Table 1; Data Summary

∑f=80

Calculations

The sample arithmetic mean

x ̅=( ∑▒〖(MI)×f〗)/(∑▒f) =1535/80=19.19

The median

x ̅ =〖 L〗_(1 )+( (N/2-∑▒〖fmed )×c〗)/fmed =15+((40-27))/((22))×5 = 15 + 2.95 = 17.95

〖 L〗_(1 )=Lower class limit of the median

N=Total Frequency

∑▒█(fmed )=Sum of the frequencies of classes below the median class@) fmed =Frequency of the median class

c=Width of the median class

The mode

(X ) ̂=〖 L〗_(1 )+ (f_(0-f_1 )/(2 f_0-f_(1 - ) f_2 ))×h =15+(22-19)/(2(22)-19-13) ×5

=15+3/10×5

=16.25

Where

L_(1 )= Lower boundary of class containing mode

f_(0= )Frequency of modal class

f_(1 )= Frequency of class preceding modal class

f_( 2)= Frequency of class after modal class

h= Class width

The inter-quartile range

X_(25=) 10+((20-8) )/((27-8) ) × 5

=10+12/19 ×5

=13.16

X_(50=) 15+((40-27) )/((49-27) )×5

=15+(13 )/(22 )×5

=17.95

X_(75=) 20+((60-49) )/((62-49) )×5

=20+((11) )/((13) )×5

=24.23

Interquartile range = X_(75 ) - X_25= 24.23-13.16=11.07

Lower quartile is define as X_(25 )and upper quartileX_75.

The standard deviation

S= √( (〖(MI)〗^2×f)/(n-1)-x^(-2)=√(34400/79)) -〖19〗^2

=√(435.443-361)

=√74.443

=8.62

Question 2 Part (b)

REPORT ON THE SUMMURISED DATA

1.0 Introduction

The Board of management of a not-for-profit organisation has commissioned a study to determine the length of time (in months) that clients using it counselling services have been unemployed. 2.0 Methodology

A random sample consisting of 80 service recipients were selected for the study and each of the recipients was asked how many months they have been unemployed. The response received were tabulated and analysed using measures of central tendency, standard deviation, and interquartile range. The resulting characteristics of the data are summarised in this report. 3.0 Summary of data

. The data collected in the study were analysed and summarisedin table 1. Table 1

Number of participants80

Mean 19.19

Median 17.95

Mode16.25

Standard Deviation8.62

3.0 Interpretation of data

On average the service recipients have been unemployed for 19 months In relation to the eighty recipients the three measures of central tenden indicate slightly different results. The mean and median are greater than the mode and lie on the right of the mode. This indicates a slightly positive skewed distribution as shown in figure 1. As for the median the value indicates that half of the value lie above and the other half below as shown in figure 2. The mode indicates the months that occurs most frequently and the mean is derived by adding up all the frequency and divide by the total number of frequency. The three values have been projected as interpolation.

Whereas for measure of dispersion, interquartile range is the distance between the top of the lower quartile 25th percentile , 50th represent the median and the bottom of the upper quartile 75th percentile to produce the box and whisker plot as shown in figure 4. The interquartile is calculated by minus the maximum value X75 with minimum X25 and the median score is 15.15.

Figure 1: Sample frequency polygon

Figure 2: Sample Ogive

Figure 3: Sample Histogram

Figure 4: Box of whisker plot for number of unemployed individual.The box and whisker plot, left side of the box represent the minimum value and the right sides of the box represent maximum values. The rectangular box represents the interquartile and the middle (X50) represent the median. Another way to measure depression is based on the average difference of the value from the mean is known as the standard deviation. The value that has been attained is 8.41. Approximately 95% of all scores in the sample lie with two standard deviation of the mean, showing that are no evidence of outlier values in the sample.

Therefore...