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A PAM source generates four symbols 3 V, 1 V, -1 V and -3 V with probability of p(3) = p(-3) = 0.2 and p(1) = p(-1) = 0.3 respectively. The variance for the source will be
- a)4.2 V
- b)3.2 V
- c)3.6 V
- d)4.6 V
Correct answer is option 'A'. Can you explain this answer?
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Most Upvoted Answer
A PAM source generates four symbols 3 V, 1 V, -1 V and -3 V with proba...
Concept:
For discrete random variable x:
Mean = E[x] = ∑I xi p (xi)
Mean square value (MSQ) = E[x2] = ∑I xi2 p (xi)
Variance (σ2) = E[x2] – {E(x)}2
Standard deviation (σ) =
Calculation:
Given data: for the given PAM source
Mean = E[x] = ∑i xi p (xi)
= (-3)(0.2) + (-1)(0.3) + (1)(0.3) + (3)(0.2)
= 0
MSQ = E[x2] = ∑I xi2 p (xi)
= (-3)2(0.2) + (-1)2(0.3) + (1)2(0.3) + (3)2(0.2) V
= 4.2 V
Variance = E[x2] – {E[x]}2
= 4.2 – 0
= 4.2 V
For discrete random variable x:
Mean = E[x] = ∑I xi p (xi)
Mean square value (MSQ) = E[x2] = ∑I xi2 p (xi)
Variance (σ2) = E[x2] – {E(x)}2
Standard deviation (σ) =
Calculation:
Given data: for the given PAM source
Mean = E[x] = ∑i xi p (xi)
= (-3)(0.2) + (-1)(0.3) + (1)(0.3) + (3)(0.2)
= 0
MSQ = E[x2] = ∑I xi2 p (xi)
= (-3)2(0.2) + (-1)2(0.3) + (1)2(0.3) + (3)2(0.2) V
= 4.2 V
Variance = E[x2] – {E[x]}2
= 4.2 – 0
= 4.2 V
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Community Answer
A PAM source generates four symbols 3 V, 1 V, -1 V and -3 V with proba...
Variance is a measure of how spread out the values of a random variable are around the mean. In this case, we are given a PAM (Pulse Amplitude Modulation) source that generates four symbols with different probabilities. We need to calculate the variance for this source.
The formula to calculate the variance of a discrete random variable is:
Var(X) = Σ(xi - μ)^2 * P(xi)
where Var(X) is the variance, xi is each possible value of the random variable, μ is the mean, and P(xi) is the probability of each value.
Let's calculate the variance step by step:
1. Calculate the mean:
The mean (μ) of a discrete random variable can be calculated by multiplying each value with its probability and summing them up:
μ = Σ(xi * P(xi))
In this case, the mean is:
μ = (3 V * 0.2) + (1 V * 0.3) + (-1 V * 0.3) + (-3 V * 0.2)
= 0.6 V + 0.3 V - 0.3 V - 0.6 V
= 0 V
2. Calculate the variance:
Using the formula mentioned earlier, we can calculate the variance:
Var(X) = Σ(xi - μ)^2 * P(xi)
Substituting the values, we get:
Var(X) = (3 V - 0 V)^2 * 0.2 + (1 V - 0 V)^2 * 0.3 + (-1 V - 0 V)^2 * 0.3 + (-3 V - 0 V)^2 * 0.2
= 9 V^2 * 0.2 + 1 V^2 * 0.3 + 1 V^2 * 0.3 + 9 V^2 * 0.2
= 1.8 V^2 + 0.3 V^2 + 0.3 V^2 + 1.8 V^2
= 4.2 V^2
Therefore, the variance for the given PAM source is 4.2 V^2.
Since the question asks for the variance in volts (V), we need to take the square root of the variance to get the variance in volts:
√(4.2 V^2) = 2.049 V
So, the correct option is A) 4.2 V.
The formula to calculate the variance of a discrete random variable is:
Var(X) = Σ(xi - μ)^2 * P(xi)
where Var(X) is the variance, xi is each possible value of the random variable, μ is the mean, and P(xi) is the probability of each value.
