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Standard deviationIB MYP Maths Extended: Revision notes

Section 1

What the standard deviation measures

The mean tells you where the centre of a data set is. Two data sets can have the same mean but be very different, so you also need a measure of spread. The standard deviation (written σ\sigma) measures how far, on average, the values are from the mean.

  • A small standard deviation means the values are close together and close to the mean (consistent).
  • A large standard deviation means the values are spread out (less consistent).
  • If all the values are equal, the standard deviation is 00. It is measured in the same units as the data.
Key termsspreadstandard deviation
Common mistake

Thinking a large standard deviation means a large mean. The standard deviation says nothing about the size of the mean.

Section 2

Calculating the standard deviation of a list

In MYP you find the standard deviation using technology (a graphic display calculator or spreadsheet). Enter the values into a list, then choose the one-variable statistics and read the value of σx\sigma_x, the population standard deviation. Example: 4,6,7,9,144,6,7,9,14. The mean is 405=8\frac{40}{5}=8. The calculator gives σ=3.41\sigma=3.41 minutes (to 3 s.f.). What the calculator does: it finds the distance of each value from the mean (−4,−2,−1,1,6-4,-2,-1,1,6), squares them (16,4,1,1,3616,4,1,1,36), finds the mean of the squares (585=11.6\frac{58}{5}=11.6, the variance) and takes the square root.

Key termsvariancepopulation standard deviation
Exam tip

A calculator shows two similar values: σx\sigma_x (divides by nn) and sxs_x (divides by n−1n-1). Use σx\sigma_x unless the question says otherwise.

Section 3

Standard deviation from a frequency table

For a frequency table, enter the values in one list and the frequencies in a second list, then ask for one-variable statistics using the frequencies. Example: 2 students read 00 books, 5 read 11, 6 read 22, 4 read 33 and 3 read 44. The total is n=20n=20. The mean is ∑fxn=0(2)+1(5)+2(6)+3(4)+4(3)20=4120=2.05\frac{\sum fx}{n}=\frac{0(2)+1(5)+2(6)+3(4)+4(3)}{20}=\frac{41}{20}=2.05. Technology gives σ=1.20\sigma=1.20. Check by hand that the frequencies add to the number of students and that the mean lies within the range of the data.

Key termsfrequencyfrequency table
Common mistake

Typing only the values and forgetting the frequency list. The calculator then treats every value as occurring once.

Section 4

Interpreting the standard deviation

Always link the number to the context. Say what the spread means for the situation.

  • A bus company with a small standard deviation in its journey times is reliable: you can plan using the mean.
  • A test with a large standard deviation means students' marks are very different from each other. You can also find the interval one standard deviation either side of the mean, xˉ−σ\bar{x}-\sigma to xˉ+σ\bar{x}+\sigma. For the books example this is 0.850.85 to 3.253.25, so the values 1,2,31,2,3 lie inside it (15 of the 20 students). Adding a value equal to the mean keeps the mean the same but reduces the standard deviation, because the new value has zero distance from the mean.
Key termsconsistentreliable
Exam tip

Write your conclusion in words about the context, for example: 'Company Y is more consistent because its standard deviation is smaller'.

Section 5

Comparing two distributions

To compare two data sets, always make two comparisons: one using an average (the mean) and one using the spread (the standard deviation). Example: Company X and Company Y both have mean 2828 minutes. Company X has σ=5.66\sigma=5.66 and Company Y has σ=1.41\sigma=1.41.

  • Same mean: on average the two take equally long.
  • Smaller σ\sigma for Y: its times are less spread out, so Y is more consistent and predictable. Different means give another comparison: for example, a higher mean score is better in a test, but a lower mean time is better in a race. Finish with a reasoned conclusion that uses both measures.
Key termscomparedistribution
Common mistake

Comparing only the means. A comparison that ignores the standard deviation does not describe the spread.

That's the notes covered.

Carry on to the next subtopic.

Exam questions on Standard deviation

  1. Five students in Singapore record how many minutes it takes them to solve a puzzle: 4, 6, 7, 9, 144,\ 6,\ 7,\ 9,\ 14. Use technology and the population standard deviation σ\sigma where needed.
    A sixth student takes exactly 88 minutes. State the new mean and find the new standard deviation. Say whether the spread has increased or decreased.2 marks
  2. A teacher investigates how the standard deviation changes as values spread out. Each list below has five values and mean 1010. List P is 8,9,10,11,128,9,10,11,12 with standard deviation 1.411.41. List Q is 6,8,10,12,146,8,10,12,14 with standard deviation 2.832.83. List R is 4,7,10,13,164,7,10,13,16 with standard deviation 4.244.24. (All standard deviations are given to 3 significant figures.)
    List T is 5,7.5,10,12.5,155,7.5,10,12.5,15. Use the pattern to predict its standard deviation, then verify your prediction using technology.2 marks
  3. Two courier companies in Nairobi each make five deliveries to the same shop. The delivery times in minutes are: Company X: 20, 24, 28, 32, 3620,\ 24,\ 28,\ 32,\ 36. Company Y: 26, 27, 28, 29, 3026,\ 27,\ 28,\ 29,\ 30.
    Find the mean and the standard deviation of the delivery times for each company.3 marks
See the full worksheet

Written by the Exaim team, led by Shaun Daswani (Head of Upper Secondary, Improve ME Institute; MSc Financial Mathematics, Imperial College London; BSc, UCL) and Jason Daswani (operational lead, Improve ME Institute; LSE).