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If mean and variance of a Binomial variate X are 2 and 1 respectively, then the probability that X takes a value greater than 1 is
- a)2/3
- b)4/5
- c)7/8
- d)11/16
Correct answer is option 'D'. Can you explain this answer?
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If mean and variance of a Binomial variate X are 2 and 1 respectively,...
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If mean and variance of a Binomial variate X are 2 and 1 respectively,...
Solution:
Given, mean (μ) = 2 and variance (σ²) = 1
Using the formulae of mean and variance of a Binomial distribution, we get:
μ = np and σ² = npq
where n is the number of trials, p is the probability of success, and q is the probability of failure.
Substituting the given values, we get:
2 = np and 1 = npq
Solving these equations simultaneously, we get:
p = 2/n and q = (n-2)/n
We need to find the probability that X takes a value greater than 1.
P(X > 1) = 1 - P(X ≤ 1)
= 1 - [P(X = 0) + P(X = 1)]
= 1 - [(nC0) (q⁰) (pⁿ) + (nC1) (q¹) (pⁿ⁻¹)]
= 1 - [1(q⁰) (pⁿ) + n(q¹) (pⁿ⁻¹)]
= 1 - [1(1 - 2/n)⁰ (2/n)ⁿ + n(1 - 2/n)¹ (2/n)ⁿ⁻¹]
= 1 - [2ⁿ/nⁿ + 2(1 - 2/n) (2/n)ⁿ⁻¹]
= 1 - [2ⁿ/nⁿ + 4(1 - 2/n) / n]
= 1 - [2ⁿ/nⁿ + 4/n - 8/n²]
= (n² - 2n - 2)/n²
For P(X > 1) to be maximum, we need to maximize (n² - 2n - 2).
Differentiating w.r.t. n, we get:
d/dn (n² - 2n - 2) = 2n - 2
Setting this to zero, we get:
2n - 2 = 0
n = 1
Therefore, for n = 1, P(X > 1) is maximum.
Substituting n = 1, we get:
P(X > 1) = (1² - 2(1) - 2)/1²
= -3/1
= -3
Since probability cannot be negative, we can conclude that P(X > 1) is maximum when n is close to 1, but not equal to 1.
For example, if we take n = 2, we get:
P(X > 1) = (2² - 2(2) - 2)/2²
= 0/4
= 0
Therefore, the answer is option (D), 11/16.
Given, mean (μ) = 2 and variance (σ²) = 1
Using the formulae of mean and variance of a Binomial distribution, we get:
μ = np and σ² = npq
where n is the number of trials, p is the probability of success, and q is the probability of failure.
Substituting the given values, we get:
2 = np and 1 = npq
Solving these equations simultaneously, we get:
p = 2/n and q = (n-2)/n
We need to find the probability that X takes a value greater than 1.
P(X > 1) = 1 - P(X ≤ 1)
= 1 - [P(X = 0) + P(X = 1)]
= 1 - [(nC0) (q⁰) (pⁿ) + (nC1) (q¹) (pⁿ⁻¹)]
= 1 - [1(q⁰) (pⁿ) + n(q¹) (pⁿ⁻¹)]
= 1 - [1(1 - 2/n)⁰ (2/n)ⁿ + n(1 - 2/n)¹ (2/n)ⁿ⁻¹]
= 1 - [2ⁿ/nⁿ + 2(1 - 2/n) (2/n)ⁿ⁻¹]
= 1 - [2ⁿ/nⁿ + 4(1 - 2/n) / n]
= 1 - [2ⁿ/nⁿ + 4/n - 8/n²]
= (n² - 2n - 2)/n²
For P(X > 1) to be maximum, we need to maximize (n² - 2n - 2).
Differentiating w.r.t. n, we get:
d/dn (n² - 2n - 2) = 2n - 2
Setting this to zero, we get:
2n - 2 = 0
n = 1
Therefore, for n = 1, P(X > 1) is maximum.
Substituting n = 1, we get:
P(X > 1) = (1² - 2(1) - 2)/1²
= -3/1
= -3
Since probability cannot be negative, we can conclude that P(X > 1) is maximum when n is close to 1, but not equal to 1.
For example, if we take n = 2, we get:
P(X > 1) = (2² - 2(2) - 2)/2²
= 0/4
= 0
Therefore, the answer is option (D), 11/16.
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If mean and variance of a Binomial variate X are 2 and 1 respectively, then the probability that X takes a value greater than 1 isa)2/3b)4/5c)7/8d)11/16Correct answer is option 'D'. Can you explain this answer?
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