AI HL · 13.3
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Data, sampling and averages
13.3.1
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Menu > Statistics > Stat Calculations > One-Variable Statistics
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A company has 840 staff: 360 in the office, 300 in the factory and 180 drivers. A survey needs 56 staff. The 56 are chosen at random, with the same proportion taken from each of the three groups. (a) Name this sampling method. (b) How many drivers are in the sample?
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Stratified sampling
The population is split into groups (strata) and a random sample is taken from every group, in proportion to the group's size.
\(\frac{56}{840}=\frac{1}{15}\)
The sample is this fraction of the whole population, so it is this fraction of every group.
\(\frac{1}{15}\times 180=12\)
The sample has 12 drivers. Check: 24 + 20 + 12 = 56.
A factory makes 1200 bottles in a shift. An inspector tests the 7th bottle and then every 25th bottle after it (the 7th, 32nd, 57th, …). (a) How many bottles are tested? (b) The machine is fitted with a faulty valve that fails on every 25th bottle starting from bottle 25. Explain why this sampling method would miss the fault.
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Start at 7, then add 25 each time
A systematic sample takes items at a fixed interval from a list.
\(7+25k \le 1200,\ \text{so } k=0,1,\dots,47\)
There are 48 values of k, so 48 bottles are tested.
The tested bottles are never numbers 25, 50, 75, …
A fixed interval can line up with a pattern in the process. Bottles 7, 32, 57, … all differ from the faulty bottles, so the sample is biased and misses the fault.
A leisure centre posts this question on its own social media page: "Do you agree that our new pool is excellent?" 214 people reply and 190 say yes. The centre announces that "almost 89% of local residents think the pool is excellent". (a) Find the percentage of the replies that said yes, to 3 significant figures. (b) Give two reasons why the announcement may not be reliable. (c) Suggest a better way to find out what residents think.
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\(\frac{190}{214}\times 100=88.8\%\)
Only the people who replied are counted, so divide by 214.
Self-selected replies from the centre's own followers
People chose to answer, and followers of the centre are more likely to like it. The sample is biased: it is not a random sample of residents.
"Do you agree that our new pool is excellent?"
A leading question pushes people towards yes. The source also gains from a good result.
Random sample of residents, neutral question
Choose residents at random, for example from the electoral register. Ask "How would you rate the new pool?" and follow up the people who do not reply.
The time t, in minutes, that 53 students take to cycle to school is recorded to the nearest minute. (a) Find how many students there were in total. (b) Complete the grouped table for the classes 4.5 ≤ t < 7.5, 7.5 ≤ t < 10.5 and 10.5 ≤ t < 13.5. (c) A student's time is recorded as 9 minutes. Between which values could the actual time lie?
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\(2 + 5 + 9 + 12 + 10 + 7 + 4 + 3 + 1 = 53\)
Add every frequency to find the number of students.
4.5 ≤ t < 7.5: 2 + 5 + 9 = 16; 7.5 ≤ t < 10.5: 12 + 10 + 7 = 29; 10.5 ≤ t < 13.5: 4 + 3 + 1 = 8
Time is continuous. A recorded 5, 6 or 7 minutes lies in 4.5 ≤ t < 7.5. The boundaries sit halfway between recorded values.
\(16+29+8=53\)
The grouped frequencies must still add to the total.
8.5 ≤ t < 9.5
Rounded to the nearest minute: half a minute either side of 9.
The numbers of text messages sent by a student on eight days were: 14, 9, 17, 12, 9, 20, 15, 9. Find the mean, the median and the mode.
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\(\bar{x}=\frac{14+9+17+12+9+20+15+9}{8}=\frac{105}{8}=13.125\)
Add the values and divide by how many there are.
9, 9, 9, 12, 14, 15, 17, 20
The median needs the values in order.
\(\frac{12+14}{2}=13\)
There are 8 values, so the median is halfway between the 4th and 5th.
