Final answer:
The initial data set's mean was 10066.67 SSP, and the median was 1800 SSP. After removing the outlier (68000), the new mean was 2825 SSP, and the median remained the same at 1800 SSP. The median is considered more appropriate as a measure for this data set because it is less influenced by the extreme outlier.
Step-by-step explanation:
a) To find the mean of the given data, we add all the data values together and divide by the number of data values.
The given data values in SSP are: 745, 2000, 1500, 68000, 461, 549, 3750, 1800, 4795.
Their sum is 745 + 2000 + 1500 + 68000 + 461 + 549 + 3750 + 1800 + 4795 = 90600.
Since there are 9 households, we divide 90600 by 9 to get the mean, which is 90600 / 9 = 10066.67 SSP.
To find the median, we arrange the data values in ascending order and locate the middle value. If there is an even number of data values, we take the average of the two middle values. The ordered data values are: 461, 549, 745, 1500, 1800, 2000, 3750, 4795, 68000.
The middle value is 1800, hence the median is 1800 SSP.
b) It seems that the value 68000 is an outlier, as it is much larger than the rest of the data. Excluding this value, we get the new sum as 90600 - 68000 = 22600.
The new mean with the outlier removed is 22600 / 8 = 2825 SSP, and the new median without the outlier (since 68000 was the last value) remains 1800 SSP.
Both the mean and median changed, but the mean changed more significantly when the outlier was dropped, indicating that the mean is more sensitive to outliers.
c) For this series, considering the presence of an outlier, the median is more appropriate as a measure of central tendency because it is less affected by extreme values.