Measures of centre for data in classes

Statistics and probability, grade 10 (4 units) Lesson 8 of 20Learning

Measures of centre for data in classes

Data in classes

A continuous variable, such as height, takes a different value in almost every item. So the data are sorted into classes of equal width, and the data in each class are counted.

The heights of 14 students in a class, in cm:
148, 153, 155, 158, 162, 163, 163, 167, 169, 174, 177, 178, 185, 188.

Height (cm)	140–150	150–160	160–170	170–180	180–190
Frequency	1	3	5	3	2

The class 150–160 is “from 150 up to less than 160”: a student 160 cm tall is counted in the class 160–170.
The table shows the overall picture, but no longer the data themselves.
In the histogram the bars touch, because each class begins where the one before ends:

The class middle and the estimated mean

From data in classes only an estimated mean can be worked out: assume all the data in a class lie at the class middle, and work it out as from a frequency table.

\displaystyle \bar{x}\approx\frac{m_1f_1+m_2f_2+\dots+m_kf_k}{f_1+f_2+\dots+f_k}

$\displaystyle m_1, \dots, m_k$ – the classes’ middles, $\displaystyle f_1, \dots, f_k$ – their frequencies.

The middle of the class 140–150 is $\displaystyle {\frac{140+150}{2}}={145}$ , and the middles of the other classes are 155, 165, 175 and 185.

\bar{x}\approx\frac{145\cdot1+155\cdot3+165\cdot5+175\cdot3+185\cdot2}{14}=\frac{2330}{14}\approx166.43

The mean of the list and the mean of the classes

From the list of 14 heights the mean is:

\bar{x}=\frac{2340}{14}\approx167.14

The two means are close, but not equal.

In the working from the classes each height was replaced by its class middle: 148 counted as 145, and 169 as 165.
Some data grew in the replacement and some shrank, so the difference is small – but it is 0 only by chance.
When there is only a table in classes, the estimated mean is the best that can be done.

The modal class

The modal class is the class with the highest frequency – the tallest bar of the histogram.

In the heights the modal class is 160–170 cm, with five students.
In the list itself the mode is 163, which is indeed in this class. But the list’s mode depends on chance: two students of exactly the same height. The modal class shows what matters – where the data are crowded.

An open end class

Sometimes the first or the last class has no bound: “less than 20”, “60 and over”. This is an open end class.

The ages of the members of a club:

Age	20–30	30–40	40–50	50 and over
Frequency	7	12	9	5

The class “50 and over” has no middle: perhaps all of them are 52, perhaps 80. So no mean can be worked out, not even an estimate.
The modal class is 30–40 – it can always be found.

Exercise 1

40 employees were asked how many minutes their journey to work takes:

Minutes	0–10	10–20	20–30	30–40
Frequency	6	14	12	8

Work out the estimated mean length of the journey.

/ – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – check and a new line · * – times · ^ – power, ↓ – end of the power

/ – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – a new line · * – times · ^ – power, ↓ – end of the power

What is the modal class?

Exercise 2

At the post office 25 parcels were weighed:

Weight (kg)	0–2	2–4	4–6	6–8
Frequency	4	9	7	5

Work out the estimated mean weight of a parcel.

/ – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – a new line · * – times · ^ – power, ↓ – end of the power

Exercise 3

The histogram shows the noon temperature on 30 days:

Work out the estimated mean temperature.

/ – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – a new line · * – times · ^ – power, ↓ – end of the power

Exercise 4

The marks of 25 students in a test:

Mark	40–60	60–80	80–100
Frequency	3	12	10

Work out the estimated mean mark.

/ – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – a new line · * – times · ^ – power, ↓ – end of the power

The teacher worked out the mean from the list of marks, and got 76.2. Why does her mean differ from the one you worked out?

Exercise 5

The waiting time of 50 callers at a call centre:

Minutes	0–5	5–10	10–15	15 and over
Frequency	18	15	10	7

Can the estimated mean waiting time be worked out? Can the modal class be found? Explain.

