Question: [Maximum mark: 6]
A discrete random variable, X, has the following probability distribution:
| x | 0 | 1 | 2 | 3 |
| P(X = x) | 0.41 | k – 0.28 | 0.46 | 0.29 – 2k2 |
(a) Show that 2k2 – k + 0.12 = 0.
(b) Find the value of k , giving a reason for your answer.
(c) Hence, find E(X).
▶️Answer/Explanation
Ans:
(a) 0.41 + k – 0.28 + 0.46 + 0.29 – 2k2 = 1 OR k – 2k2 + 0.01 = 0.13 (or equivalent)
2k2 – k + 0.12 = 0
(b) one of 0.2 OR 0.3
k = 0.3
reasoning to reject k = 0.2 eg P(1) = k – 0.28 0 ≥ 0 therefore k ≠ 0.2
(c) attempting to use the expected value formula
E(X) = 0 × 0.41 + 1 × (0.3 – 0.28) + 2 × 0.46 + 3 ×(0.29 – 2 × 0.32)
= 1.27
Note: Award M1A0 if additional values are given.
Question
A fisherman notices that in any hour of fishing, he is equally likely to catch exactly two fish, as he is to catch less than two fish. Assuming the number of fish caught can be modelled by a Poisson distribution, calculate the expected value of the number of fish caught when he spends four hours fishing.
▶️Answer/Explanation
Markscheme
\(X \sim {\text{Po}}(m)\)
\({\text{P}}(X = 2) = {\text{P}}(X < 2)\) (M1)
\(\frac{1}{2}{m^2}{{\text{e}}^{ – m}} = {{\text{e}}^{ – m}}(1 + m)\) (A1)(A1)
\(m = 2.73 {\text{ }}\left( {1 + \sqrt 3 } \right)\) A1
in four hours the expected value is 10.9\(\,\,\,\,\left( {4 + 4\sqrt 3 } \right)\) A1
Note: Value of m does not need to be rounded.
[5 marks]
Examiners report
Many candidates did not attempt this question and many others did not go beyond setting the equation up. Among the ones who attempted to solve the equation, once again, very few candidates took real advantage of GDC use to obtain the correct answer.
Question
The probability density function of a continuous random variable X is given by
\(f(x) = \frac{1}{{1 + {x^4}}}\), \(0\) \(”\) \(x\) \(”\) \(a\) .
a.Find the value of a .[3]
b.Find the mean of X .[2]
▶️Answer/Explanation
Markscheme
\(\int_0^a {\frac{1}{{1 + {x^4}}}{\text{d}}x = 1} \) M2
a = 1.40 A1
[3 marks]
\({\text{E}}(X) = \int_0^a {\frac{x}{{1 + {x^4}}}{\text{d}}x} \) M1
\(\left( { = \frac{1}{2}\arctan ({a^2})} \right)\)
= 0.548 A1
[2 marks]