Question

Find the mean, median, and mode using the data in the frequency tables Tall Mark 4 Frequency 4 4 10I Number of Cups of Coffee 0-3 4-7 Tally Frequency 8 /// 12-15 16 19

Answer 1

1)

Marks(x)	frequency(f)	x*f	Cumulative frequency
4	2	8	2
5	2	10	4
6	4	24	8
7	5	35	13
8	4	32	17
9	2	18	19
10	1	10	20

Here,

Mean

=sum(x*f)/sum(f)

=137/20

=6.85.

Mode

=the value of marks with highest frequency

=7.

Median

Here,we arrange the sample marks values in order,keeping in mind their order.

The raw data becomes,

4 4 5 5 6 6 6 6 7 7 7 7 7 8 8 8 8 9 9 10.

The middle values of these 20 observations are the 10th and 11th values,which are 7 and 7.

So,the median is

=(7+7)/2

=14/2

=7.

2)

Class interval	Class limit	Class value(x)	Frequency(f)	x*f	Cumulative frequency
0-3	0-3.5	1.5	2	3	2
4-7	3.5-7.5	5.5	3	16.5	5
8-11	7.5-11.5	9.5	8	76	13
12-15	11.5-15.5	13.5	3	40.5	16
16-19	15.5-19.5	17.5	2	35	18
			18	171

Mean

=sum(x*f)/sum(f)

=171/18

=9.5.

Median

Here,n/2=9,

The cumulative frequency 13 just exceeds 9.

So,8-11 is the median group.

Estimated Median = L + ( (n/2) − B) × w/G

where:

• L is the lower class boundary of the group containing the median
• n is the total number of values
• B is the cumulative frequency of the groups before the median group
• G is the frequency of the median group
• w is the group width.

For our example:

• L = 8
• n = 18
• B = 5
• G = 8
• w = 4

So,estimated median

=8+4(9-5)/8

=8+2

=10.

Mode

Group with the highest frequency is the modal group.

here,the modal group is 8-11.

Estimated Mode = L + ( f_m − f_m-1)/((f_m − f_m-1) + (f_m − f_m+1) )× w

where:

• L is the lower class boundary of the modal group
• f_m-1 is the frequency of the group before the modal group
• f_m is the frequency of the modal group
• f_m+1 is the frequency of the group after the modal group
• w is the group width.

In this example:

• L = 8
• f_m-1 = 3
• f_m = 8
• f_m+1 = 3
• w = 4

So,estimated mode

=8+(8-3)*4/(16-6)

=8+2

=10.