List s and list t each contain 5 positive integers, and for each list the average (arithmetic mean) of the integers in the list is 40. if the integers 30, 40, and 50 are in both lists, is the standard deviation of the integers in list s greater than the standard deviation of the integers in list t? (1) the integer 25 is in list s. (2) the integer 45 is in list t.

Yes, SDS > SDT

Step-by-step explanation:

List S

25, 30, 40, 50, XS

Average list S = 40

So, we could write,

Average list S = 40 = (25 + 30 + 40 + 50 + XS) / 5

Solving for XS

XS = 40 x 5 – 25 – 30 – 40 – 50 = 200 – 145 = 55  

SDs = SD (25, 30, 40, 50, 55) = 12.74

List T

30, 40, 45, 50, XT

Average list T = 40

So, we could write,

Average list T = 40 = (30 + 40 + 45 + 50 + XT) / 5

Solving for XT

XT = 40 x 5 – 30 – 40 – 45 – 50 = 200 – 165 = 35  

SDT = SD (30, 35, 40, 45, 50) = 7.1

Even at first sight SDS > SDT  because 25 is out of the range 30-50, while 45 is within that range.

List S is more spread than list T.

answer: y=4000x+14000

step-by-step explanation: