# 1a. The first five points are 1, 2, 3,4 and 5 and the sixth point

Statistics 3

1a.The first five points are 1, 2, 3,4 and 5 and the sixth point usedis 12, which is far away from the original data points. In the firstcase, standard deviation is 1.58 and by adding the sixth point,standard deviation becomes 4.5. By creating the sixth point far awayfrom the original point, which is an outlier, the mean is affected. This is because the extreme points are changed. Hence the standarddeviation is also affected since this is the average distance fromthe centre.

1b.The data points are 5,6,7,8,10,11,11.5 and 12 originally .Thesecond data points are 5, 7, 9 ,10, 11, 12, 13, 14 .The standarddeviation is increased by choosing the new point in such a way thatthe absolute value of the difference between new point and 10 isgreater than 1(√(1+1/8).

2.Standard deviation is measures how spread out the data points are.Since all the data points are the same here, there is no spread andhence standard deviation is zero. The mean of the data is 50 and thedistance from the mean to al data points is zero.

3.The maximum extent of spread is the same, which is 90in all thethree cases since the minimum and maximum values are the same (100and 10). At the same time, the average distance from the centre,which is the median is different and hence standard deviation isdifferent in three cases. Standard deviation increase as the averagedistance from the median increases. Hence it is the largest for thefirst data set.

4.Outliersare defined as data points lying at a distance, which is abnormalfrom the other values in a population`s sample.

Figure1:Scatterplot

Figure1 shows that 47 (state AZ)and 49(state CA) are outliers based on thedefinition of outliers.

5.Thedata points 109 ,108, 106 and 111 are questionable since the maximumofficially recorded temperature is 102 in US cities.