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2.4 Probability Distributions

2.4 Probability Distributions

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Question 66

A laboratory synthesises organic protein filaments whose lengths depend on specific cultivation conditions. The length LLL, in millimetres (mm), of a filament is modelled by a continuous random variable with probability density function

g(l)={3250(10l−l2)0≤l≤50otherwise g(l) = \begin{cases} \frac{3}{250}(10l - l^2) & 0 \le l \le 5 \\ 0 & \text{otherwise} \end{cases} g(l)={2503​(10l−l2)0​0≤l≤5otherwise​
a.

Use algebraic integration to determine the mean length of a filament. State your answer in millimetres and micrometres (1 mm=1000 μm1 \text{ mm} = 1000 \text{ }\mu\text{m}1 mm=1000 μm).

[4]
b.

Show that the probability of a randomly selected filament having a length between 1 mm and 4 mm is 81125\displaystyle \frac{81}{125}12581​.

[3]
c.

A researcher examines a batch of 200 filaments grown under these conditions.

Using a suitable approximation, calculate the probability that at least 140 of these filaments have a length between 1 mm and 4 mm.

[5]
Markscheme

2.4 Probability Distributions Questions

  1. A Level
  2. /Maths
  3. /2.4 Probability Distributions

390 exam-style questions on OCR (MEI) A Level Maths 2.4 Probability Distributions, covering 2.4.1 Recognise binomial situations, 2.4.2 Probability of success p, 2.4.3 Calculate binomial probabilities, 2.4.4 Mean of the binomial distribution, 2.4.5 Expected frequencies for binomial, 2.4.6 Probability functions and discrete random variables, 2.4.7 Numerical probabilities for a simple distribution, 2.4.8 Normal distribution as a model (A-level only), 2.4.9 Shape of the Normal curve (A-level only), 2.4.10 Linear transformation and standardising (A-level only), 2.4.11 Symmetry and inflection of Normal curve (A-level only), 2.4.12 Calculate probabilities from a Normal distribution (A-level only), 2.4.13 Model with probability distributions, and 2.4 Probability Distributions. Each one has a worked solution and a mark scheme showing where the marks go.

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