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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
Question 6
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.
Answer
To determine the value of k, we start with the cumulative distribution function at the maximum value of x, which is 4:
Given that:
[ F(4) = 1 ]
Substituting this into the equation:
[ F(4) = \frac{(4+k)^2}{25} = 1 ]
Multiplying both sides by 25 gives:
[ (4+k)^2 = 25 ]
Taking the square root of both sides:
[ 4+k = 5 \quad \text{or} \quad 4+k = -5 ]
From the first equation:
[ k = 5 - 4 = 1 ]
Since k is a positive integer, we discard the second equation, leading to:
[ k = 1 ]
Step 2
Find the probability distribution of X.
Answer
With k = 1, we can now compute the cumulative distribution function for each value of X:
- For x = 2:
[ F(2) = \frac{(2+1)^2}{25} = \frac{3^2}{25} = \frac{9}{25} ]
- For x = 3:
[ F(3) = \frac{(3+1)^2}{25} = \frac{4^2}{25} = \frac{16}{25} ]
- For x = 4:
[ F(4) = \frac{(4+1)^2}{25} = \frac{5^2}{25} = 1 ]
Now to find the probabilities:
- (P(X=2) = F(2) = \frac{9}{25})
- (P(X=3) = F(3) - F(2) = \frac{16}{25} - \frac{9}{25} = \frac{7}{25})
- (P(X=4) = F(4) - F(3) = 1 - \frac{16}{25} = \frac{9}{25})
Thus the probability distribution of X is:
| x | P(X=x) |
|---|---|
| 2 | (\frac{9}{25}) |
| 3 | (\frac{7}{25}) |
| 4 | (\frac{9}{25}) |