Unit 15 Review Exercises
Statistics
These questions mix the ideas from the whole unit. Paper 1 means no calculator; Paper 2 means a GDC is allowed.
1 Paper 1 A company has 80 managers, 360 office staff and 560 factory workers. A stratified sample of 50 employees is to be taken.
(a) Find the number of each type of employee in the sample.
(b) Explain why stratified sampling is suitable here.
(a) Total \(= 1000\). Managers \(\dfrac{80}{1000} \times 50 = 4\), office staff \(\dfrac{360}{1000} \times 50 = 18\), factory workers \(\dfrac{560}{1000} \times 50 = 28\). (\(4 + 18 + 28 = 50\) ✓)
(b) The three groups may have different opinions; stratified sampling makes sure each group is represented in proportion. See 15.1.
2 Paper 1 The table shows the number of times, \(x\), that some students visited the library in a week. The mean is \(2.5\).
| \(x\) | 0 | 1 | 2 | 3 | 4 | 5 |
|---|---|---|---|---|---|---|
| Frequency | 2 | 5 | \(q\) | 8 | 4 | 3 |
(a) Find \(q\). \(\qquad\) (b) Find the median and the mode.
(a) \(\sum f = 22 + q\), \(\sum fx = 0 + 5 + 2q + 24 + 16 + 15 = 60 + 2q\). \(\dfrac{60 + 2q}{22 + q} = 2.5 \Rightarrow 60 + 2q = 55 + 2.5q \Rightarrow q = 10\).
(b) \(n = 32\): median between the 16th and 17th values. Running totals \(2, 7, 17, \ldots\), so both are \(2\): median \(= 2\). Mode \(= 2\) (frequency 10). See 15.2.
3 Paper 2 The lengths, \(L\) mm, of 60 insects are shown.
| Length, \(L\) (mm) | \(10 \le L < 14\) | \(14 \le L < 18\) | \(18 \le L < 22\) | \(22 \le L < 26\) | \(26 \le L < 30\) |
|---|---|---|---|---|---|
| Frequency | 7 | 13 | 22 | 12 | 6 |
(a) Write down the modal class.
(b) Use your GDC to estimate the mean and the standard deviation.
4 Paper 1 The marks in a test have mean \(54\) and standard deviation \(8\). The teacher increases every mark by \(10\%\) and then adds \(3\). Find the new mean and the new variance.
New mark \(= 1.1x + 3\). New mean \(= 1.1 \times 54 + 3 = 62.4\). New s.d. \(= 1.1 \times 8 = 8.8\), so the new variance is \(8.8^2 = 77.44\). See 15.3.
5 Paper 1 The numbers of messages received by a student on 13 days were:
\[ 4,\ 9,\ 12,\ 7,\ 10,\ 31,\ 8,\ 11,\ 6,\ 9,\ 13,\ 10,\ 8 \]
(a) Find the median and the quartiles.
(b) Show that \(31\) is an outlier.
(c) Describe the box plot of the data.
(a) In order: \(4, 6, 7, 8, 8, 9, \mathbf{9}, 10, 10, 11, 12, 13, 31\). Median \(= 9\) (7th value). Lower half \(4, 6, 7, 8, 8, 9\): \(Q_1 = 7.5\). Upper half \(10, 10, 11, 12, 13, 31\): \(Q_3 = 11.5\).
(b) IQR \(= 4\); upper boundary \(11.5 + 6 = 17.5 < 31\), so \(31\) is an outlier.
(c) Box from \(7.5\) to \(11.5\) with the median at \(9\); lower whisker to \(4\); upper whisker to \(13\) (the largest non-outlier); a cross at \(31\). See 15.5 and 15.6.
6 Paper 1 The heights of plants given fertiliser F have five-number summary \(12, 20, 26, 31, 38\) cm. Plants with no fertiliser have summary \(10, 14, 18, 21, 33\) cm. Compare the two groups.
The median with fertiliser (\(26\) cm) is higher than without (\(18\) cm), so plants with fertiliser are usually taller. IQR: \(11\) cm against \(7\) cm, so the fertilised plants are more spread out. The group without fertiliser has a long upper whisker (\(12\) against \(4\)), so it is positively skewed; the fertilised group is roughly symmetric. See 15.5 and 15.6.
7 Paper 1 The cumulative frequency table shows the ages of 120 people at a concert.
| Age less than | 10 | 20 | 30 | 40 | 50 | 60 |
|---|---|---|---|---|---|---|
| Cumulative frequency | 0 | 18 | 30 | 60 | 90 | 120 |
(a) Write down the median and the IQR.
(b) Find the number of people aged 40 or over.
(c) Write down the frequency of the class \(20 \le a < 30\).
(a) \(\frac{120}{2} = 60\): median \(= 40\). \(\frac{120}{4} = 30\): \(Q_1 = 30\); \(\frac{3 \times 120}{4} = 90\): \(Q_3 = 50\). IQR \(= 20\).
(b) \(120 - 60 = 60\). (c) \(30 - 18 = 12\). See 15.5.
8 Paper 1 A reporter wants to know what residents of a city think about a new railway line. She asks the first 100 people who come out of one railway station.
(a) Name this sampling method.
(b) Explain why the sample is likely to be biased.
(c) Suggest a better method.
(a) Convenience sampling.
(b) People who use that station already use trains and live or work near it; their views are not typical of all residents.
(c) For example, stratified sampling by district, choosing residents at random from a list in each district. See 15.1.
9 Paper 1 For a set of 20 values, \(\sum x = 360\) and \(\sum x^2 = 6800\).
(a) Find the mean and the variance.
(b) Each value is halved and then \(4\) is added. Find the new mean and the new standard deviation.
(a) Mean \(= 18\). Variance \(= \dfrac{6800}{20} - 18^2 = 340 - 324 = 16\).
(b) New mean \(= \dfrac{18}{2} + 4 = 13\). Old s.d. \(= 4\), new s.d. \(= \dfrac{4}{2} = 2\). See 15.3.
10 Paper 2 The times, in seconds, of 12 swimmers in a race were:
\[ 62.1,\ 58.4,\ 60.3,\ 59.7,\ 61.5,\ 57.9,\ 63.0,\ 60.8,\ 59.2,\ 74.6,\ 61.1,\ 58.8 \]
(a) Use your GDC to find the mean and the standard deviation.
(b) Find the quartiles and show that 74.6 is an outlier.
(c) The swimmer with 74.6 s lost a goggle during the race. Find the mean and the standard deviation of the other 11 times, and comment.
(a) \(\bar{x} = 61.45\) s, \(\sigma = 4.23\) s (3 s.f.).
(b) In order: \(57.9, 58.4, 58.8, 59.2, 59.7, 60.3, 60.8, 61.1, 61.5, 62.1, 63.0, 74.6\). \(Q_1 = \dfrac{58.8 + 59.2}{2} = 59.0\), \(Q_3 = \dfrac{61.5 + 62.1}{2} = 61.8\), IQR \(= 2.8\). Upper boundary \(61.8 + 4.2 = 66.0 < 74.6\), so 74.6 is an outlier.
(c) Without 74.6: \(\bar{x} = 60.3\) s, \(\sigma = 1.54\) s (3 s.f.). The outlier is a genuine time but was caused by an accident, so it is reasonable to leave it out when describing typical times. Removing it lowers the mean a little and makes the standard deviation much smaller. See 15.3 and 15.6.