# 6.3 The central limit theorem for sums

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Suppose $X$ is a random variable with a distribution that may be known or unknown (it can be any distribution) and suppose:

• ${\mu }_{X}=$ the mean of $X$
• ${\sigma }_{X}=$ the standard deviation of $X$
If you draw random samples of size $n$ , then as $n$ increases, the random variable $\mathrm{\Sigma X}$ which consists of sums tends to be normally distributed and

$\Sigma X$ ~ $N\left(n\cdot {\mu }_{X},\sqrt{n}\cdot {\sigma }_{X}\right)$

The Central Limit Theorem for Sums says that if you keep drawing larger and larger samples and taking their sums, the sums form their own normal distribution (the sampling distribution) which approaches a normal distribution as the sample size increases. The normal distribution has a mean equal to the original mean multiplied by the sample size and a standard deviationequal to the original standard deviation multiplied by the square root of the sample size.

The random variable $\Sigma X$ has the following z-score associated with it:

• $\mathrm{\Sigma x}$ is one sum.
• $z=\frac{\Sigma x-n\cdot {\mu }_{X}}{\sqrt{n}\cdot {\sigma }_{X}}$
• $n\cdot {\mu }_{X}=$ the mean of $\mathrm{\Sigma X}$
• $\sqrt{n}\cdot {\sigma }_{X}=$ standard deviation of $\mathrm{\Sigma X}$

An unknown distribution has a mean of 90 and a standard deviation of 15. A sample of size 80 is drawn randomly from the population.

• Find the probability that the sum of the 80 values (or the total of the 80 values) is more than 7500.
• Find the sum that is 1.5 standard deviations above the mean of the sums.

Let $X$ = one value from the original unknown population. The probability question asks you to find a probability for the sum (or total of) 80 values.

$\mathrm{\Sigma X}$ = the sum or total of 80 values. Since ${\mu }_{X}=90$ , ${\sigma }_{X}=15$ , and $n=80$ , then

$\Sigma X$ ~ $N\left(\mathrm{80}\cdot \mathrm{90},\sqrt{\mathrm{80}}\cdot \mathrm{15}\right)$

• mean of the sums = $n\cdot {\mu }_{X}=\left(80\right)\left(90\right)=7200$
• standard deviation of the sums = $\sqrt{n}\cdot {\sigma }_{X}=\sqrt{80}\cdot 15$
• sum of 80 values = $\mathrm{\Sigma x}=7500$

• Find $(P\left(\mathrm{\Sigma x}, 7500)\right)\phantom{\rule{20pt}{0ex}}$

$(P\left(\mathrm{\Sigma x}, 7500)\right)=0.0127$

normalcdf (lower value, upper value, mean of sums, stdev of sums)

The parameter list is abbreviated (lower, upper, $n\cdot {\mu }_{X},\sqrt{n}\cdot {\sigma }_{X}$ )

normalcdf (7500,1E99, $80\cdot 90,\sqrt{80}\cdot 15\right)=0.0127$

Reminder: $\mathrm{1E99}={10}^{99}$ . Press the EE key for E.

• Find $\mathrm{\Sigma x}$ where $z$ = 1.5:

$\mathrm{\Sigma x}$ = $n\cdot {\mu }_{X}$ + $z\cdot \sqrt{n}\cdot {\sigma }_{X}$ = (80)(90) + (1.5)( $\sqrt{80}$ ) (15)= 7401.2

#### Questions & Answers

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research.net
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how did you get the value of 2000N.What calculations are needed to arrive at it
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