The Mean of Grouped Data

Estimating the mean of grouped data from the mid-values of the classes, from a table or a histogram; why the answer is an estimate; a missing frequency; and the modal class beside the mean.

What You Should Already Know

Lesson 12.4: grouped data are counted in classes such as $10 \le t < 20$ (10 included, 20 not); the modal class is the class with the highest frequency.
Lesson 12.7: the mean of a frequency table is $\bar{x} = \dfrac{\Sigma fx}{\Sigma f}$ .

Why the Mean Can Only Be Estimated

A grouped frequency table tells us how many values lie in each class, but not what they are. Six values in the class $10 \le t < 20$ could be 10, 11, 11, 12, 13 and 19, or six values of 18. Without the values we cannot find their sum, so we cannot find the exact mean.

We estimate it instead, by taking every value in a class to be at the middle of the class, its mid-value.

The estimated mean of grouped data

Mid-value of a class $=$ (lower boundary $+$ upper boundary) $\div$ 2

Estimated mean $= \dfrac{\Sigma fx}{\Sigma f}$ , where $f$ is the frequency of each class and $x$ its mid-value.

The answer is an estimate, and the working should say so.

Example 1: Mid-values

Find the mid-value of each class. (a) $10 \le t < 20$ (b) $0 \le m < 5$ (c) $150 \le h < 170$ (d) $15 < t \le 20$

Solution

(a) $\dfrac{10 + 20}{2} = 15$ (b) $\dfrac{0 + 5}{2} = 2.5$ (c) $\dfrac{150 + 170}{2} = 160$

(d) Some books write classes the other way round, with the upper boundary included: $15 < t \le 20$ holds 20 but not 15. The mid-value is found in the same way: $\dfrac{15 + 20}{2} = 17.5$ .

(a) 15 (b) 2.5 (c) 160 (d) 17.5

Estimating the Mean from a Table

Add two columns to the table: the mid-value $x$ of each class, and $fx$ . The column of frequencies is headed $f$ . To keep the data and the mid-value apart, the data are named by another letter, such as $t$ for a time; $x$ is kept for the mid-value.

Example 2: An estimated mean and the modal class

The table shows the time, $t$ minutes, that 50 pupils took to travel to school. (a) Estimate the mean time. (b) State the modal class.

Time, $t$ min	Frequency
$0 \le t < 10$	$6$
$10 \le t < 20$	$15$
$20 \le t < 30$	$17$
$30 \le t < 40$	$8$
$40 \le t < 50$	$4$
Total	$50$

Solution

(a) The mid-values are $\dfrac{0 + 10}{2} = 5$ , then 15, 25, 35 and 45.

Time, $t$ min	$f$	$x$	$fx$
$0 \le t < 10$	$6$	$5$	$5 \times 6 = 30$
$10 \le t < 20$	$15$	$15$	$15 \times 15 = 225$
$20 \le t < 30$	$17$	$25$	$25 \times 17 = 425$
$30 \le t < 40$	$8$	$35$	$35 \times 8 = 280$
$40 \le t < 50$	$4$	$45$	$45 \times 4 = 180$
Total	$50$		$1140$

$\Sigma f = 50$ and $\Sigma fx = 30 + 225 + 425 + 280 + 180 = 1140\text{.}$

Estimated mean $= \dfrac{\Sigma fx}{\Sigma f} = \dfrac{1140}{50} = 22.8$ minutes.

(b) The highest frequency is 17, in $20 \le t < 30$ . That is the most common class, but 17 of 50 pupils is not most of them.

(a) The mean time is about 22.8 minutes (an estimate). (b) $20 \le t < 30$

Example 3: The estimate and the true mean

The stem-and-leaf diagram shows the times, in seconds, that 20 pupils took to run 200 m. (a) Find the mean time. (b) Group the times into the classes $25 \le t < 30$ , $30 \le t < 35$ , …, $45 \le t < 50$ and estimate the mean from the table. (c) Compare the two answers.

Solution

(a) Sum of the 20 times $= 735$ (row by row: $57 + 420 + 258$ ). Mean $= 735 \div 20 = 36.75$ s.

(b) 30, 35, 40 and 45 each start a class.

Time, $t$ s	$f$	$x$	$fx$
$25 \le t < 30$	$2$	$27.5$	$27.5 \times 2 = 55$
$30 \le t < 35$	$5$	$32.5$	$32.5 \times 5 = 162.5$
$35 \le t < 40$	$7$	$37.5$	$37.5 \times 7 = 262.5$
$40 \le t < 45$	$4$	$42.5$	$42.5 \times 4 = 170$
$45 \le t < 50$	$2$	$47.5$	$47.5 \times 2 = 95$
Total	$20$		$745$

Estimated mean $= \dfrac{745}{20} = 37.25$ s.

