# Standard Deviation Worksheet - Page 3

Standard Deviation Worksheet
• Page 3
21.
If $b$ = $a$ + 5, $c$ = $b$ + 5, then find the standard deviation of the set of numbers $a$, $b$, and $c$.
 a. 4.1 b. 2.5 c. 2.2 d. $a$ + 5

#### Solution:

Mean x = 1n i=1n xi

= 13 [a + b + c]

= 13 [a + a + 5 + b + 5]
[Substitute; b = a + 5, c = b + 5.]

= 13 [ a + a + 5 + a + 5 + 5]

= 13(3 a + 15) = a + 5

Variance σ2 = 1n i=1n (xi - x)2

= (13)[(a-(a+5))2 +(b-(a+5))2 +(c-(a+5))2]

= (13)[(a-a-5)2 +(a+5-a-5)2 +(a+5+5-a-5)2]
[Substitute; b = a + 5, c = b + 5 = a + 5 + 5.]

= (13)[25+ 0 + 25]

= 503 » 4.1

22.
If $a$ = $b$ - 2 and $c$ = $b$ + 2, then find the mean and variance of the set of numbers $a$, $b$ and $c$.
 a. Mean = 3$b$; variance = 2.7 b. Mean = 0; variance = 1.4 c. Mean = 0; variance = 2 d. Mean = $b$; variance = 2.7

#### Solution:

Mean x = 1n i=1n xi

= 13[ a + b + c]

= [b - 2 + b + b + 2]
[Substitute: a = b - 2; c = b + 2.]

= 13 (3b) = 3b

Variance σ2 = 1n i=1n (xi - x)2

= 13 [(a - b)2 + (b - b)2 + (c - b)2]

= 13 [(b - 2 - b)2 + 0 + (b + 2 - b)2

= 13 [4 + 0 + 4]

= 13 (8) » 2.7

23.
The statement "If each value in a set of data is multiplied by a constant '$a$' and is decreased by a constant '$b$', then the mean of the new set of data is equal to '$a$' times the mean of the original set of data decreased by a constant '$b$' ", is _______.Use following Rules wherever required.

 a. False b. True

#### Solution:

Let the data be x1, x2, x3... xn

Mean x = 1n i=1n xi
[Mean of the data.]

Now set of data ax1 - b, a x2 - b, ... ,axn - b.
[Each value is multiplied by a constant Ã¢â‚¬ËœaÃ¢â‚¬â„¢ and decreased by a constant Ã¢â‚¬ËœbÃ¢â‚¬â„¢.]

X = 1n i=1n (axi - b)
[Mean of the new set of data.]

= 1n[ i=1naxi - i=1nb ]

= 1n[ a i=1nxi - nb ]
[Use: i=1naxi = a i=1nxi and i=1nb = nb;]

= a 1n i=1nxi - b

= ax - b
[Substitute x = 1ni=1nxi, x is the mean of the original set of data.]

Therefore, mean of the new set of data is Ã¢â‚¬ËœaÃ¢â‚¬â„¢ times the mean of the original set of data decreased by a constant Ã¢â‚¬ËœbÃ¢â‚¬â„¢

So, the statement it True.