S1 January 2013 Q5
5. A survey of 100 households gave the following results for weekly income £\(y\).
| Income \(y\) (£) | Mid-point | Frequency \(f\) |
|---|---|---|
| \(0 \leqslant y \lt 200\) | 100 | 12 |
| \(200 \leqslant y \lt 240\) | 220 | 28 |
| \(240 \leqslant y \lt 320\) | 280 | 22 |
| \(320 \leqslant y \lt 400\) | 360 | 18 |
| \(400 \leqslant y \lt 600\) | 500 | 12 |
| \(600 \leqslant y \lt 800\) | 700 | 8 |
(You may use \(\sum fy^2 = 12\,452\,800\))
A histogram was drawn and the class \(200 \leqslant y \lt 240\) was represented by a rectangle of width 2 cm and height 7 cm.
One measure of skewness is \(\dfrac{3(\text{mean} - \text{median})}{\text{standard deviation}}\).
Katie suggests using the random variable \(X\) which has a normal distribution with mean 320 and standard deviation 150 to model the weekly income for these data.
| Scheme | Marks |
|---|---|
| Width = 4 (cm) | B1 |
| Area of 14 cm\(^2\) represents frequency 28 and area of \(4h\) represents 18 | M1 |
| Or \(\dfrac{4h}{18} = \dfrac{14}{28}\) (o.e.) \(\boldsymbol{h} = \underline{\mathbf{2.25}}\) (cm) | A1 |
| (3) |
Notes
B1 for width (ignore units)
M1 for clear method using area and frequency or their width \(\times\) their height = 9
e.g. seeing both fd of 0.7 and 0.225 (may see fd in the table) [Must use correct interval]
| Scheme | Marks |
|---|---|
| \(m = (240) + \dfrac{10}{22}\times 80\) (o.e.) | M1 |
| \(= 276.36\ldots\) (\(\frac{3040}{11}\)) ((£)276 \(\leqslant m \lt\) (£)276.5) | A1 |
| (2) |
Notes
M1 for \(\dfrac{10}{22}\times 80\) or \(\dfrac{10.5}{22}\times 80\) (o.e.). Allow use of \((n + 1)\) leading to £278.18… or [278, 278.5)
A1 Do not award if incorrect end-point seen but answer only is 2/2
| Scheme | Marks |
|---|---|
| \(\sum fy = 31600\) leading to \(\underline{\bar{y} = 316}\) | M1A1 |
| \(\sigma_y = \sqrt{\dfrac{12452800}{100} - (\bar{y})^2}\) \(= 157.07\ldots\) (awrt 157) Allow \(s = 157.86\ldots\) | M1A1 |
| (4) |
Notes
1st M1 attempt at \(\sum fy\) with at least 3 correct products or ans. that rounds to 30 000 (to 1 sf) &/100
2nd M1 for correct expression including \(\sqrt{\ }\). Follow through \(\bar{y}\). Need \(\sum fy^2\) correct but condone a minor transcription error e.g. 12458200.
| Scheme | Marks |
|---|---|
| Skewness \(= 0.764\ldots\) (awrt 0.76 or 0.75) [If \(n + 1\) used in (b) and \(m\) = £278 accept awrt 0.73 or 0.72] | B1 |
| Positive skew | B1ft |
| (2) |
Notes
1st B1 for awrt 0.76/0.75 for \(m\) = £276 or awrt 0.73/0.72 for \(m\) = £278
2nd B1ft for a correct description of their skew based on their measure or if no measure given based on their values of mean and median. (correlation is B0)
| Scheme | Marks |
|---|---|
| \(z = \pm\dfrac{80}{150}\) | M1 |
| \(\mathrm{P}(240 \lt X \lt 400) = \underline{\mathbf{0.40 \sim 0.41}}\) | A1 |
| (2) |
Notes
M1 for an attempt to standardise using the 320 and 150 and either 240 or 400 (implied by 0.53)
A1 for answer in range [0.40, 0.41] (tables gives 0.4038, calculator 0.40619...) Ans only 2/2
| Scheme | Marks |
|---|---|
| (e) suggests a reasonable fit for this range BUT (d) since skew it will not be a good fit overall | B2/1/0 |
| (2) | |
| (15 marks) |
Notes
For B2 we need 2 comments that make reference to each of part (e) and part (d)
One comment should suggest it is not good since skew. The other it is since matches range in (e)
1st B1 for one relevant comment
2nd B1 for both comments NB Do not use B0B1