(c) The estimate is 0.5 s more than the true mean. In $25 \le t < 30$ the times are 28 and 29, but the estimate takes both as 27.5; in $35 \le t < 40$ the seven times add up to 257, but the estimate takes them as $7 \times 37.5 = 262.5$ . Such differences partly cancel, so the estimate is close, but it is not exact.

(a) 36.75 s (b) 37.25 s (an estimate) (c) The estimate is close to the true mean but 0.5 s more, because each time was taken as its class's mid-value.

Why only an estimate

The estimate assumes that the values in each class are spread evenly about its mid-value, so that their mean is the mid-value. Real data are rarely spread exactly so. Narrower classes usually give an estimate closer to the true mean; when the raw data are known, find the exact mean from them instead.

Estimating the Mean from a Histogram

Example 4: Reading the frequencies first

The histogram shows how much money, $ $a$ , each of 60 pupils spent at a school book fair. (a) Estimate the mean amount spent. (b) State the modal class.

Solution

(a) Read each bar's height from the frequency axis: 8, 18, 20, 10, 4. The classes are 20 dollars wide, so the mid-values are 10, 30, 50, 70 and 90.

Amount, $ $a$	$f$	$x$	$fx$
$0 \le a < 20$	$8$	$10$	$10 \times 8 = 80$
$20 \le a < 40$	$18$	$30$	$30 \times 18 = 540$
$40 \le a < 60$	$20$	$50$	$50 \times 20 = 1000$
$60 \le a < 80$	$10$	$70$	$70 \times 10 = 700$
$80 \le a < 100$	$4$	$90$	$90 \times 4 = 360$
Total	$60$		$2680$

Estimated mean $= \dfrac{2680}{60} = 44.666\ldots\text{,}$ which is $44.67 to the nearest cent.

(b) The tallest bar is $40 \le a < 60$ , with 20 pupils, a third of them.

(a) About $44.67 (b) $40 \le a < 60$

A Missing Frequency

Example 5: Finding $k$ from an estimated mean

The table shows the number of hours, $h$ , that some pupils spent exercising last week. An estimate of the mean is 7.6 hours. Find $k$ .

Hours, $h$	Frequency
$0 \le h < 4$	$6$
$4 \le h < 8$	$10$
$8 \le h < 12$	$k$
$12 \le h < 16$	$4$

Solution

The mid-values are 2, 6, 10 and 14.

$\Sigma f = 6 + 10 + k + 4 = 20 + k\text{.}$

$\Sigma fx = 2 \times 6 + 6 \times 10 + 10k + 14 \times 4 = 12 + 60 + 10k + 56 = 128 + 10k\text{.}$

$\dfrac{128 + 10k}{20 + k} = 7.6\text{,}$ so $128 + 10k = 152 + 7.6k\text{.}$

$2.4k = 24$ , so $k = 10$ .

Check: $\Sigma f = 30$ , $\Sigma fx = 228$ , and $228 \div 30 = 7.6$ .

$k = 10$

Spotting Errors

Example 6: Three wrong estimates

Three pupils estimated the mean travel time from the table of Example 2 (frequencies 6, 15, 17, 8, 4 in classes of 10 minutes from 0 to 50). Find each one's mistake. (a) Ben used 0, 10, 20, 30, 40 as $x$ and got 17.8 minutes. (b) Chloe used 10, the class width, as $x$ for every class and got 10 minutes. (c) Dev divided 1140 by 5 and got 228 minutes.

Solution

(a) Ben used the lower boundaries: $\Sigma fx = 0 + 150 + 340 + 240 + 160 = 890\text{,}$ and $890 \div 50 = 17.8$ . This takes every time as the smallest it could be, so the estimate is too low. Use the mid-values.

(b) The class width is the same for every class and says nothing about where the class lies. Chloe's answer would be 10 whatever the times were.

(c) 5 is the number of classes, not the number of pupils. Divide by $\Sigma f = 50$ . A mean of 228 minutes is far above the largest class, which ends at 50.

The estimate uses the mid-values as $x$ and divides by $\Sigma f$ : 22.8 minutes.

Common Mistakes

Using the lower or upper boundary, or the class width, as $x$ instead of the mid-value.
Dividing $\Sigma fx$ by the number of classes instead of by $\Sigma f$ .
Calling the answer the mean. From grouped data it is an estimate of the mean.
Using one letter for both the data and the mid-value.
Saying that most of the data lie in the modal class. It is only the class with the highest frequency.
Reading the frequencies of a histogram wrongly: read each bar's height against the frequency axis before working out $fx$ .

Practice Questions

Try each question first, then open the solution to check your working. Give an answer that is not exact to 3 significant figures, and money to the nearest cent.

