34 7.2 The Central Limit Theorem for Sums
Suppose X is a random variable with a distribution that may be known or unknown (it can be any distribution) and suppose:
- μ = the mean of Χ
- σ = the standard deviation of X
If you draw random samples of size n, then as n increases, the random variable Σx consisting of sums tends to be normally distributed such that [latex]\displaystyle\sum{x}{\sim}{N}[{{({n})}{({\mu})},{(\sqrt{{n}})}{({\sigma})}}][/latex].
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 deviation equal to the original standard deviation multiplied by the square root of the sample size.
The random variable Σx has the following z-score associated with it:
- Σx is one sum.
- [latex]{z}=\frac{{\sum{x} - {({n})}{({\mu})}}}{{{(\sqrt{{n}})}{({\sigma})}}}[/latex] or [latex]\sum{x} = (n)({{\mu}})+({z})({{\sigma}})(\sqrt{n})[/latex]
(Do not memorize both formula. They are same! )
- [latex]{({n})}{({\mu})}[/latex] = [latex]\displaystyle{\mu}_{\sum{x}}[/latex] , the mean of [latex]\sum{x}[/latex]
- ([latex]\sqrt{n}[/latex])(σ) = [latex]\displaystyle{\sigma}_{\sum{x}}[/latex] , the standard deviation of [latex]\sum{x}[/latex]
Guide for TI-CalculatorTo find probabilities for sums on the calculator, follow these steps.
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Example 1
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 7,500.
- Find the sum that is 1.5 standard deviations above the mean of the sums.
Solution
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.
[latex]\displaystyle\sum{x}[/latex] = the sum or total of 80 values, [latex]\displaystyle{\mu}=90,{\sigma}=15[/latex], and n = 80,
- mean of the sums, [latex]\displaystyle{\mu}_{\sum{x}}[/latex] = (n)(μ) = (80)(90) = 7,200
- standard deviation of the sums, [latex]\displaystyle{\sigma}_{\sum{x}}[/latex] = [latex]\displaystyle{(\sqrt{{n}})}{({\sigma})}={(\sqrt{{80}})}{({15})}[/latex]
Therefore, [latex]\sum{x}[/latex]~ N((80)(90), ([latex]\displaystyle\sqrt{{80}}[/latex])(15)).
- Probability that the sum of the 80 values (or the total of the 80 values) is more than 7,500
= P(Σx > 7,500)
= Shaded area
TI-Calculator: normalcdf (7500, 1E99, (80)(90), [latex]\displaystyle{(\sqrt{{80}})}[/latex](15)) = 0.0127
Therefore, P(Σx > 7,500) = 0.0127 - Find Σx where z = 1.5.
[latex]\displaystyle{\sum{x}}={(n)}{({\mu})}+{(z)}{(\sqrt{n})}{({\sigma}})=(80)(90)+(1.5)(\sqrt{80})(15)=7401.2[/latex]
Try It
An unknown distribution has a mean of 45 and a standard deviation of 8.