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The discrete random variable Y has probability distribution | y | 1 | 2 | 3 | 4 | |---|---|---|---|---| | P(Y = y) | a | b | 0.3 | c | where a, b and c are constants - Edexcel - A-Level Maths Statistics - Question 3 - 2011 - Paper 2

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The discrete random variable Y has probability distribution | y | 1 | 2 | 3 | 4 | |---|---|---|---|---| | P(Y = y) | a | b | 0.3 | c | where a, b and c are constan... show full transcript

Worked Solution & Example Answer:The discrete random variable Y has probability distribution | y | 1 | 2 | 3 | 4 | |---|---|---|---|---| | P(Y = y) | a | b | 0.3 | c | where a, b and c are constants - Edexcel - A-Level Maths Statistics - Question 3 - 2011 - Paper 2

Step 1

Find the value of a, the value of b, the value of c and the value of d.

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Answer

To find the values of a, b, c, and d, we start with the probability distribution. Since the sum of all probabilities must equal 1:

a + b + 0.3 + c = 1 \tag{1}

From the cumulative distribution function F(y), we have:

  • For y = 1, F(1) = P(Y \leq 1) = 0.1, which implies ( a = 0.1 )
  • For y = 2, F(2) = P(Y \leq 2) = 0.5, thus we have: F(2) = a + b = 0.5 \tag{2} Substituting (1) into (2), we get:

0.1 + b = 0.5 \Rightarrow b = 0.4 \tag{3}

Next, substituting values of a and b into (1):

0.1 + 0.4 + 0.3 + c = 1 \Rightarrow c = 0.2 \tag{4}

Finally, for d, since F(4) = 1, F(4) = 0.1 + 0.4 + 0.3 + d = 1 \Rightarrow d = 1 - 0.8 = 0.2 \tag{5}

In summary:

  • a = 0.1
  • b = 0.4
  • c = 0.2
  • d = 0.2

Step 2

Find P(3Y + 2 ≥ 8).

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Answer

We start by rearranging the inequality:

3Y+283Y6Y23Y + 2 \geq 8 \Rightarrow 3Y \geq 6 \Rightarrow Y \geq 2

Hence, we are interested in P(Y \geq 2). This can be computed as:

P(Y2)=1P(Y1)P(Y \geq 2) = 1 - P(Y \leq 1)

From the cumulative distribution function, P(Y \leq 1) = F(1) = 0.1:

Thus:

P(Y2)=10.1=0.9P(Y \geq 2) = 1 - 0.1 = 0.9

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