Formulas: statistics and probability, grade 10 (4 units)

Open as a page

Presenting data

Relative frequency – $\displaystyle f$ a value’s frequency, $\displaystyle n$ the number of all the data
$\displaystyle \frac{f}{n}$
Relative frequency in percent
$\displaystyle \frac{f}{n}\cdot100\%$
The angle of a slice in a pie chart
$\displaystyle \frac{f}{n}\cdot360^{\circ}$

Measures of centre and position

The mean of $\displaystyle n$ data
$\displaystyle \bar{x}=\frac{x_1+x_2+\dots+x_n}{n}$
The mean from a frequency table – $\displaystyle x_1, \dots, x_k$ the values, $\displaystyle f_1, \dots, f_k$ their frequencies
$\displaystyle \bar{x}=\frac{x_1f_1+x_2f_2+\dots+x_kf_k}{f_1+f_2+\dots+f_k}$
A weighted mean – $\displaystyle w_1, \dots, w_k$ the weights of the values $\displaystyle x_1, \dots, x_k$
$\displaystyle \bar{x}=\frac{w_1x_1+w_2x_2+\dots+w_kx_k}{w_1+w_2+\dots+w_k}$
The mean of data in classes – $\displaystyle m_1, \dots, m_k$ the classes’ middles, $\displaystyle f_1, \dots, f_k$ their frequencies
$\displaystyle \bar{x}\approx\frac{m_1f_1+m_2f_2+\dots+m_kf_k}{f_1+f_2+\dots+f_k}$
The median’s place in an ordered list of $\displaystyle n$ data
$\displaystyle \frac{n+1}{2}$
The lower decile’s place in an ordered list; the upper decile’s – $\displaystyle 0.9\cdot n$
$\displaystyle 0.1\cdot n$

Measures of spread

The range – the greatest value minus the least
$\displaystyle x_{max}-x_{min}$
The variance – the mean of the squared deviations from the mean
$\displaystyle \sigma^2=\frac{(x_1-\bar{x})^2+(x_2-\bar{x})^2+\dots+(x_n-\bar{x})^2}{n}$
The standard deviation – the square root of the variance
$\displaystyle \sigma=\sqrt{\sigma^2}$
The variance from a frequency table
$\displaystyle \sigma^2=\frac{(x_1-\bar{x})^2f_1+\dots+(x_k-\bar{x})^2f_k}{f_1+\dots+f_k}$

The rules of probability

Equally likely outcomes – $\displaystyle k$ the number of outcomes in the event $\displaystyle A$ , $\displaystyle n$ the number of outcomes in the sample space
$\displaystyle P(A)=\frac{k}{n}$
The probability of any event; $\displaystyle {P(\emptyset)}={0}$ , and $\displaystyle {P(U)}={1}$ for the sample space $\displaystyle U$
$\displaystyle 0\le P(A)\le1$
The complement $\displaystyle \overline{A}$ – $\displaystyle A$ does not happen
$\displaystyle P(\overline{A})$ $\displaystyle =$ $\displaystyle 1-P(A)$
The union – $\displaystyle A$ or $\displaystyle B$ (or both); $\displaystyle A\cap B$ – the intersection, both $\displaystyle A$ and $\displaystyle B$
$\displaystyle P(A\cup B)$ $\displaystyle =$ $\displaystyle P(A)+P(B)-P(A\cap B)$
The union of disjoint events
$\displaystyle P(A\cup B)$ $\displaystyle =$ $\displaystyle P(A)+P(B)$ $\displaystyle A\cap B=\emptyset$

Conditional probability and independence

The probability of $\displaystyle A$ given $\displaystyle B$
$\displaystyle P(A|B)=\frac{P(A\cap B)}{P(B)}$
A path in a tree – the product of the probabilities on its branches
$\displaystyle P(A\cap B)$ $\displaystyle =$ $\displaystyle P(B)\cdot P(A|B)$
The law of total probability – a weighted mean of the conditional probabilities
$\displaystyle P(A)=P(B)\cdot P(A|B)+P(\overline{B})\cdot P(A|\overline{B})$
Bayes’ formula – from $\displaystyle P(B|A)$ to $\displaystyle P(A|B)$
$\displaystyle P(A|B)\cdot P(B)$ $\displaystyle =$ $\displaystyle P(B|A)\cdot P(A)$
Independent events
$\displaystyle P(A\cap B)$ $\displaystyle =$ $\displaystyle P(A)\cdot P(B)$
Independent events – knowing that $\displaystyle B$ happened does not change the probability of $\displaystyle A$
$\displaystyle P(A|B)$ $\displaystyle =$ $\displaystyle P(A)$