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The Central Limit Theorem for Sums

Module by: Susan Dean, Barbara Illowsky, Ph.D.. E-mail the authors

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

  • a. μ X = μ X = the mean of XX
  • b. σ X = σ X = the standard deviation of XX
If you draw random samples of size nn, then as nn increases, the random variable ΣXΣX which consists of sums tends to be normally distributed and

Σ XΣX ~ N ( n ⋅ μ X , n ⋅ σ X ) N(n⋅ μ X , n ⋅ σ X )

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 Σ XΣX has the following z-score associated with it:

  • a. ΣxΣx is one sum.
  • b. z = Σ x - n ⋅ μ X n ⋅ σ X z= Σ x - n ⋅ μ X n ⋅ σ X
  • a. n ⋅ μ X = n⋅ μ X = the mean of ΣXΣX
  • b. n ⋅ σ X = n ⋅ σ X = standard deviation of ΣXΣX

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.

Problem 1

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

Solution

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

ΣXΣX = the sum or total of 80 values. Since μ X = 90 μ X =90, σ X = 15 σ X =15, and n = 80 n=80, then

Σ XΣX ~ N ( 80 ⋅ 90 , 80 ⋅ 15 ) N(80⋅90, 80 ⋅15)

  • •. mean of the sums = n ⋅ μ X = ( 80 ) ( 90 ) = 7200 n⋅ μ X =(80)(90)=7200
  • •. standard deviation of the sums = n ⋅ σ X = 80 ⋅ 15 n ⋅ σ X = 80 ⋅15
  • •. sum of 80 values = Σx = 7500 Σx=7500

  • a: Find P ( Σx > 7500 ) P ( Σx 7500 )

P ( Σx > 7500 ) = 0.0127 P ( Σx 7500 )=0.0127

Normal distribution curve of sum X with the values of 7200 and 7500 on the x-axis. A vertical upward line extends from point 7500 on the x-axis up to the curve. The probability area occurs from point 7500 and to the end of the curve.

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

The parameter list is abbreviated (lower, upper, n ⋅ μ X , n ⋅ σ X n⋅ μ X , n ⋅ σ X )

normalcdf(7500,1E99, 80 ⋅ 90 , 80 ⋅ 15 ) = 0.0127 80⋅90, 80 ⋅15)=0.0127

Reminder: 1E99 = 10 99 1E99= 10 99 . Press the EE key for E.

  • b: Find ΣxΣx where zz = 1.5:


ΣxΣx = n ⋅ μ X n⋅ μ X + z ⋅ n ⋅ σ X z⋅ n ⋅ σ X = (80)(90) + (1.5)( 80 80 ) (15) = 7401.2

Glossary

Central Limit Theorem:
Given a random variable (RV) with known mean μμ and known standard deviation σσ. We are sampling with size n and we are interested in two new RVs - the sample mean, X¯X¯, and the sample sum, ΣXΣX. If the size nn of the sample is sufficiently large, then X¯X¯ size 12{ { bar {X}}} {}∼ N μ σ n N μ σ n and ΣXΣX size 12{X} {} ∼ N ( nμ , n σ )N(nμ,nσ). If the size n of the sample is sufficiently large, then the distribution of the sample means and the distribution of the sample sums will approximate a normal distribution regardless of the shape of the population. The mean of the sample means will equal the population mean and the mean of the sample sums will equal n times the population mean. The standard deviation of the distribution of the sample means, σ n σ n , is called the standard error of the mean.
Normal Distribution:
A continuous random variable (RV) with pdf f(x)=1σ2πe−(x−μ)2/2σ2f(x)=1σ2πe−(x−μ)2/2σ2 size 12{ ital "pdf"= { {1} over {σ sqrt {2π} } } e rSup { size 8{ - \( x - μ \) rSup { size 6{2} } /2σ rSup { size 6{2} } } } } {}, where μμ is the mean of the distribution and σσ is the standard deviation. Notation: XX ~ N μ σ N μ σ . If μ=0μ=0 and σ=1σ=1, the RV is called the standard normal distribution.

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