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If the standard deviation of a Poisson variate X is 2, what is P (1.5 < X < 2.9)?
- a)0.231.
- b)0.158.
- c)0.15.
- d)0.144.
Correct answer is option 'D'. Can you explain this answer?
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If the standard deviation of a Poisson variate X is 2, what is P (1.5 ...
Given: Standard deviation of a Poisson variate X is 2.
To find: P(1.5 < x="" />< />
Solution:
Step 1: Convert the given Poisson distribution to a normal distribution using the central limit theorem.
- Mean of the Poisson distribution (μ) = Standard deviation of the Poisson distribution^2 = 2^2 = 4
- Variance of the Poisson distribution (σ^2) = Standard deviation of the Poisson distribution = 2
- Since the sample size is not given, we assume it to be large enough for the central limit theorem to hold.
Step 2: Standardize the normal distribution using z-scores.
z = (x - μ) / σ
- z1 = (1.5 - 4) / √2 = -1.76
- z2 = (2.9 - 4) / √2 = -0.70
Step 3: Find the probability using the standard normal table or calculator.
P(1.5 < x="" />< 2.9)="P(-1.76" />< z="" />< />
= P(z < -0.70)="" -="" p(z="" />< />
= 0.2417 - 0.0978
= 0.1439
Therefore, the probability that 1.5 < x="" />< 2.9="" is="" 0.144,="" which="" is="" closest="" to="" option="" d.="" 2.9="" is="" 0.144,="" which="" is="" closest="" to="" option="" />
To find: P(1.5 < x="" />< />
Solution:
Step 1: Convert the given Poisson distribution to a normal distribution using the central limit theorem.
- Mean of the Poisson distribution (μ) = Standard deviation of the Poisson distribution^2 = 2^2 = 4
- Variance of the Poisson distribution (σ^2) = Standard deviation of the Poisson distribution = 2
- Since the sample size is not given, we assume it to be large enough for the central limit theorem to hold.
Step 2: Standardize the normal distribution using z-scores.
z = (x - μ) / σ
- z1 = (1.5 - 4) / √2 = -1.76
- z2 = (2.9 - 4) / √2 = -0.70
Step 3: Find the probability using the standard normal table or calculator.
P(1.5 < x="" />< 2.9)="P(-1.76" />< z="" />< />
= P(z < -0.70)="" -="" p(z="" />< />
= 0.2417 - 0.0978
= 0.1439
Therefore, the probability that 1.5 < x="" />< 2.9="" is="" 0.144,="" which="" is="" closest="" to="" option="" d.="" 2.9="" is="" 0.144,="" which="" is="" closest="" to="" option="" />
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If the standard deviation of a Poisson variate X is 2, what is P (1.5 ...
We know that in poison distribution variance =mean
variance = (standard deviation)^2
=(2)^2=4= mean
probability (x) =e^-4 *(4)^2/2!=0.146( closest to 0.144)
here we taken x =2 because it is integer lying between 1.5 and 2.9
we cannot take X value in points
variance = (standard deviation)^2
=(2)^2=4= mean
probability (x) =e^-4 *(4)^2/2!=0.146( closest to 0.144)
here we taken x =2 because it is integer lying between 1.5 and 2.9
we cannot take X value in points
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If the standard deviation of a Poisson variate X is 2, what is P (1.5 < X < 2.9)?a)0.231.b)0.158.c)0.15.d)0.144.Correct answer is option 'D'. Can you explain this answer?
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