Question

A class of $32$ students was randomly divided into two equal groups, A and B. Each group was taught a new mathematics topic using different methods. Their scores in a common test on that topic are as follows.

Group A

$74$

$29$

$74$

$64$

$30$

$96$

$42$

$70$

$68$

$53$

$85$

$42$

$53$

$66$

$30$

Group B

$69$

$72$

$80$

$32$

$72$

$45$

$27$

$61$

$60$

$45$

$60$

$32$

$55$

$72$

$80$

$66$

Find the mean and median scores of Group A.

Accepted Answer

Given,

Group A
$74$

$29$

$74$

$64$

$30$

$96$

$42$

$70$

$68$

$68$

$53$

$85$

$42$

$53$

$66$

$30$

Group B
$69$

$72$

$80$

$32$

$72$

$45$

$27$

$61$

$60$

$45$

$60$

$32$

$55$

$72$

$80$

$66$

A mean in maths is the average of a data set, found by adding all numbers together and then dividing the sum of the numbers by the number of numbers.

Mean $= \frac{S u m o f o b s e r v a t i o n s}{T o t a l o b s e r v a t i o n s}$

Mean

$= \frac{74 + 74 + 30 + 42 + 68 + 53 + 42 + 66 + 29 + 64 + 96 + 70 + 68 + 85 + 53 + 30}{16} = \frac{944}{16} = 59$

Thus, the mean of the data is $59$

The median is the middle value in a list ordered from smallest to largest.

Arrange the given data in ascending or descending order.

$29, 30, 30, 42, 42, 53, 53, 64, 66, 68, 68, 70, 74, 74, 85, 96$

If the number of observations is even then median

$= \frac{1}{2} [(\frac{n}{2}) t h o b s e r v a t i o n + (\frac{n}{2} + 1) t h o b s e r v a t i o n]$

Here, $n = 16$

Median

$= \frac{1}{2} [(\frac{16}{2}) t h o b s e r v a t i o n s + (\frac{16}{2} + 1) t h o b s e r v a t i o n s] = \frac{1}{2} [8 t h o b s e r v a t i o n s + 9 t h o b s e r v a t i o n s] = \frac{1}{2} [64 + 66] = \frac{1}{2} \times 130 = 65$

Median of the data is $65$

Number of goals	$0$	$1$	$2$	$3$
Number of teams	$5$	$6$	$3$	$y$

Vistor Chow wal keung | RC Yadav Solutions for Chapter: Data Handling, Exercise 6: REVIEW EXERCISE

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