Question 1

Find the mid-value of each class. (a) $30 \le m < 40$ (b) $1.5 \le d < 2.0$ (c) $40 < t \le 60$ (d) $0 \le h < 25$

ShowHide solution

(a) $\dfrac{30 + 40}{2} = 35$ (b) $\dfrac{1.5 + 2.0}{2} = 1.75$ (c) $\dfrac{40 + 60}{2} = 50$ (d) $\dfrac{0 + 25}{2} = 12.5$

(a) 35 (b) 1.75 (c) 50 (d) 12.5

Question 2

The table shows the masses, $m$ g, of 40 eggs. (a) Estimate the mean mass. (b) State the modal class.

Mass, $m$ g	Frequency
$40 \le m < 45$	$3$
$45 \le m < 50$	$8$
$50 \le m < 55$	$14$
$55 \le m < 60$	$11$
$60 \le m < 65$	$4$
Total	$40$

ShowHide solution

(a)

Mass, $m$ g	$f$	$x$	$fx$
$40 \le m < 45$	$3$	$42.5$	$42.5 \times 3 = 127.5$
$45 \le m < 50$	$8$	$47.5$	$47.5 \times 8 = 380$
$50 \le m < 55$	$14$	$52.5$	$52.5 \times 14 = 735$
$55 \le m < 60$	$11$	$57.5$	$57.5 \times 11 = 632.5$
$60 \le m < 65$	$4$	$62.5$	$62.5 \times 4 = 250$
Total	$40$		$2125$

Estimated mean $= \dfrac{2125}{40} = 53.125 \approx 53.1$ g (to 3 significant figures).

(b) The highest frequency is 14, in $50 \le m < 55$ .

(a) About 53.1 g (b) $50 \le m < 55$

Question 3

The table shows the time, $t$ minutes, that 30 people took to complete a walk. The classes include their upper boundaries. Estimate the mean time.

Time, $t$ min	Frequency
$60 < t \le 80$	$4$
$80 < t \le 100$	$9$
$100 < t \le 120$	$12$
$120 < t \le 140$	$5$
Total	$30$

ShowHide solution

The mid-values are found in the usual way: $\dfrac{60 + 80}{2} = 70$ , then 90, 110 and 130.

Time, $t$ min	$f$	$x$	$fx$
$60 < t \le 80$	$4$	$70$	$70 \times 4 = 280$
$80 < t \le 100$	$9$	$90$	$90 \times 9 = 810$
$100 < t \le 120$	$12$	$110$	$110 \times 12 = 1320$
$120 < t \le 140$	$5$	$130$	$130 \times 5 = 650$
Total	$30$		$3060$

Estimated mean $= \dfrac{3060}{30} = 102$ minutes.

About 102 minutes.

Question 4

The histogram shows the distances, $d$ km, that 50 cyclists rode at East Coast Park on a Sunday. (a) Estimate the mean distance. (b) State the modal class. (c) Is the estimated mean in the modal class?

ShowHide solution

(a) The frequencies are 7, 16, 14, 9, 4, and the mid-values 5, 15, 25, 35, 45.

Distance, $d$ km	$f$	$x$	$fx$
$0 \le d < 10$	$7$	$5$	$5 \times 7 = 35$
$10 \le d < 20$	$16$	$15$	$15 \times 16 = 240$
$20 \le d < 30$	$14$	$25$	$25 \times 14 = 350$
$30 \le d < 40$	$9$	$35$	$35 \times 9 = 315$
$40 \le d < 50$	$4$	$45$	$45 \times 4 = 180$
Total	$50$		$1120$

Estimated mean $= \dfrac{1120}{50} = 22.4$ km.

(b) The tallest bar is $10 \le d < 20$ , with 16 cyclists.

(c) No. 22.4 km is in $20 \le d < 30$ . The mean uses every class, and the many cyclists who rode more than 20 km pull it above the modal class.

(a) About 22.4 km (b) $10 \le d < 20$ (c) No: it lies in $20 \le d < 30$ .

Question 5

The table shows the ages, $a$ years, of the members of a residents' committee. An estimate of their mean age is 34 years. Find $p$ .

Age, $a$ years	Frequency
$10 \le a < 20$	$4$
$20 \le a < 30$	$9$
$30 \le a < 40$	$p$
$40 \le a < 50$	$8$
$50 \le a < 60$	$3$

ShowHide solution

Mid-values: 15, 25, 35, 45, 55.

$\Sigma f = 4 + 9 + p + 8 + 3 = 24 + p\text{.}$

$\Sigma fx = 60 + 225 + 35p + 360 + 165 = 810 + 35p\text{.}$

$\dfrac{810 + 35p}{24 + p} = 34\text{,}$ so $810 + 35p = 816 + 34p\text{,}$ and $p = 6$ .

Check: $\Sigma f = 30$ , $\Sigma fx = 1020$ , $1020 \div 30 = 34$ .

