Answer with explanation:
Mean of this data set
[tex]=\frac{\text{sum of all the observation}}{\text{Total number of observation}}\\\\=\frac{30+37+37+41+22+69+45+64+51+22+63+31+64+65+25+51+26+48+28+49}{20}\\\\=43.4[/tex]
Arranging the data set in ascending order
22, 22,25, 26, 28,30,31,37,37,41,45,48,49,51,51,63,64,64,65,69
Median =Middle observation ,if number of terms is odd
and if number of terms is even, divide number of terms by ,2 and then follow the procedure
[tex]=\frac{10^{th term}+11^{th term}}{2}\\\\=\frac{41+45}{2}\\\\=43[/tex]
To find the Outlier
[tex]Q_{1}=\frac{28+30}{2}=29,Q_{3}=\frac{51+63}{2}=57[/tex]
Interquartile range = 57 - 29=28
Data values are approximately equally distributed on both sides of median.
That is, 43-22=21, and ,69 -43=26
So,we can say that there will be no outliers.
Graph will be positively skewed.
Mean > Median
So,mean will be better measure of central tendency because, 43.4-22=21.4, and 69-43.4=25.6
Option A: There are no outliers.
And, Option C: The Mean is good measure of central tendency.
