c)3/4d)1/2Correct answer is option 'B'. Can you explain this answer? covers all topics & solutions for JEE 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for In a binomial distribution, mean is 18 and variance is 12 then p = . . . . . .a)2/3b)1/3c)3/4d)1/2Correct answer is option 'B'. Can you explain this answer?.
Solutions for In a binomial distribution, mean is 18 and variance is 12 then p = . . . . . .a)2/3b)1/3c)3/4d)1/2Correct answer is option 'B'. Can you explain this answer? in English & in Hindi are available as part of our courses for JEE.
Download more important topics, notes, lectures and mock test series for JEE Exam by signing up for free.
Here you can find the meaning of In a binomial distribution, mean is 18 and variance is 12 then p = . . . . . .a)2/3b)1/3c)3/4d)1/2Correct answer is option 'B'. Can you explain this answer? defined & explained in the simplest way possible. Besides giving the explanation of
In a binomial distribution, mean is 18 and variance is 12 then p = . . . . . .a)2/3b)1/3c)3/4d)1/2Correct answer is option 'B'. Can you explain this answer?, a detailed solution for In a binomial distribution, mean is 18 and variance is 12 then p = . . . . . .a)2/3b)1/3c)3/4d)1/2Correct answer is option 'B'. Can you explain this answer? has been provided alongside types of In a binomial distribution, mean is 18 and variance is 12 then p = . . . . . .a)2/3b)1/3c)3/4d)1/2Correct answer is option 'B'. Can you explain this answer? theory, EduRev gives you an
ample number of questions to practice In a binomial distribution, mean is 18 and variance is 12 then p = . . . . . .a)2/3b)1/3c)3/4d)1/2Correct answer is option 'B'. Can you explain this answer? tests, examples and also practice JEE tests.
|
|
Explore Courses for JEE exam
|
|
Signup for Free!
Signup to see your scores go up within 7 days! Learn & Practice with 1000+ FREE Notes, Videos & Tests.
|
© EduRev
|
Education Revolution
|
|
Signup to see your scores
go up within 7 days!
Access 1000+ FREE Docs, Videos and Tests
Takes less than 10 seconds to signup