Module 7: The Central Limit Theorem
The Central Limit Theorem for Sums
Barbara Illowsky & OpenStax et al.
Learning Objectives
Suppose X is a random variable with a distribution that may be known or unknown (it can be any distribution) and suppose:
- μX = the mean of Χ
- σΧ = the standard deviation of X
If you draw random samples of size n, then as n increases, the random variable [latex]sum X[/latex] consisting of sums tends to be normally distributed and
[latex]displaystyle {sum{X}{sim}{N}(n cdot mu_X ,sqrt{n}sigma_X)}[/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:
- [latex]sum x[/latex] is one sum.
- [latex]{z}=frac{{sum{x}-{({n})}{({mu}_{{X}})}}}{{{(sqrt{{n}})}{({mu}_{{X}})}}}[/latex]
- [latex]{({n})}{({mu}_{{X}})}=text{the mean of }sum{X}[/latex]
- [latex]{left(sqrt{{n}}right)}{left({sigma}_{{X}}right)} =text{the standard deviation of }sum{X}[/latex]
To find probabilities for sums on the calculator, follow these steps.
2nd
DISTR
2:normalcdf
normalcdf
(lower value of the area, upper value of the area, (n)(mean), ([latex]displaystylesqrt{{n}}[/latex])(standard deviation))
where: mean is the mean of the original distribution standard deviation is the standard deviation of the original distribution sample size = n
Example
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]displaystylesum{X}[/latex] = the sum or total of 80 values. Since [latex]displaystyle{mu}_{x}=90,{sigma}_{x}=15[/latex], and n = 80, [latex]sum{X}[/latex]~ N((80)(90), ([latex]displaystylesqrt{{80}}[/latex])(15))
- mean of the sums = (n)(μX) = (80)(90) = 7,200
- standard deviation of the sums =[latex]displaystyle{(sqrt{{n}})}{({sigma}_{{X}})}={(sqrt{{80}})}{({15})}[/latex]
- sum of 80 values = Σx = 7,500
- Find P(Σx > 7,500)
P(Σx > 7,500) = 0.0127
normalcdf
(lower value, upper value, mean of sums, stdev
of sums) The parameter list is abbreviated(lower, upper, (n)(μX, [latex]displaystyle{(sqrt{{n}})}[/latex](σX))
normalcdf
(7500, 1E99, (80)(90), [latex]displaystyle{(sqrt{{80}})}[/latex](15)) = 0.0127
Remember that 1E99 = 1099.
Press the EE
key for E.
2. Find Σx where z = 1.5.