9709/61

Mathematics 9709/61May/June 2017

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

7
questions
50
marks
75
minutes

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

Q1Representation of DataFree sample

Kadijat noted the weights, xx grams, of 30 chocolate buns. Her results are summarised by

(xk)=315,(xk)2=4022,\sum(x - k) = 315, \quad \sum(x - k)^2 = 4022,

where kk is a constant. The mean weight of the buns is 50.5 grams.

(i)

Find the value of kk.

2M
DifficultyMedium-Easy
Worked solution

Approach

Use the coded data y=xky = x - k. The mean of the coded data is found from y\sum y, and since x=y+kx = y + k, the original mean is the coded mean plus kk.

Working

Let yi=xiky_i = x_i - k. Then

yˉ=y30=31530=10.5\bar{y} = \frac{\sum y}{30} = \frac{315}{30} = 10.5

Since x=y+kx = y + k,

xˉ=yˉ+k\bar{x} = \bar{y} + k

Given xˉ=50.5\bar{x} = 50.5,

50.5=10.5+k50.5 = 10.5 + k k=40k = 40

Answer

k=40k = 40
Final answer

k = 40

Detailed explanation

Walkthrough

We are told that the sum of the coded values xkx-k is 315. Since there are 30 buns, the mean of these coded values is

yˉ=31530=10.5\bar{y} = \frac{315}{30} = 10.5

Each original weight is kk more than its coded value, so the original mean is the coded mean plus kk:

xˉ=yˉ+k\bar{x} = \bar{y} + k

The original mean is given as 50.5, so

50.5=10.5+k50.5 = 10.5 + k

and therefore k=40k = 40. Another way to see this is to multiply the mean by 30 to get x=1515\sum x = 1515, then solve 151530k=3151515 - 30k = 315.

Key Takeaways

Coding data by subtracting a constant shifts the mean by that same constant. The mean of coded data can be used to find the original mean, or to find the coding constant when the original mean is known.

Common Mistakes

  • Dividing 315 by 30 is essential; using 315 directly as the coded mean is wrong.
  • After finding 1515315=12001515 - 315 = 1200, some students give 1200 as the answer, but this is 30k30k, not kk; it must be divided by 30.
  • A sign error in (xk)=x30k\sum(x-k) = \sum x - 30k can lead to k=61k = 61 instead of 40.

Things to Be Careful About

  • Use n=30n = 30 as the denominator when finding the coded mean.
  • Remember that subtracting kk from every value lowers the mean by kk, so xˉ=yˉ+k\bar{x} = \bar{y} + k.
  • Check the answer: 50.510.5=4050.5 - 10.5 = 40 confirms the value of kk.
Techniques used
calculate the mean of coded datarelate the coded mean to the original meansolve a linear equation for the coding constant
(ii)

Find the standard deviation of xx.

2M
DifficultyMedium-Easy
Worked solution

Approach

Adding a constant to every data value does not change the standard deviation. So the standard deviation of xx equals the standard deviation of the coded values y=xky = x - k. Use the coded variance formula Var(y)=y230yˉ2\text{Var}(y) = \frac{\sum y^2}{30} - \bar{y}^2.

Working

With y=xky = x - k and k=40k = 40, yˉ=10.5\bar{y} = 10.5. Therefore

Var(y)=40223010.52\text{Var}(y) = \frac{4022}{30} - 10.5^2 Var(y)=134.0667110.25=23.8167\text{Var}(y) = 134.0667 - 110.25 = 23.8167 sd=23.8167=4.88 g (3 s.f.)\text{sd} = \sqrt{23.8167} = 4.88 \text{ g (3 s.f.)}

Answer

standard deviation=4.88 g\text{standard deviation} = 4.88 \text{ g}
Final answer

4.88 g

Detailed explanation

Walkthrough

Since x=(xk)+kx = (x-k)+k, each original weight is obtained from its coded value by adding the same constant kk. Adding a constant shifts all values but does not change how spread out they are, so the standard deviation of xx is the same as the standard deviation of xkx-k. For the coded values y=xky = x-k, the variance is

Var(y)=y2nyˉ2\text{Var}(y) = \frac{\sum y^2}{n} - \bar{y}^2

Here y2=4022\sum y^2 = 4022, n=30n = 30, and yˉ=10.5\bar{y} = 10.5. Therefore

Var(y)=40223010.52=134.0667110.25=23.8167\text{Var}(y) = \frac{4022}{30} - 10.5^2 = 134.0667 - 110.25 = 23.8167

Taking the square root gives

sd=23.8167=4.88 g\text{sd} = \sqrt{23.8167} = 4.88 \text{ g}

Alternatively, expand (xk)2=x22kx+30k2\sum(x-k)^2 = \sum x^2 - 2k\sum x + 30k^2. With k=40k = 40 and x=1515\sum x = 1515,

4022=x280(1515)+30(1600)4022 = \sum x^2 - 80(1515) + 30(1600) x2=77222\sum x^2 = 77222

Then Var(x)=772223050.52=23.8167\text{Var}(x) = \frac{77222}{30} - 50.5^2 = 23.8167, giving the same standard deviation.

Key Takeaways

Variance measures spread using squared deviations, and the formula Var=y2nyˉ2\text{Var} = \frac{\sum y^2}{n} - \bar{y}^2 can be applied to coded data. Since subtracting a constant does not affect spread, the standard deviation of coded data equals the standard deviation of the original data.

Common Mistakes

  • Using 4022/304022/30 as the variance without subtracting yˉ2\bar{y}^2.
  • Subtracting 50.5250.5^2 when using the coded formula; the coded mean is 10.510.5, not 50.5.
  • Forgetting to take the square root at the end.
  • Mixing coded sums with uncoded means in the same formula.

Things to Be Careful About

  • Use n=30n = 30 in the denominator, not 29.
  • The variance is in grams squared; the standard deviation is in grams.
  • Round only at the end to avoid rounding errors; the required value is 4.88 g.
  • When expanding the uncoded sum of squares, keep the signs correct: (xk)2=x22kx+30k2\sum(x-k)^2 = \sum x^2 - 2k\sum x + 30k^2.
Techniques used
apply the coded variance formulasubtract the squared coded meantake the square root to find the standard deviation

The rest of this paper

6 more questions
  • Q2Probability5M
  • Q3Probability6M
  • Q4Representation of Data6M
  • Q5Discrete Random Variables7M
  • Q6The Normal Distribution11M
  • Q7Permutations and Combinations11M
Loading the full paper…