Question

Each of the following data sets has a mean of x = 10. (i) 8 9 10 11 12 (ii) 7 9 10 11 13 (iii) 7 8 10 12 13 (a) Without doing any computations, order the data sets according to increasing value of standard deviations. (i), (iii), (ii) (ii), (i), (iii) (iii), (i), (ii) (iii), (ii), (i) (i), (ii), (iii) (ii), (iii), (i) (b) Why do you expect the difference in standard deviations between data sets (i) and (ii) to be greater than the difference in standard deviations between data sets (ii) and (iii)? Hint: Consider how much the data in the respective sets differ from the mean. The data change between data sets (i) and (ii) increased the squared difference ÎŁ(x - x)2 by more than data sets (ii) and (iii). The data change between data sets (ii) and (iii) increased the squared difference ÎŁ(x - x)2 by more than data sets (i) and (ii). The data change between data sets (i) and (ii) decreased the squared difference ÎŁ(x - x)2 by more than data sets (ii) and (iii). none of the above

Answer 1

Answer:

Step-by-step explanation:

Given are 3 data sets with values as:

(i) 8 9 10 11 12   ... Mean =10

(ii) 7 9 10 11 13    ... Mean =10

(iii) 7 8 10 12 13   ... Mean =10

We see that  data set shows mean deviations as

(i) -2 -1 0 1 2  

(ii) -3 -1 0 1 3    

(iii) -3 -2 0 2 3  

Since variance is the square of std deviation, we find that std deviation is larger when variance is larger.

Variance is the sum of squares of (x-mean).  Whenever x-mean increases variance increases and also std deviation.

Hence we find that without calculations also (i) has least std dev followed by (ii) and then (iii)

(i) (ii) (iii) is the order.

b) Between (i) and (ii) we find that 3 entries are the same and 2 entries differ thus increasing square by 9-4 twice.  But between (ii) and (iii) we find that

increase in square value would be 4-1 twice. Obviously the latter is less.