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Mathematics · Oct / Nov 2025 · Paper 61 · Question 4

YMA01

As printed in the paper

The question as text

4. A discrete random variable X has the probability distribution given in the table below, where a and b are constants. –1 0 1 2 3 x P(X = x) 0.05 0.1 a b 0.35 Given that E( X ) = 1.8

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  • 4(a)[3]

    0.05 + 0.1 + a + b + 0.35 = 1 [a + b = 0.5]

    –0.05 + a + 2b + 1.05 = 1.8 [a + 2b = 0.8]

    b = 0.3 a = 0.2

    [x]

    –1

    0

    1

    2

    3

  • 4(b)[2]

    0.0

    [F(x)]

    0.15

    0.35 0.65

    1

    5

  • 4(c)[1]

    0.15

    E(5

    −

    3

    X

    )

    = ×

    8 0.05

    + ×

    5 0.1

    +

    E(

    X

    2

    )

    = ×

    1 0.05 1 "0.2"

    + ×

    +

    2 "0.2" 1 "0.3" 4 0.35[

    ×

    −×

    −×

    = −

    0.4]

  • 4(d)[4]

    ×

    + ×

    =

    4 "0.3" 9 0.35[

    4.6]

    or E(5

    −

    3

    X

    )

    =

    5

    −×

    3 1.8[

    = −

    0.4]

    E(5 3

    −

    X

    )

    2

    =

    64 0.05

    ×

    +

    25 0.1

    ×

    +

    X =

    −

    =

    Var(

    )

    "4.6" 1.8

    2

    1.36

    ×

    + ×

    +

    ×

    =

    4 "0.2" 1 "0.3" 16 0.35[ 12.4]

    Var(5 – 3X) = 9 × “1.36”

    Var(5 – 3X) = “12.4” – (“–0.4”)2

    = 12.24

    = 12.24

    awrt 12.2

    Notes

    (a)

    M1

    for either correct equation in a and b

    M1

    for both correct equations in a and b

    both a = 0.2 oe and b = 0.3 oe

    A1

    correct answers score 3 out of 3 unless from obvious

    incorrect working

    In part (b) and (d) working labelled in script takes priority.

    Look out for working shown under the table which can be marked if it does not

    contradict the working in the script.

    for 3 correct values of x and their associated cumulative

    (b)

    B1

    probabilities

    for a fully correct cdf

    B1ft

    ft their a and b provided both are probabilities and a + b = 0.5

    (c)

    B1

    0.15 oe

    for method to find

    2

    with at least 3 correct (ft) nonzero

    E(

    X

    )

    products (may be implied by 4.6) only allow ft products which

    use probabilities

    (d)

    M1

    or a method to find E(5 – 3X) with at least 3 correct (ft) non

    zero products (may be implied by –0.4)

    or use of E(5 – 3X) = 5 – 3E(X)

    for a correct method to find Var(X) ft their

    E(

    X

    2

    )

    (may be

    M1

    implied by 1.36)

    or a correct method to find E(5 – 3X)2 (may be implied by 12.4)

    for use of

    X ft their Var(X)

    23 Var(

    )

    M1

    or use of E(5 – 3X)2 – (E(5 – 3X))2

    awrt 12.2 working must be shown Answer only send to

    A1

    review.

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