9
9 appears 3 times, more than any other value.
The mean of the six numbers 3, 5, a, 2a, 9 and a + 4 is 8. Find the value of a.
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\(\frac{3+5+a+2a+9+a+4}{6}=8\)
Mean = sum ÷ number of values. There are 6 values.
\(\frac{21+4a}{6}=8\)
Collect the numbers (3 + 5 + 9 + 4 = 21) and the a terms (a + 2a + a = 4a).
\(21+4a=48,\ 4a=27,\ a=27/4\)
Multiply both sides by 6, then solve.
Class P has 18 students with a mean mark of 64. Class Q has 12 students with a mean mark of 74. (a) Find the mean mark of all 30 students. (b) A mark in Class P was recorded as 56 but should have been 65. Find the correct mean mark of Class P.
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\(18\times 64=1152\) and \(12\times 74=888\)
Mean × number of values gives the total of the values.
\(\frac{1152+888}{18+12}=\frac{2040}{30}=68\)
Divide the combined total by the combined number of students.
\(1152-56+65=1161\)
Take out the wrong mark and put in the right one. The total rises by 9.
\(\frac{1161}{18}=64.5\)
The number of students has not changed, so divide by 18.
The table shows the number of goals scored by a team in 24 matches. The value y is missing. The mean number of goals is 3. (a) Find y. (b) How many matches had more than 2 goals?
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\(6+9+5+4=24\)
The problem says 24 matches, so check the table adds to 24.
\(\frac{1\times6+2\times9+4\times5+4y}{24}=3\)
Mean = Σfx ÷ Σf. Multiply each value by its frequency.
\(44+4y=72,\ 4y=28,\ y=7\)
Multiply both sides by 24, then solve.
\(5+4=9\)
The values above 2 are 4 and y = 7, both greater than 2. Their frequencies are 5 and 4.
The heights of 60 sunflower seedlings, in cm, are in the table. (a) How many seedlings are shorter than 14.5 cm? (b) State the modal group. (c) Estimate the mean height. (d) Explain why your answer to (c) is only an estimate.
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\(6+14=20\)
The first two groups end at 14.5.
14.5 ≤ h < 16.5
The highest frequency is 21. For grouped data, the mode is replaced by the modal group.
Midpoints: 11.5, 13.5, 15.5, 17.5, 19.5
Midpoint = (lower + upper) ÷ 2. Use it to stand for every value in the group.
\(\frac{11.5\times6+13.5\times14+15.5\times21+17.5\times12+19.5\times7}{60}=15.5\)
Multiply each midpoint by its frequency, add, then divide by the total frequency.
The actual values are not known
We assumed every value is at its group's midpoint. The real values are spread across each group.
The lengths of 50 phone calls to a helpline, in minutes, are grouped in the table. Estimate the mean length of a call.
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2.5, 7.5, 15, 30
The classes are not the same width, so find each midpoint separately.
\(12\times2.5+18\times7.5+15\times15+5\times30=540\)
Multiply each midpoint by its frequency.
\(\frac{540}{50}=10.8\)
Divide by the total frequency, 50.
The table shows the number of pets owned by each of 40 students. Use the calculator to find the mean and the standard deviation.
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List x: 0, 1, 2, 3, 4, 5; list f: 4, 9, 12, 8, 5, 2
Values in one column, frequencies in the next.
From the calculator: x̄ = 2.18, n = 40
Tell the calculator to use the frequency column.
σx = 1.32
Use σx, the population standard deviation. Do not use sx.
The scores of 12 students in a test are: 58, 71, 64, 82, 75, 71, 90, 66, 71, 79, 85, 69. Use the calculator to find the mean, the median and the mode.
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Type the 12 scores into one column
Check the count is 12.
From the calculator: x̄ = 73.4, n = 12
The mean is the first value shown.
Med = 71
The calculator puts the list in order for you.
Mode = 71
The calculator may not show the mode. Look for the value that appears most often.