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3.2 Finding Probabilities for Normal Distributions

3.2 Finding Probabilities for Normal Distributions

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Question 27

In a study of professional basketball players, the wing span of a player has a mean of 2.052.052.05 metres and a standard deviation of 0.120.120.12 metres.

The wing spans of 95%95\%95% of these players are found to be between 1.811.811.81 metres and 2.292.292.29 metres.

a.

Comment on whether a normal distribution may be suitable to model the wing span of a professional basketball player in this study.

[2]
bi.

You may assume that the wing span of a professional basketball player may be modelled by a normal distribution with mean 2.052.052.05 metres and standard deviation 0.120.120.12 metres.

Find the probability that the wing span of a randomly selected player is exactly 2.102.102.10 metres.

[1]
bii.

Find the probability that the wing span of a randomly selected player is between 1.951.951.95 metres and 2.152.152.15 metres.

[2]
biii.

Two players are chosen at random. Calculate the probability that both of their wing spans are between 1.951.951.95 metres and 2.152.152.15 metres.

[2]
c.

The summarised data for the wing spans, www metres, of a random sample of 505050 amateur basketball players is given below:

∑w=92.5and∑(w−wˉ)2=0.98 \sum w = 92.5 \quad \text{and} \quad \sum(w - \bar{w})^2 = 0.98 ∑w=92.5and∑(w−wˉ)2=0.98

Use this data to calculate estimates of the mean and standard deviation of the wing spans of amateur basketball players.

[2]
d.

Using your answers from part (c), compare the wing spans of professional basketball players and amateur basketball players.

[2]
Markscheme

3.2 Finding Probabilities for Normal Distributions Questions

  1. A Level
  2. /Maths
  3. /3.2 Finding Probabilities for Normal Distributions

80 exam-style questions on Edexcel A Level Maths 3.2 Finding Probabilities for Normal Distributions. Each one has a worked solution and a mark scheme showing where the marks go.

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