### Measures of dispersion - (16.4-16.6)

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

• (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.
• (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.

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

• (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
• (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 $$\bar x - 2\sigma$$
$$\begin{array}{l} = 50\% - 34\% - 13.5\% \\ = {\underline{\underline {2.5\% }} ^{}}\end{array}$$