9709/61

Mathematics 9709/61May/June 2013

Cambridge AS Level · Probability & Statistics 1 (S1) · worked solutions for every part, with the mark scheme

7
questions
50
marks
75
minutes

Topics Representation of Data · The Normal Distribution · Discrete Random Variables · Permutations and Combinations · Probability

Q1Representation of DataFree sample

A summary of 30 values of xx gave the following information:

Σ(xc)=234,Σ(xc)2=1957.5,\Sigma(x - c) = 234, \quad \Sigma(x - c)^2 = 1957.5,

where cc is a constant.

(i)

Find the standard deviation of these values of xx.

2M
DifficultyMedium-Easy
Worked solution

Approach

Since adding a constant cc to each value does not change the spread of the data, the standard deviation of xx is the same as the standard deviation of xcx-c. With n=30n = 30 and the given sums, use

sd2=Σ(xc)2n(Σ(xc)n)2\text{sd}^2 = \frac{\Sigma (x - c)^2}{n} - \left(\frac{\Sigma (x - c)}{n}\right)^2

and then take the square root.

Working

sd2=1957.530(23430)2\text{sd}^2 = \frac{1957.5}{30} - \left(\frac{234}{30}\right)^2 sd2=65.257.82\text{sd}^2 = 65.25 - 7.8^2 sd2=65.2560.84=4.41\text{sd}^2 = 65.25 - 60.84 = 4.41 sd=4.41=2.1\text{sd} = \sqrt{4.41} = 2.1

Answer

sd=2.1\text{sd} = 2.1
Final answer

2.1

Detailed explanation

Walkthrough

We have n=30n = 30 values of xx, but only information about the coded values xcx - c. The constant cc shifts all data by the same amount, so it changes the mean but not the spread. Therefore the standard deviation of xx equals the standard deviation of xcx - c.

For any values y1,,yny_1, \ldots, y_n, the variance can be written as

Var(y)=Σy2n(Σyn)2.\text{Var}(y) = \frac{\Sigma y^2}{n} - \left(\frac{\Sigma y}{n}\right)^2.

Here y=xcy = x - c, so substitute the given sums:

sd2=1957.530(23430)2.\text{sd}^2 = \frac{1957.5}{30} - \left(\frac{234}{30}\right)^2.

Now compute each piece:

1957.530=65.25,23430=7.8.\frac{1957.5}{30} = 65.25, \qquad \frac{234}{30} = 7.8.

Thus

sd2=65.257.82=65.2560.84=4.41.\text{sd}^2 = 65.25 - 7.8^2 = 65.25 - 60.84 = 4.41.

Finally, standard deviation is the positive square root:

sd=4.41=2.1.\text{sd} = \sqrt{4.41} = 2.1.

Key Takeaways

  • The variance formula can be used with Σy\Sigma y and Σy2\Sigma y^2 without needing the raw data.
  • Adding or subtracting a constant from every value does not change variance or standard deviation.
  • For coded data, the variance is still the mean of the squares minus the square of the mean of the coded values.

Common Mistakes

  • Forgetting to divide both sums by 30 before combining them.
  • Evaluating the formula correctly but forgetting to take the square root at the end.
  • Squaring 23430\frac{234}{30} incorrectly; note 7.82=60.847.8^2 = 60.84.

Things to Be Careful About

  • The mark scheme awards M1 for correct substitution into the variance formula and A1 for the accurate value 2.12.1 (or 2.102.10).
  • Standard deviation is always non-negative, so choose the positive square root.
  • Keep the working exact: 4.414.41 is exactly 2.122.1^2, so 2.12.1 is the exact answer.
Techniques used
use the coded-data variance formulasubstitute the given sums into the formulatake the square root to find the standard deviation
(ii)

Given that the mean of these values is 86, find the value of cc.

2M
DifficultyMedium-Easy
Worked solution

Approach

The mean of xx is found from the coded values by adding the constant cc back:

xˉ=c+Σ(xc)n.\bar{x} = c + \frac{\Sigma (x - c)}{n}.

Substitute xˉ=86\bar{x} = 86 and n=30n = 30, then solve for cc.

Working

86=c+2343086 = c + \frac{234}{30} 86=c+7.886 = c + 7.8 c=867.8=78.2c = 86 - 7.8 = 78.2

Answer

c=78.2c = 78.2
Final answer

c = 78.2

Detailed explanation

Walkthrough

The coded values are xcx - c. Their mean is

Σ(xc)n=23430=7.8.\frac{\Sigma (x - c)}{n} = \frac{234}{30} = 7.8.

If we add cc to every coded value, we recover the original values xx. Adding a constant to every value adds the same constant to the mean, so

xˉ=Σ(xc)n+c=7.8+c.\bar{x} = \frac{\Sigma (x - c)}{n} + c = 7.8 + c.

We are told xˉ=86\bar{x} = 86. Substitute this into the equation:

86=c+7.8.86 = c + 7.8. c=867.8=78.2.c = 86 - 7.8 = 78.2.

Key Takeaways

  • The mean of xx can be recovered from coded data by adding the coding constant to the mean of the coded data.
  • The relation xˉ=c+Σ(xc)n\bar{x} = c + \frac{\Sigma (x - c)}{n} is the direct link between the original mean and the coded sums.

Common Mistakes

  • Writing 86=23430c86 = \frac{234}{30} - c or using the wrong sign between the two terms.
  • Quoting c=78.2c = 78.2 without any method; the mark scheme requires the 23430\frac{234}{30} step to be seen.
  • Confusing the sum Σ(xc)=234\Sigma(x-c) = 234 with its mean 7.87.8.

Things to Be Careful About

  • The mark scheme gives M1 as soon as 23430\frac{234}{30} is seen in the equation, then A1 for c=78.2c = 78.2.
  • Remember that subtracting the coding constant would give the opposite answer; adding it back is correct because x=(xc)+cx = (x-c) + c.
  • Since cc is a constant, 867.8=78.286 - 7.8 = 78.2 exactly; no rounding is needed.
Techniques used
compute the mean of the coded valuesuse the mean-shift relation for a constant added to datasolve the resulting linear equation

The rest of this paper

6 more questions
  • Q2The Normal Distribution5M
  • Q3Representation of Data5M
  • Q4The Normal Distribution7M
  • Q5Discrete Random Variables9M
  • Q6Permutations and Combinations9M
  • Q7Probability11M
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