Question

Find the median
Class interval 0-8 8-16 16-24 24-32 32-40 40-48
Frequency 8 10 16 24 15 7

Answer 1

Median =26

Step-by-step explanation:-

SOLUTION:-

The given frequency distribution table is

$\begin{gathered}\boxed{\begin{array}{c|c} \bf \underline{Classes}&\bf \underline{Frequency} \\ \sf{} & \sf{} & \sf{} \\ 0-8&8\\8 - 16&10 \\16 - 24 & 16 \\ 24 - 32 & 24 \\32 - 40 &15 \\ 40 - 48 &7\end{array}}\end{gathered}$

Let's prepare the cumulative frequency table, as given below

$\begin{gathered}\boxed{\begin{array}{c|c|c} \tt{Classes}&\tt{Frequency}&\tt{Cumulative\;frequency} \\ \sf{} & \sf{} & \sf{} \\ 0-8&8&8\\8 - 16&10&18 \\ 16 - 24 & 16&34\\ 24-32 & 24 &58\\ 32 - 40&15&73\\40 - 48 &7 &80 \\ \sf{} & \overline{\;\;\;\;\sf{N = \Sigma f_1=80}\;\;\;\;} & \sf{} \end{array}}\end{gathered}$

Now ,

N=80

=N/2=40

The cumulative frequency just greater than 40 is 58 and the corresponding classes are 24-32

Thus , the median class is 24-32

✦Lower limit (l)=24

✦Height of the class(h)=8

✦Frequency (f)=24

✦Cumulative frequency of preceding class=34

✦N/2=40

Using below formula, Finding median

$\sf\bigstar \boxed{\sf Median, M_e=l+ \bigg\{h× \dfrac{ \frac{N}{2} -cf}{f} \bigg\}}$

$\implies \sf 24 + \bigg \{8 \times \dfrac{(40 - 34}{24} \bigg \}$

$\implies \sf 24 + \bigg \{\cancel{8} \times \dfrac{6}{\cancel{24}\:\:3} \bigg \}$

$\implies \sf 24 + \bigg \{\dfrac{\cancel{6}\:\:2}{\cancel{3}} \bigg \}$

$\implies \sf(24 + 2)$

$\implies \bf26$

Therefore,

•Required Median=26

Answer 2

Step-by-step explanation:

The given frequency distribution table is

\begin{gathered}\begin{gathered}\boxed{\begin{array}{c|c} \bf \underline{Classes}&\bf \underline{Frequency} \\ \sf{} & \sf{} & \sf{} \\ 0-8&8\\8 - 16&10 \\16 - 24 & 16 \\ 24 - 32 & 24 \\32 - 40 &15 \\ 40 - 48 &7\end{array}}\end{gathered}\end{gathered}

Classes

0−8

8−16

16−24

24−32

32−40

40−48

Frequency

8

10

16

24

15

7

Let's prepare the cumulative frequency table, as given below

\begin{gathered}\begin{gathered}\boxed{\begin{array}{c|c|c} \tt{Classes}&\tt{Frequency}&\tt{Cumulative\;frequency} \\ \sf{} & \sf{} & \sf{} \\ 0-8&8&8\\8 - 16&10&18 \\ 16 - 24 & 16&34\\ 24-32 & 24 &58\\ 32 - 40&15&73\\40 - 48 &7 &80 \\ \sf{} & \overline{\;\;\;\;\sf{N = \Sigma f_1=80}\;\;\;\;} & \sf{} \end{array}}\end{gathered}\end{gathered}

Classes

0−8

8−16

16−24

24−32

32−40

40−48

Frequency

8

10

16

24

15

7

N=Σf

1

=80

Cumulativefrequency

8

18

34

58

73

80

N=80

=N/2=40

The cumulative frequency just greater than 40 is 58 and the corresponding classes are 24-32

Thus , the median class is 24-32

✦Lower limit (l)=24

✦Height of the class(h)=8

✦Frequency (f)=24

✦Cumulative frequency of preceding class=34

✦N/2=40

Using below formula, Finding median

\sf\bigstar \boxed{\sf Median, M_e=l+ \bigg\{h× \dfrac{ \frac{N}{2} -cf}{f} \bigg\}}★

Median,M

e

=l+{h×

f

2

N

−cf

}

\implies \sf 24 + \bigg \{8 \times \dfrac{(40 - 34}{24} \bigg \}⟹24+{8×

24

(40−34

}

\implies \sf 24 + \bigg \{\cancel{8} \times \dfrac{6}{\cancel{24}\:\:3} \bigg \}⟹24+{

8

×

24

3

6

}

\implies \sf 24 + \bigg \{\dfrac{\cancel{6}\:\:2}{\cancel{3}} \bigg \}⟹24+{

3

6

2

}

\implies \sf(24 + 2)⟹(24+2)

\implies \bf26⟹26

Therefore,

•Required Median=26

Now ,