$p = 6$

Question 6

In a grouped table, the class $20 \le t < 30$ has frequency 5. The five values in it were really 21, 22, 22, 23 and 24. (a) What does the estimate of the mean take their sum to be? (b) What is their real sum? (c) For this class, is the estimate too high or too low?

ShowHide solution

(a) Each value is taken as the mid-value 25: $5 \times 25 = 125$ .

(b) $21 + 22 + 22 + 23 + 24 = 112\text{.}$

(a) 125 (b) 112 (c) Too high.

Question 7

The table shows the time, $t$ minutes, that 25 pupils spent on homework one evening. (a) Sam used 0, 30, 60, 90 as $x$ and got 48 minutes. Tina found $\Sigma fx = 1575$ and divided by 4. What did each do wrong? (b) Estimate the mean correctly. (c) A newsletter says, "Most pupils spent 60 to 90 minutes on homework." Is this right?

Time, $t$ min	Frequency
$0 \le t < 30$	$3$
$30 \le t < 60$	$8$
$60 \le t < 90$	$10$
$90 \le t < 120$	$4$
Total	$25$

ShowHide solution

(a) Sam used the lower boundaries, which makes the estimate too low ( $0 + 240 + 600 + 360 = 1200\text{,}$ and $1200 \div 25 = 48$ ). Tina divided by the number of classes, 4, getting 393.75 minutes, more than any pupil could have spent; she should divide by 25.

(b)

Time, $t$ min	$f$	$x$	$fx$
$0 \le t < 30$	$3$	$15$	$15 \times 3 = 45$
$30 \le t < 60$	$8$	$45$	$45 \times 8 = 360$
$60 \le t < 90$	$10$	$75$	$75 \times 10 = 750$
$90 \le t < 120$	$4$	$105$	$105 \times 4 = 420$
Total	$25$		$1575$

Estimated mean $= \dfrac{1575}{25} = 63$ minutes.

(a) Sam used lower boundaries; Tina divided by the number of classes. (b) About 63 minutes (c) No: 10 of 25 is not most.

Question 8

The lengths, in minutes, of 16 phone calls were: 12, 3, 17, 8, 22, 5, 14, 9, 2, 15, 11, 27, 6, 18, 12, 7. (a) Find the mean length. (b) Group the lengths into the classes $0 \le t < 5$ , $5 \le t < 10$ , …, $25 \le t < 30$ and estimate the mean. (c) By how much does the estimate differ from the mean?

ShowHide solution

(a) Sum $= 188$ , so the mean $= 188 \div 16 = 11.75$ minutes.

(b) 5, 15 start their classes.

Length, $t$ min	$f$	$x$	$fx$
$0 \le t < 5$	$2$	$2.5$	$2.5 \times 2 = 5$
$5 \le t < 10$	$5$	$7.5$	$7.5 \times 5 = 37.5$
$10 \le t < 15$	$4$	$12.5$	$12.5 \times 4 = 50$
$15 \le t < 20$	$3$	$17.5$	$17.5 \times 3 = 52.5$
$20 \le t < 25$	$1$	$22.5$	$22.5 \times 1 = 22.5$
$25 \le t < 30$	$1$	$27.5$	$27.5 \times 1 = 27.5$
Total	$16$		$195$

Estimated mean $= \dfrac{195}{16} = 12.1875 \approx 12.2$ minutes (to 3 significant figures).

(a) 11.75 minutes (b) About 12.2 minutes (c) About 0.44 minutes more.

Key Points

Grouped data do not show the single values, so their mean can only be estimated.
Mid-value $=$ (lower boundary $+$ upper boundary) $\div$ 2, for $a \le x < b$ and $a < x \le b$ alike.
Estimated mean $= \dfrac{\Sigma fx}{\Sigma f}$ , with $x$ the mid-value of each class; say that it is an estimate.
The estimate takes every value of a class as its mid-value; narrower classes usually give a closer estimate.
From a histogram, read the frequencies from the bar heights first.
A missing frequency is found by writing $\dfrac{\Sigma fx}{\Sigma f}$ equal to the given estimate and solving.
The modal class is the class with the highest frequency, not where most of the data lie; the estimated mean need not lie in it.

What You Should Already Know

Why the Mean Can Only Be Estimated

The estimated mean of grouped data

Example 1: Mid-values

Estimating the Mean from a Table

Example 2: An estimated mean and the modal class

Example 3: The estimate and the true mean

Why only an estimate

Estimating the Mean from a Histogram

Example 4: Reading the frequencies first

A Missing Frequency

Example 5: Finding kkk from an estimated mean

Spotting Errors

Example 6: Three wrong estimates

Common Mistakes

Practice Questions

Question 1

Question 2

Question 3

Question 4

Question 5

Question 6

Question 7

Question 8

Key Points

Example 5: Finding $k$ from an estimated mean