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The discrete random variable X can take only the values 2, 3 or 4 - Edexcel - A-Level Maths Statistics - Question 6 - 2008 - Paper 2

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The discrete random variable X can take only the values 2, 3 or 4. For these values the cumulative distribution function is defined by $$F(x) = \frac{(x+k)^2}{25}... show full transcript

Worked Solution & Example Answer:The discrete random variable X can take only the values 2, 3 or 4 - Edexcel - A-Level Maths Statistics - Question 6 - 2008 - Paper 2

Step 1

Find k.

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Answer

To find the value of kk, we first need to ensure that the cumulative distribution function F(x)F(x) satisfies the property that F(4)=1F(4) = 1 when x=4x = 4:

F(4)=(4+k)225=1F(4) = \frac{(4+k)^2}{25} = 1

This gives us:

(4+k)2=25.(4+k)^2 = 25.

Taking the square root of both sides:

4+k=5or4+k=54+k = 5 \quad \text{or} \quad 4+k = -5

Since kk is a positive integer, we discard 4+k=54+k = -5 and solve:

4+k=5k=1.4+k = 5 \Rightarrow k = 1.

Thus, we find that k=1k = 1.

Step 2

Find the probability distribution of X.

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Answer

Now, we can use the value of kk to find the probability distribution of XX.

Using the cumulative distribution function:

  1. For x=2x = 2:
    F(2)=(2+1)225=3225=925.F(2) = \frac{(2+1)^2}{25} = \frac{3^2}{25} = \frac{9}{25}.
    Thus,
    P(X=2)=F(2)=925.P(X=2) = F(2) = \frac{9}{25}.

  2. For x=3x = 3:
    F(3)=(3+1)225=4225=1625.F(3) = \frac{(3+1)^2}{25} = \frac{4^2}{25} = \frac{16}{25}.
    Thus,
    P(X=3)=F(3)F(2)=1625925=725.P(X=3) = F(3) - F(2) = \frac{16}{25} - \frac{9}{25} = \frac{7}{25}.

  3. For x=4x = 4:
    F(4)=(4+1)225=5225=1.F(4) = \frac{(4+1)^2}{25} = \frac{5^2}{25} = 1.
    Thus,
    P(X=4)=F(4)F(3)=11625=925.P(X=4) = F(4) - F(3) = 1 - \frac{16}{25} = \frac{9}{25}.

Finally, we summarize the probability distribution of XX:

xP(X=x)
29/25
37/25
49/25

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