Help with mean and standard deviation question.

**Jay Jay** · March 28th 2008, 05:46 AM

A summary of 24 observations of x gave the following information:
∑(x-a)=-73.2 and ∑(x-a)2=2115
The mean of these values of x is 8.95.

(i) Find the value of the constant a.
(ii) Find the standard deviation of these values of x.

HELP PLZ!

**mathceleb** · March 28th 2008, 09:41 AM

Originally Posted by Jay Jay

A summary of 24 observations of x gave the following information:
?(x-a)=-73.2 and ?(x-a)2=2115
The mean of these values of x is 8.95.

(i) Find the value of the constant a.
(ii) Find the standard deviation of these values of x.

HELP PLZ!

Since no probability is given, we denote the mean of x as $\mu = \Sigma x/n$

We know n is 24, and $\mu = 8.95$ . Therefore, $\Sigma x = (24)(8.95) = 214.8$

Your problem gives $\Sigma (x - a) = -73.2$

That can be rewritten as $\Sigma x - a = -73.2$ .
$a = 214.8 + 73.2$ <------ $\Sigma x = 214.8$ from above
$a = 288$

I'll take a stab at part (ii) as well.

From above, for your second equation, substitute our value of 288 for a, we get $\Sigma (x - 288)^2 = 2115$

Expanding and simplifying, we get:
$\Sigma x^2 - 576\Sigma x + 82944 = 2115$
$\Sigma x^2 - 576(214.8) + 82944 = 2115$ <------ remember that $\Sigma x = 214.8$ from part i
$\Sigma x^2 = 42895.8$

Now, we know that Variance, denoted as $\sigma ^ 2 = \Sigma (x - \mu)^2$

Plugging in our value for $\mu = 8.95$ and simplifying, we get:

$\sigma^2 = \Sigma x^2 - 17.9\Sigma x + \Sigma 80.1025$
$\sigma^2 = 42895.8 - 17.9(214.8) + 80.1025$ <---- we figured out earlier $\Sigma x$ and $\Sigma x ^2$
$\sigma^2 = 42895.8 - 3844.92 + 80.1025$
$\sigma^2 = 39130.98$

Now calculate standard deviation, denoted as $\sigma$
$\sigma = \sqrt{\sigma^2}$
$\sigma = \sqrt{39130.98}$
$\sigma = 197.8155$

**BabyMilo** · November 18th 2009, 10:53 PM

um... which one is right?

http://www.mathhelpforum.com/math-he...tml#post407749

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