Question

The table below shows the time in minutes that customers

   spent waiting in a bank.

Time (min.)	Frequency
5	7
6-10	8
11-15	9
16-20	4
21-25	6

1. Calculate the mean, mode and median       [4+4 + 4]

(b) Calculate the variance.                                    [5]

(c) Calculate the standard deviation.                  [2]

(d) Calculate the coefficient of skewness           [3]

(e) Calculate the coefficient of variation.

Answer 1

Answer:-

Given that:-

Time(min)	Frequency
(1-5)	7
(6-10)	8
(11-15)	9
(16-20)	4
(21-25)	6

For class intervals are not continuous , first it is to be converted in to continuous class by following method

If d is the gap between the upper limit of any class and lower of the succeding class, the class boundries for any class are then given by:

Upper class boundary = Upper class limit +

Lower class boundary = Lower class limit -

For given data d=1

Continuous classes	Mid values	Frequency	Cumulatative (c.f)
0.5-5.5	3	7	7	21	63
5.5-10.5	8	8	15	64	512
10.5-15.5	13	9	24	117	1521
15.5-20.5	18	4	20	72	1296
20.5-25.5	23	6	34	130	3174
Total	65	34	-	412	6566

(a)(i) Mean

(ii)Mode for cpntinuous frequency distribution

Modewhere l= lower limit of modal class

h= mognitude of modal class

= frequency of modal class

and are frequencyes of the classes precrding and succeding the modal class respectively

Modal class is a class which has maximum frequency.

for given data class(10.5-15.5) has maximum frequency 9 So this become modal class of given data

l = 10.5

h= 5 (which of class interval i.e., 15.5-10.5=5)

Mode

(iii) Median

where

l = is the lower limit of the median class

f= is the frequency of the median class

h = is the magnitude of the median class

'C'= is the c.f(Cumulative frequency ) of the class preceding the median class

and

The median class is the class corresponding to the c.f just greater than

for given data

is median class

l = 10.5

f = 9

N= 34

C= 15

h = 5

Median

(b) Variance=

  

(c) Standard deviation

  

=6.80

(d) Coefficient of skewness

=Mean-mode/Standard deviation

sistribution is positively skewed.

(e) coefficient of variance

  = standard deviation/mean