Section –III
2. The random variable X has the following probability distribution
X 0 1 2 3
P(x) 0.22 0.38 0.1 0.3
A. Prove that the given table satisfies the two properties needed for a probability distribution
Ans:
a discrete random variable x takes a set of separate values (such as 0, 1, 2...). Its probability distribution assigns a probability P(x)to each possible value x. For each x, the probability P(x) falls between 0 and 1 inclusive and the sum of the probabilities for all the possible x values equals to 1.
1. For each xx, 0≤P(x)≤1.
2. P(x0)+P(x1)+P(x2)+…+P(xn)=1Px0+Px1+Px2+…+Pxn=1.
0.22 is between 0 and 1 inclusive, which meets the first property of the probability distribution.
0.38 is between 0 and 1 inclusive
0.1 is between 0 and 1 inclusive, which meets the first property of the probability distribution.
0.3 is between 0 and 1 inclusive
For each x, the probability P(x)Px falls between 0 and 1 inclusive, which meets the first property of the probability distribution.
0≤P(x)≤1 for all x values
Find the sum of the probabilities for all the possible x values.
0.22+0.38+0.1+0.3
The sum of the probabilities for all the possible xx values is 0.22+0.38+0.1+0.3=1.
The table satisfies the two properties of a probability distribution:
Property 1: 0≤P(x)≤1 for all x values
Property 2: 0.22+0.38+0.1+0.3=1
B. Find the mean & the Variance.
The expectation mean of a distribution is the value expected if trials of the distribution could continue indefinitely. This is equal to each value multiplied by its discrete probability.
Mean: µ= (0*0.22)+(1*0.38)+(2*0.1)+(3*0.3) = 1.48
The variance is
σ=∑[X^2*P(x)]-μ^2
= [(0^2*0.22)+(1^2*0.38)+(2^2*0.1)+(3^2*0.3)] - (1.48)^2
= 3.48 -2.19
= 1.29
محاولة حل للسوال اي أحد عنده تصحيح لا يبخل علينا
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