SOLUTION: For the scores 60, 49, 58, 45, 55, 62, 49 find the following: a.) mean - 54 b.) median - 55 c.) mode - 49 d.) range- 9 e.) interquartile range- 11 f.) standard deviation- (t

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Question 257452: For the scores 60, 49, 58, 45, 55, 62, 49 find the following:
a.) mean - 54
b.) median - 55
c.) mode - 49
d.) range- 9
e.) interquartile range- 11
f.) standard deviation-
(that i am stuck on) I hope I did the rest correct. Could you please explain how to get that standard deviation?
Answer by Edwin McCravy(20056)   (Show Source): You can put this solution on YOUR website!

List them in a column, like this

60
49
58 
45 
55 
62 
49

Subtract the mean, 54, from each one in the next column, like this

60   60 - 54 =  6
49   49 - 54 = -5
58   58 - 54 =  4 
45   45 - 54 = -9 
55   55 - 54 =  1
62   62 - 54 =  8 
49   49 - 54 = -5


Square these results in the next column

60   60 - 54 =  6   (6)2 = 36
49   49 - 54 = -5  (-5)2 = 25  
58   58 - 54 =  4   (4)2 = 16 
45   45 - 54 = -9  (-9)2 = 81  
55   55 - 54 =  1   (1)2 =  1
62   62 - 54 =  8   (8)2 = 64 
49   49 - 54 = -5  (-5)2 = 25

Add those squares up, like this

60   60 - 54 =   6   (6)2 = 36
49   49 - 54 =  -5  (-5)2 = 25  
58   58 - 54 =   4   (4)2 = 16 
45   45 - 54 = -10  (-9)2 = 81  
55   55 - 54 =   1   (1)2 =  1
62   62 - 54 =   8   (8)2 = 64 
49   49 - 54 =  -5  (-5)2 = 25
                         -----
                           248

There are 7 numbers.  Subtract 1, getting 6

Divide 248 by 6, getting 41.33333333

Take the square root of 41.33333333, getting 6.429100507

That's the standard deviation

Edwin
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