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A discrete random variable X has the following probability distribution: P(X=0)=a, P(X=1)=a, P(X=2)=a, P(X=3)=b, P(X=4)=b. If P(X>=2) = 3P(X<2), find the values of a and b, and the expected value of X

Mathematics

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Answers

4 parts
a

a=18,b=516a = \frac{1}{8}, b = \frac{5}{16}

b

E(X)=4116E(X) = \frac{41}{16}

c

σX=47916\sigma_X = \frac{\sqrt{479}}{16}

d

E(Y)=25316E(Y) = \frac{253}{16}

Step-by-step solution

3 steps

Here's how to solve the problem:

Step 1: Set up equations for aa and bb.
The sum of all probabilities in a distribution must equal 1:

P(X=0)+P(X=1)+P(X=2)+P(X=3)+P(X=4)=1P(X=0) + P(X=1) + P(X=2) + P(X=3) + P(X=4) = 1 a+a+a+b+b=1a + a + a + b + b = 1 3a+2b=1(Equation 1)3a + 2b = 1 \quad (\text{Equation 1})

The given condition is P(X≥2)=3P(X<2)P(X \geq 2) = 3P(X < 2).
Calculate P(X≥2)P(X \geq 2):

P(X≥2)=P(X=2)+P(X=3)+P(X=4)=a+b+b=a+2bP(X \geq 2) = P(X=2) + P(X=3) + P(X=4) = a + b + b = a + 2b

Calculate P(X<2)P(X < 2):

P(X<2)=P(X=0)+P(X=1)=a+a=2aP(X < 2) = P(X=0) + P(X=1) = a + a = 2a

Substitute these into the given condition:

a+2b=3(2a)a + 2b = 3(2a) a+2b=6aa + 2b = 6a 2b=5a(Equation 2)2b = 5a \quad (\text{Equation 2})

Step 2: Solve for aa and bb.
Substitute Equation 2 into Equation 1:

3a+(5a)=13a + (5a) = 1 8a=18a = 1 a=18a = \frac{1}{8}

Now substitute the value of aa back into Equation 2 to find bb:

2b=5(18)2b = 5 \left(\frac{1}{8}\right) 2b=582b = \frac{5}{8} b=516b = \frac{5}{16}

Part a· 1 step

The values of the probabilities aa and bb are

a=18,b=516\boxed{a = \frac{1}{8}, b = \frac{5}{16}}

  1. Calculate the expected value of XX, E(X)E(X).

    The formula for the expected value is E(X)=∑x⋅P(X=x)E(X) = \sum x \cdot P(X=x).

    E(X)=(0⋅a)+(1⋅a)+(2⋅a)+(3⋅b)+(4⋅b)E(X) = (0 \cdot a) + (1 \cdot a) + (2 \cdot a) + (3 \cdot b) + (4 \cdot b) E(X)=0+a+2a+3b+4bE(X) = 0 + a + 2a + 3b + 4b E(X)=3a+7bE(X) = 3a + 7b

    Substitute the values of a=18a = \frac{1}{8} and b=516b = \frac{5}{16}:

    E(X)=3(18)+7(516)E(X) = 3 \left(\frac{1}{8}\right) + 7 \left(\frac{5}{16}\right) E(X)=38+3516E(X) = \frac{3}{8} + \frac{35}{16} E(X)=616+3516E(X) = \frac{6}{16} + \frac{35}{16} E(X)=4116E(X) = \frac{41}{16}

Part b

E(X)=4116E(X) = \frac{41}{16}

The expected value of XX is

E(X)=4116\boxed{E(X) = \frac{41}{16}}

  1. Calculate the standard deviation of XX, σX\sigma_X.

    First, calculate E(X2)E(X^2):

    E(X2)=∑x2⋅P(X=x)E(X^2) = \sum x^2 \cdot P(X=x) E(X2)=(02⋅a)+(12⋅a)+(22⋅a)+(32⋅b)+(42⋅b)E(X^2) = (0^2 \cdot a) + (1^2 \cdot a) + (2^2 \cdot a) + (3^2 \cdot b) + (4^2 \cdot b) E(X2)=0+a+4a+9b+16bE(X^2) = 0 + a + 4a + 9b + 16b E(X2)=5a+25bE(X^2) = 5a + 25b

    Substitute the values of a=18a = \frac{1}{8} and b=516b = \frac{5}{16}:

    E(X2)=5(18)+25(516)E(X^2) = 5 \left(\frac{1}{8}\right) + 25 \left(\frac{5}{16}\right) E(X2)=58+12516E(X^2) = \frac{5}{8} + \frac{125}{16} E(X2)=1016+12516E(X^2) = \frac{10}{16} + \frac{125}{16} E(X2)=13516E(X^2) = \frac{135}{16}

    Next, calculate the variance, Var(X)=E(X2)−[E(X)]2Var(X) = E(X^2) - [E(X)]^2:

    Var(X)=13516−(4116)2Var(X) = \frac{135}{16} - \left(\frac{41}{16}\right)^2 Var(X)=13516−1681256Var(X) = \frac{135}{16} - \frac{1681}{256} Var(X)=135⋅16256−1681256Var(X) = \frac{135 \cdot 16}{256} - \frac{1681}{256} Var(X)=2160256−1681256Var(X) = \frac{2160}{256} - \frac{1681}{256} Var(X)=479256Var(X) = \frac{479}{256}

    Finally, calculate the standard deviation, σX=Var(X)\sigma_X = \sqrt{Var(X)}:

    σX=479256\sigma_X = \sqrt{\frac{479}{256}} σX=47916\sigma_X = \frac{\sqrt{479}}{16}

Part c

σX=47916\sigma_X = \frac{\sqrt{479}}{16}

The standard deviation of XX is

σX=47916\boxed{\sigma_X = \frac{\sqrt{479}}{16}}

  1. Calculate the expected value of YY where Y=5X+3Y = 5X + 3.

    Using the property E(cX+d)=cE(X)+dE(cX + d) = cE(X) + d:

    E(Y)=E(5X+3)=5E(X)+3E(Y) = E(5X + 3) = 5E(X) + 3

    Substitute the value of E(X)=4116E(X) = \frac{41}{16}:

    E(Y)=5(4116)+3E(Y) = 5 \left(\frac{41}{16}\right) + 3 E(Y)=20516+3E(Y) = \frac{205}{16} + 3 E(Y)=20516+4816E(Y) = \frac{205}{16} + \frac{48}{16} E(Y)=25316E(Y) = \frac{253}{16}

Part d

E(Y)=25316E(Y) = \frac{253}{16}

The expected value of YY where Y=5X+3Y = 5X + 3 is

E(Y)=25316\boxed{E(Y) = \frac{253}{16}}
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