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If X ~ P (m) and its coefficient of variation is 50, what is the probability that X would assume only non-zero values?
- a)0.018.
- b)0.982.
- c)0.989.
- d)0.976.
Correct answer is option 'B'. Can you explain this answer?
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If X ~ P (m) and its coefficient of variation is 50, what is the proba...
Given: X ~ P(m) and coefficient of variation = 50
To find: Probability that X assumes only non-zero values
Solution:
Coefficient of Variation (CV) = Standard Deviation / Mean
CV = σ / µ
For Poisson distribution, µ = m and σ = sqrt(m)
So, CV = sqrt(m) / m = 1 / sqrt(m)
Given, CV = 50
50 = 1 / sqrt(m)
sqrt(m) = 1 / 50
m = 1 / (50)^2
Now we can find the probability that X assumes only non-zero values:
P(X > 0) = 1 - P(X = 0)
P(X = 0) = e^(-m) = e^(-(1/50)^2) ≈ 0.018
P(X > 0) ≈ 1 - 0.018 = 0.982
Therefore, the probability that X assumes only non-zero values is 0.982, which is option B.
To find: Probability that X assumes only non-zero values
Solution:
Coefficient of Variation (CV) = Standard Deviation / Mean
CV = σ / µ
For Poisson distribution, µ = m and σ = sqrt(m)
So, CV = sqrt(m) / m = 1 / sqrt(m)
Given, CV = 50
50 = 1 / sqrt(m)
sqrt(m) = 1 / 50
m = 1 / (50)^2
Now we can find the probability that X assumes only non-zero values:
P(X > 0) = 1 - P(X = 0)
P(X = 0) = e^(-m) = e^(-(1/50)^2) ≈ 0.018
P(X > 0) ≈ 1 - 0.018 = 0.982
Therefore, the probability that X assumes only non-zero values is 0.982, which is option B.
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If X ~ P (m) and its coefficient of variation is 50, what is the probability that X would assume only non-zero values?a)0.018.b)0.982.c)0.989.d)0.976.Correct answer is option 'B'. Can you explain this answer?
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