A-LevelMathematicsProbability and statistics9709/62 · Feb/Mar 2025 · Q310 Marks

The random variable X has the distribution Po(1.5).

(a)Find P(X≥ 3).[2]
(b)Find the probability that the sum of three independent values of X is between 3 and 5 inclusive.[3]
(c)The sum of a large number, n, of values of X is denoted by T. Using a suitable approximation, it was found that P(T > 330) = 0.0391, correct to 3 significant figures. Find the value of n.[5]

Mark scheme

Marking points from the official mark scheme (Feb/Mar 2025). The mark or mark type is shown in brackets.

[M1] 1 – e-1.5(1 + 1.5 + 1.5²/2!) = 1- e-1.5(1 + 1.5 +1.125) = 1-(0.22313 +0.334695 +0.25102) Allow one end error. Accept fully correct sigma notation.
[A1] = 0.191 SC unsupported correct answer scores B1.
[B1] λ = 4.5
[M1] e-4.5(4.5³/3! + 4.5⁴/4! + 4.5⁵/5!) = e-4.5 (15.1875 +17.0859 + 15.3773) = 0.1687 + 0.1898+0.1708 Any λ. Allow one end error. Accept fully correct sigma notation.
[A1] = 0.529 SC. Unsupported correct answer scores B1 B1.
[B1] T~ N(1.5n, 1.5n) May be implied.
[M1] Ф-1(1 – 0.0391) [= 1.761] Attempted.
[M1*] 330.5-1.5n / √1.5n = 1.761 Attempt to standardise and = Ф¯¹(1 – 0.0391). Allow no or incorrect cc instead of 330.5. Note: (330.5-1.5n)/√1.5 scores either B1M0* or B0M1*
[M1dep] [1.5n +1.761√1.5n - 330.5 = 0 or 1.5n + 2.1568 √n - 330.5 = 0 or 2.25n² - 996.1516n +109230.25 = 0] Correctly forming and valid attempt to solve a quadratic equation in √n or n or √(1.5n) OE. SOI by correct answer.
[A1] [√n = 14.14 [or -15.58] ] ⇒ n = 200 (only) Dep 330.5 used.

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