S1 June 2016 Q5
5. A midwife records the weights, in kg, of a sample of 50 babies born at a hospital. Her results are given in the table below.
| Weight (\(w\) kg) | Frequency (f) | Weight midpoint (\(x\)) |
|---|---|---|
| \(0 \leqslant w \lt 2\) | 1 | 1 |
| \(2 \leqslant w \lt 3\) | 8 | 2.5 |
| \(3 \leqslant w \lt 3.5\) | 17 | 3.25 |
| \(3.5 \leqslant w \lt 4\) | 17 | 3.75 |
| \(4 \leqslant w \lt 5\) | 7 | 4.5 |
[You may use \(\sum \mathrm{f}x^2 = 611.375\)]
A histogram has been drawn to represent these data.
The bar representing the weight \(2 \leqslant w \lt 3\) has a width of 1 cm and a height of 4 cm.
Shyam decides to model the weights of babies born at the hospital, by the random variable \(W\), where \(W \sim \mathrm{N}(3.43, 0.65^2)\)
A newborn baby weighing 3.43 kg is born at the hospital.
| Scheme | Marks |
|---|---|
| Width = 0.5 (cm) | B1 |
| e.g 4 [cm\(^2\)] represents 8 babies or frequency densities are 8 and 34 | M1 |
| Height = 17 (cm) | A1 |
| (3) |
Notes
M1 for clear representation of area with frequency or height\(\times\)width = 8.5
A1 for 17 (cm) [Must be clear it is height not frequency] (Ans only must satisfy \(h \times w = 8.5\))
| Scheme | Marks |
|---|---|
| \([Q_2 =]\ \ \{3\} + \dfrac{(25 - 9)}{(26 - 9)} \times 0.5\), or \(\{3.5\} - \dfrac{(25 - 24)}{(41 - 24)} \times 0.5 = \) awrt 3.47 (allow \(\frac{59}{17}\)) | M1, A1 |
| (2) |
Notes
M1 for \(\dfrac{16}{17} \times 0.5\) or if using \(n + 1\) for \(\dfrac{16.5}{17} \times 0.5\) May see \(-\dfrac{1}{17} \times 0.5\) if working down
A1 for awrt 3.47 (or \(\frac{59}{17}\)) [check from correct working] or (if using (\(n\) + 1) for 3.485 or awrt 3.49)
| Scheme | Marks |
|---|---|
| (i) \(\sum \mathrm{f}x = 1 \times 1 + 2.5 \times 8 + 3.25 \times 17 + 3.75 \times 17 + 4.5 \times 7 = 171.5\), \(\bar{x} = \dfrac{171.5}{50} = (3.43)\) (*) | B1cso |
| (ii) \(\sqrt{\dfrac{611.375}{50} - 3.43^2}\) ,\(= 0.680147\ldots\) = awrt 0.680 (Accept 0.68) | M1, A1 |
| (3) |
Notes
(i) B1cso for \(\Sigma \mathrm{f}x\) (at least 3 correct & no incorrect products seen) and correct \(\dfrac{\sum \mathrm{f}x}{50}\) or \(\dfrac{171.5}{50}\)
(ii) M1 for a correct expression including square root. Must use 3.43 no ft
A1 for awrt 0.680 (accept 0.68). Allow use of \(s\) = awrt 0.687 (Ans only 2/2)
| Scheme | Marks |
|---|---|
| \(\left[\mathrm{P}(W \lt 3) = \mathrm{P}\left(Z \lt \dfrac{-0.43}{0.65}\right)\right] = \mathrm{P}(Z \lt -0.6615..)\) | M1 |
| = 1 – 0.7454 (tables) | M1 |
| = 0.2546 awrt 0.254~0.255 | A1 |
| (3) |
Notes
1st M1 for an attempt to standardise with 3, 3.43 and 0.65. Allow \(\pm\) and also use of their sd
2nd M1 for 1 – \(p\) where \(0.74 \lt p \lt 0.75\) NB calculator gives 0.7458665…
A1 for awrt 0.254 or 0.255
| Scheme | Marks |
|---|---|
| (b) and (c)(i) mean \(\neq\) med or skew or mean~ median or no skew and comment | B1 |
| (d) = 0.254 or 0.255 compare data = 0.18 (or 12.7 compared with 9) | B1 |
| 0.18 different from 0.25 so normal not good or 0.18 similar to 0.25 so normal is OK | dB1 |
| (3) |
Notes
1st B1 for a statement about mean/median and compatible comment about normal
2nd B1 for statement comparing their (d) with data (sight of 0.18 or 12.7 and 9 required)
3rd dB1 dep on 2nd B1 for conclusion about normal compatible with 2nd statement
| Scheme | Marks |
|---|---|
| (i) No change in mean (since weight is the same) | B1 |
| (ii) s.d. will decrease (Extra value is at “centre” so data more concentrated) | B1 |
| Both statements correct and correct reasons for each | dB1 |
| (3) | |
| (17 marks) |
Notes
1st B1 for no change in mean {send a correct argument for decrease to review}
2nd B1 for s.d. decreases
3rd dB1 dep on 1st and 2nd Bs for a correct reason for both mean and sd
e.g. “new mean the same so within 1 s.d. of old mean”