Measures of dispersion - (16.4-16.6)

Understand the concept of standard deviation for both grouped and ungrouped data sets

  • Theory
  • Examples
    (a) Find the standard deviation of each of the following sets of data.
    (Give your answers correct to 3 significant figures.)
    (i) 3, 5, 7, 9, 11, 13
    (ii) 1, 2, 2, 3, 5

    (b) A fitness centre records the number of visits to the centre from
    some members last week. The following shows the number of visits from
    two groups of members aged 18 - 30 and 31 - 40.

    Members aged 18 - 30
    Number of visits Frequency
    1 13
    2 40
    3 20
    4 48
    5 39
    Members aged 31 - 40
    Number of visits Frequency
    1 38
    2 48
    3 51
    4 18
    5 5
    (i) Find the mean and standard deviation of the number of visits from
    each group of members. (Give your answers correct to 3 sig. fig. if necessary.)
    (ii) Which group of members have a more consistent number of visits to
    the centre? Explain briefly.
  • Solutions
    (a)(i) Mean
    \(\begin{array}{l} = \frac{{3 + 5 + 7 + 9 + 11 + 13}}{6}\\ = 8\end{array}\)
    Standard deviation
    \(\begin{array}{l} = \sqrt {\frac{{{{(3 - 8)}^2} + {{(5 - 8)}^2} + {{(7 - 8)}^2} + {{(9 - 8)}^2} + {{(11 - 8)}^2} + {{(13 - 8)}^2}}}{6}} \\ = \sqrt {\frac{{25 + 9 + 1 + 1 + 9 + 25}}{6}} \\ = \sqrt {\frac{{70}}{6}} \end{array}\)
    \( = \underline{\underline {3.42}} \) (corr. to 3 sig. fig.)

    (ii) Mean
    \(\begin{array}{l} = \frac{{1 + 2 + 2 + 3 + 5}}{5}\\ = 2.6\end{array}\)
    Standard deviation
    \(\begin{array}{l} = \sqrt {\frac{{{{(1 - 2.6)}^2} + {{(2 - 2.6)}^2} + {{(2 - 2.6)}^2} + {{(3 - 2.6)}^2} + {{(5 - 2.6)}^2}}}{5}} \\ = {\sqrt {\frac{{2.56 + 0.36 + 0.36 + 0.16 + 5.76}}{5}} ^{}}\\ = {\sqrt {\frac{{9.2}}{5}} ^{}}\end{array}\)
    \( = \underline{\underline {1.36}} \) (corr. to 3 sig. fig.)


    (b)(i) For members aged 18 - 30,
    mean\( = \underline{\underline {3.375}} \)
    standard deviation\( = \underline{\underline {1.31}} \) (corr. to 3 sig. fig.)
    For members aged 31 - 40,
    mean\( = \underline{\underline {2.4}} \)
    standard deviation\( = \underline{\underline {1.06}} \) (corr. to 3 sig. fig.)

    (ii)∵ standard deviation of the number of visits from members aged 31 - 40
    is smaller than that from members aged 18 - 30
    ∴members aged 31 - 40 have a more consistent number of visits to the centre.

Compare the dispersions of different sets of data using appropriate measures

  • Theory
  • Examples
  • Solutions

Understand the applications of standard deviation to real-life problems involving standard scores and the normal distribution

  • Theory
  • Examples
    (a) The following table shows the means and standard deviations of the time
    taken by Tony to complete swimming 100 m and 400 m.
    Swimming event Mean Standard deviation
    100 m 4.35 min 0.45 min
    400 m 17.5 min 1.5 min
    Yesterday, the time taken by Tony to complete swimming 100 m and
    400 m were 4.7 min and 18.1 min respectively. In which swimming event
    did Tony perform better yesterday?


    (b)The lengths of a batch of strings are normally distributed with a
    mean of 50 mm and a standard deviation of 2 mm. Find the percentage
    of strings with lengths in each of the following ranges.
    (i) Between 48 mm and 50 mm
    (ii) Less than 46 mm
  • Solutions
    (a) Standard score of the time taken by Tony to complete swimming 100 m \( = \frac{{4.7 - 4.35}}{{0.45}}\)
    \( = 0.778\) (corr. to 3 sig. fig.)
    Standard score of the time taken by Tony to complete swimming 400 m
    \(\begin{array}{l} = \frac{{18.1 - 17.5}}{{1.5}}\\ = {0.4^{}}\end{array}\)
    ∵Standard score for the time taken to complete swimming 400 m
    is greater than standard score for the time taken to complete swimming 100 m
    ∴ Tony performed better in the swimming event of 400 m.


    (b)(i)It is given that \(\bar x = 50{\rm{ mm}}\) and \(\sigma = 2{\rm{ mm}}\).
    \(\because \bar x - \sigma = 48{\rm{ mm}}\)
    ∴Percentage of strings with lengths between 48 mm and 50 mm
    =Percentage of data between \(\bar x - \sigma \) and \(\bar x\)
    \( = \underline{\underline {34\% }} \)

    (ii)\(\because \bar x - 2\sigma = 46{\rm{ mm}}\)
    ∴Percentage of strings with lengths less than 46 mm
    =Percentage of data below \(\bar x - 2\sigma \)
    \(\begin{array}{l} = 50\% - 34\% - 13.5\% \\ = {\underline{\underline {2.5\% }} ^{}}\end{array}\)