Question

From the following data, compute the mode, median, Standard Deviation and Skewness. Plot the data Polygon curve by Excel sheet. 15-19 20-24 25-29 30-34 40-44 35-39 Class Frequency 9 14 33 30 14 8

solution and the plot by excel please

Answer 1

Since the given Frequency Distribution is a Inclusive Type of Class Interval Series; So, we have to Convert it in to Exclusive type of Class Interval Series to make Calculations.

To convert Inclusive type of Class Interval Series as Exclusive; we use the following formulae.

d LB = LL- 2

d UB=UL + 2

Where LL = Lower Limit

UL = Upper Limit

LB = Lower Boundary

UB = Upper Boundary

d = Differences between the Lower Limit of Present Class Interval and Upper Limit of Previous Class Interval.

Therefore If you Observe the given frequency distribution; d = 1

=0.5

Therefore to get the Exclusive type of Class Interval Series; we have to Subtract 0.5 from each Lower Limit of the class interval and have to Add 0.5 to each Upper Limit of the Class Interval.

Therefore the Renewed Frequency Distribution is

C. I	f
14.5 - 19.5	9
19.5 - 24.5	14
24.5 - 29.5	33
29.5 - 34.5	30
34.5 - 39.5	14
39.5 - 44.5	8
TOTAL	108

(a) MODE (Z):

Z=1+ f - fi 2f - f1 – f2 *C

Where l = Lower Boundary of Modal Class

f = Frequency of the Modal Class ( Highest Frequency )

f1 = Preceeding frequency of the Modal Class

f2 = Succeeding frequency of the Modal Class

C = Length of class interval of the Modal Class ( i.e find the difference betwween the UB and LB of class interval of the Modal Class )

C. I	f
14.5 - 19.5	9
19.5 - 24.5	14   fi
24.5 - 29.5	33	MODAL CLASS
29.5 - 34.5	30 $f_{2}$
34.5 - 39.5	14
39.5 - 44.5	8

$Z = 24.5 + \left \{ \frac{33 - 14}{2*33 - 14 - 30} \right \}*5$

19 Z = 24.5 + * 5 22

$\Rightarrow Z = 24.5 + 4.3182$

$\Rightarrow Z = 28.8182$

****************************************************

(b) MEDIAN (M):

$M = l + \left \{ \frac{\frac{N}{2} -m}{f} \right \}*C$

Where l = Lower Boundary of the Median Class

f = frequency of the Median Class

m = Cumulative frequency upto the Median Class

C = Class Interval

N/2 = Median Item ( Main Part of the Formula )

Working Rule:

To find the Median; first we want to find the Less than Cumulative Frequenc (LCF). Next we want to find N/2 and have to Search this N/2 in LCF. With that We will Declare the MEDIAN CLASS.

C. I	f	LCF
14.5 - 19.5	9	9
19.5 - 24.5	14	23 m
(l) 24.5 - 29.5	33 f	56	MEDIAN CLASS
29.5 - 34.5	30	86
34.5 - 39.5	14	100
39.5 - 44.5	8	108
TOTAL	108
N	108
N/2	54

$\Rightarrow M = 24.5 + \left \{ \frac{54 -23}{33} \right \}*5$

$\Rightarrow M = 24.5 + \left \{ 0.9394 \right \}*5$

$\Rightarrow M = 24.5 + 4.6970$

$\Rightarrow M = 29.1970$

**********************************************************

(c) MEAN and STANDARD DEVIATION:

C. I	f	x		fxx
14.5 - 19.5	9	17	153	2601
19.5 - 24.5	14	22	308	6776
24.5 - 29.5	33	27	891	24057
29.5 - 34.5	30	32	960	30720
34.5 - 39.5	14	37	518	19166
39.5 - 44.5	8	42	336	14112
TOTAL	108		3166	97432
N	108
$\Sigma fx$	3166
MEAN $\left \{ \bar{x} = \frac{\Sigma fx}{N} \right \}$	29.3148
$(\bar{x})^{2}$	859.3584
$\Sigma fx^{2}$	97432
$\frac{\Sigma fx^{2}}{N}$	902.1481
VARIANCE $\left \{ \sigma ^{2} = \frac{\Sigma fx^{2}}{N} -( \bar{x})^{2} \right \}$	42.7898
S.D $\left \{ \sigma = \sqrt{VARIANCE} \right \}$	6.5414

(d) Pearson's Coefficient of Skewness:

$P_{sk} = \frac{\tilde{x} - Z}{\sigma }$

$\Rightarrow P_{sk} = \frac{29.3148 - 28.8182}{6.5414 }$

$\Rightarrow P_{sk} = \frac{0.4966}{6.5414 }$

$\Rightarrow P_{sk} =0.0759$

; So, there exist a Postive (Right Skewed ) and the given distribution is a Positively Skewed Distribution.

(e) Frequency Polygon: