SENAPREP
Binomial distribution Questions
Chapter: Statistics and Probability | Total Available: 13 Real Questions
Consider a random variable X which follows Binomial distribution with parameters n = 10 and p = 1/5. Then Y = 10 − X follows Binomial distribution with parameters n and p respectively given by
Solution
Option (d) is correct.
Explanation:
A binomial random variable \( X \) follows the distribution with parameters \( n = 10 \) and \( p = \frac{1}{5} \).
The complementary variable \( Y = 10 - X \) counts the number of failures in \( 10 \) trials:
Where \( p' = 1 - p \):
Thus, \( Y \) follows a Binomial distribution with:
Final parameters for \( Y \) are \( n = 10 \) and \( p = \frac{4}{5} \).
- Binomial Distribution: \( P(X = k) = \binom{n}{k} p^k (1-p)^{n-k} \)
- Complementary probability: \( p' = 1 - p \)
A diagram showing two binomial distributions for \( X \) and \( Y \) on a graph could illuminate how \( Y \) represents the distribution of failures while \( X \) represents successes.
Verified against NCERT Class XII Mathematics, Chapter 13 (Probability).
For a Binomial distribution with mean 6 and standard deviation √2, what is the value of P(X = 0)?
Solution
Option (a) is correct.
Explanation:
Given a binomial distribution with mean \( \mu \) and standard deviation \( \sigma \): \[ \mu = n \cdot p = 6 \] \[ \sigma = \sqrt{n \cdot p \cdot (1 - p)} = \sqrt{2} \] Squaring the standard deviation: \[ n \cdot p \cdot (1 - p) = 2 \] We have two equations: \[ 1. n \cdot p = 6 \quad (1) \] \[ 2. n \cdot p \cdot (1 - p) = 2 \quad (2) \] From equation (1), we can express \( n \): \[ n = \frac{6}{p} \] Substituting into equation (2): \[ \frac{6}{p} \cdot p \cdot \left( 1 - p \right) = 2 \] \[ 6(1 - p) = 2 \] \[ 6 - 6p = 2 \] \[ 6p = 4 \\ p = \frac{2}{3} \] Now substituting \( p \) back to find \( n \): \[ n = \frac{6}{\frac{2}{3}} = 9 \] Thus, \( n = 9 \) and \( p = \frac{2}{3} \). Now calculating \( P(X = 0) \): \[ P(X = 0) = (1 - p)^n = \left( \frac{1}{3} \right)^9 \] \[ = \frac{1}{19683} \] In conclusion: \[ P(X = 0) = \left( \frac{1}{3} \right)^9 \quad (Final) \]- Binomial Mean: \( \mu = n \cdot p \)
- Binomial Variance: \( \sigma^2 = n \cdot p \cdot (1 - p) \)
- Probability for Binomial Distribution: \( P(X = k) = \binom{n}{k} p^k (1 - p)^{n-k} \)
A flowchart depicting how to compute the mean and standard deviation for a binomial distribution, along with probability calculations would aid understanding.
Verified against NCERT Class XII Mathematics, Chapter on Probability.
Let X be a random variable following binomial distribution with parameters n = 5 and p = k. Further, P(X = 1)= 0.4096 and P(X = 2) = 0.2048. What is the value of k?
Under which of the following conditions may binomial distribution be used?
I. The number of trials is infinite and not fixed.
II. The trials are independent. III. Each trial has two possible outcomes.
Select the correct answer using the code given below.
Direction for Questions (97-98): Consider the following for the two (02) items that follow: Let X be a random variable following binomial distribution with parameters n = 6 and p = k. Further, 9P(X = 4) = P(X = 2).
What is the value of P(X = 3)?
Direction for Questions (97-98): Consider the following for the two (02) items that follow: Let X be a random variable following binomial distribution with parameters n = 6 and p = k. Further, 9P(X = 4) = P(X = 2).
What is the value of k?
Let X be a random variable following binomial distribution whose mean and variance are 200 and 160, respectively. What is the value of the number of trials (n)?
In a binomial distribution, if the mean is 6 and the standard deviation is √2, then what are the values of the parameters n and p, respectively?
If a random variable (x) follows binomial distribution with mean 5 and variance 4 and 523P(X = 3) = λ4λ, then what is the value of λ?
In a Binomial distribution B(n, p), n = 6 and 9P(X =4) = P(X = 2). What is p equal to ?
