Name: Mean of Grouped Data: Example
Uploaded: 2019-07-19
Duration: 5 min 16 s
Description: Want to find out how to calculate the mean of grouped data when the class intervals are different? Watch this interesting video and learn more!

Question

The following table shows the ages of the patients admitted in a hospital during a year:

Age (in years)	$5 - 15$	$15 - 25$	$25 - 35$	$35 - 45$	$45 - 55$	$55 - 65$
Number of patients	$6$	$11$	$21$	$23$	$14$	$5$

Find the mode and the mean of the data given above. Compare and interpret the two measures of central tendency.

Accepted Answer

We will use shortcut or assumed mean method.

First of all, we have to find class mark for each interval.

We know that,

$x_{i} = \frac{u p p e r c l a s s l i m i t + low e r c l a s s l i m i t}{2}$

Now, let us take $23$ as assumed mean $(a)$ , we can calculate $d_{i}$ and $f_{i} d_{i}$ as following.

Age (in years) Number of patients $f_{i}$ Class mark $x_{i}$ $d_{i} = x_{i} - 30$ $f_{i} d_{i}$

$5 - 15$ $6$ $10$ $- 20$ $- 120$

$15 - 25$ $11$ $20$ $- 10$ $- 110$

$25 - 35$ $21$ $30$ $0$ $0$

$35 - 45$ $23$ $40$ $10$ $230$

$45 - 55$ $14$ $50$ $20$ $280$

$55 - 65$ $5$ $60$ $30$ $150$

From the table, we may observe that

$\sum_{} f_{i} = 80$

$\sum_{} f_{i} d_{i} = 430$

Mean $\bar{x} = a + \frac{\sum_{} f_{i} d_{i}}{\sum_{} Σ f_{i}}$

$= 30 + (\frac{430}{80})$

$= 30 + 5.375$

$= 35.375$

$≃ 35.38$

Clearly, mean of this data is $35.38$ which means that the average age of a patient admitted to hospital was $35 .38$ years.

As we can see that maximum class frequency is $23$ which belongs to class interval $35 - 45 .$

So, modal class $=$ $35 - 45$

Lower limit ( $l$ ) of modal class $= 35$

Frequency ( $f_{1}$ ) of modal class $= 23$

Class size ( $h$ ) $= 10$

Frequency ( $f_{0}$ ) of class preceeding the modal class $= 21$

Frequency ( $f_{2}$ ) of class succeeding the modal class $= 14$

Now, mode $= l + (\frac{f_{1} - f_{0}}{2 f_{1} - f_{0} - f_{2}}) \times h$

$= 35 + (\frac{23 - 21}{2 (23) - 21 - 14}) \times 10$

$= 35 + \frac{20}{11}$

$= 35 + 1.81$

$= 36.8$

Clearly, mode is $36.8 .$

It represents that maximum number of patients admitted in hospital were of $36.8$ years.

Expenditure (in ₹)	Number of families
$1000 - 1500$	$24$
$1500 - 2000$	$40$
$2000 - 2500$	$33$
$2500 - 3000$	$28$
$3000 - 3500$	$30$
$3500 - 4000$	$22$
$4000 - 4500$	$16$
$4500 - 5000$	$7$

Number of students per teacher	Numbers of states/UT
$15 - 20$	$3$
$20 - 25$	$8$
$25 - 30$	$9$
$30 - 35$	$10$
$35 - 40$	$3$
$40 - 45$	$0$
$45 - 50$	$0$
$50 - 55$	$2$

Number of students per teacher	Number of states/UT
$3000 - 4000$	$4$
$4000 - 5000$	$18$
$5000 - 6000$	$9$
$6000 - 7000$	$7$
$7000 - 8000$	$6$
$8000 - 9000$	$3$
$9000 - 10000$	$1$
$10000 - 11000$	$1$

Number of cars	$0 - 10$	$10 - 20$	$20 - 30$	$30 - 40$	$40 - 50$	$50 - 60$	$60 - 70$	$70 - 80$
Frequency	$7$	$14$	$13$	$12$	$20$	$11$	$15$	$8$

Exercise 14.2

Jharkhand Board Mathematics Solutions for Exercise 14.2

Questions from Exercise 14.2 with Hints & Solutions

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