Let's calculate the variance step by step:
1. Calculate the mean:
The mean (μ) of a discrete random variable can be calculated by multiplying each value with its probability and summing them up:
μ = Σ(xi * P(xi))
In this case, the mean is:
μ = (3 V * 0.2) + (1 V * 0.3) + (-1 V * 0.3) + (-3 V * 0.2)
= 0.6 V + 0.3 V - 0.3 V - 0.6 V
= 0 V
2. Calculate the variance:
Using the formula mentioned earlier, we can calculate the variance:
Var(X) = Σ(xi - μ)^2 * P(xi)
Substituting the values, we get:
Var(X) = (3 V - 0 V)^2 * 0.2 + (1 V - 0 V)^2 * 0.3 + (-1 V - 0 V)^2 * 0.3 + (-3 V - 0 V)^2 * 0.2
= 9 V^2 * 0.2 + 1 V^2 * 0.3 + 1 V^2 * 0.3 + 9 V^2 * 0.2
= 1.8 V^2 + 0.3 V^2 + 0.3 V^2 + 1.8 V^2
= 4.2 V^2
Therefore, the variance for the given PAM source is 4.2 V^2.
Since the question asks for the variance in volts (V), we need to take the square root of the variance to get the variance in volts:
√(4.2 V^2) = 2.049 V
So, the correct option is A) 4.2 V.
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A PAM source generates four symbols 3 V, 1 V, -1 V and -3 V with probability of p(3) = p(-3) = 0.2 and p(1) = p(-1) = 0.3 respectively. The variance for the source will bea)4.2 Vb)3.2 Vc)3.6 Vd)4.6 VCorrect answer is option 'A'. Can you explain this answer?
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A PAM source generates four symbols 3 V, 1 V, -1 V and -3 V with probability of p(3) = p(-3) = 0.2 and p(1) = p(-1) = 0.3 respectively. The variance for the source will bea)4.2 Vb)3.2 Vc)3.6 Vd)4.6 VCorrect answer is option 'A'. Can you explain this answer? for Electronics and Communication Engineering (ECE) 2024 is part of Electronics and Communication Engineering (ECE) preparation. The Question and answers have been prepared according to the Electronics and Communication Engineering (ECE) exam syllabus. Information about A PAM source generates four symbols 3 V, 1 V, -1 V and -3 V with probability of p(3) = p(-3) = 0.2 and p(1) = p(-1) = 0.3 respectively. The variance for the source will bea)4.2 Vb)3.2 Vc)3.6 Vd)4.6 VCorrect answer is option 'A'. Can you explain this answer? covers all topics & solutions for Electronics and Communication Engineering (ECE) 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A PAM source generates four symbols 3 V, 1 V, -1 V and -3 V with probability of p(3) = p(-3) = 0.2 and p(1) = p(-1) = 0.3 respectively. The variance for the source will bea)4.2 Vb)3.2 Vc)3.6 Vd)4.6 VCorrect answer is option 'A'. Can you explain this answer?.
A PAM source generates four symbols 3 V, 1 V, -1 V and -3 V with probability of p(3) = p(-3) = 0.2 and p(1) = p(-1) = 0.3 respectively. The variance for the source will bea)4.2 Vb)3.2 Vc)3.6 Vd)4.6 VCorrect answer is option 'A'. Can you explain this answer? for Electronics and Communication Engineering (ECE) 2024 is part of Electronics and Communication Engineering (ECE) preparation. The Question and answers have been prepared according to the Electronics and Communication Engineering (ECE) exam syllabus. Information about A PAM source generates four symbols 3 V, 1 V, -1 V and -3 V with probability of p(3) = p(-3) = 0.2 and p(1) = p(-1) = 0.3 respectively. The variance for the source will bea)4.2 Vb)3.2 Vc)3.6 Vd)4.6 VCorrect answer is option 'A'. Can you explain this answer? covers all topics & solutions for Electronics and Communication Engineering (ECE) 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A PAM source generates four symbols 3 V, 1 V, -1 V and -3 V with probability of p(3) = p(-3) = 0.2 and p(1) = p(-1) = 0.3 respectively. The variance for the source will bea)4.2 Vb)3.2 Vc)3.6 Vd)4.6 VCorrect answer is option 'A'. Can you explain this answer?.
Solutions for A PAM source generates four symbols 3 V, 1 V, -1 V and -3 V with probability of p(3) = p(-3) = 0.2 and p(1) = p(-1) = 0.3 respectively. The variance for the source will bea)4.2 Vb)3.2 Vc)3.6 Vd)4.6 VCorrect answer is option 'A'. Can you explain this answer? in English & in Hindi are available as part of our courses for Electronics and Communication Engineering (ECE).
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