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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:

  1. μ X = the mean of Χ
  2. σ Χ = the standard deviation of X

If you draw random samples of size n , then as n increases, the random variable Σ X consisting of sums tends to be normally distributed and Σ Χ ~ N (( n )( μ Χ ), ( n )( σ Χ )).

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 X has the following z -score associated with it:

  1. Σ x is one sum.
  2. z  =  Σ x ( n ) ( μ X ) ( n ) ( σ X )
    1. ( n )( μ X ) = the mean of Σ X
    2. ( n ) ( σ X ) = standard deviation of Σ X

To find probabilities for sums on the calculator, follow these steps.

2 nd DISTR
2: normalcdf
normalcdf (lower value of the area, upper value of the area, ( n )(mean), ( n )(standard deviation))

where:

  • mean is the mean of the original distribution
  • standard deviation is the standard deviation of the original distribution
  • sample size = n

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.

  1. Find the probability that the sum of the 80 values (or the total of the 80 values) is more than 7,500.
  2. 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.

Σ X = the sum or total of 80 values. Since μ X = 90, σ X = 15, and n = 80, Σ X ~ N ((80)(90),
( 80 )(15))

  • mean of the sums = ( n )( μ X ) = (80)(90) = 7,200
  • standard deviation of the sums = ( n )( σ X ) = ( 80 ) (15)
  • sum of 80 values = Σx = 7,500

a. Find P x >7,500)

P x >7,500) = 0.0127

This is a normal distribution curve. The peak of the curve coincides with the point 7200 on the horizontal axis. The point 7500 is also labeled. A vertical line extends from point 7500 to the curve. The area to the right of 7500 below the curve is shaded.

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

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

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

Reminder

1E99 = 10 99 .

Press the EE key for E.

b. Find Σ x where z = 1.5.

Σ x = ( n )( μ X ) + ( z ) ( n ) ( σ Χ ) = (80)(90) + (1.5)( 80 )(15) = 7,401.2

Try it

An unknown distribution has a mean of 45 and a standard deviation of eight. A sample size of 50 is drawn randomly from the population. Find the probability that the sum of the 50 values is more than 2,400.

0.0040

To find percentiles for sums on the calculator, follow these steps.

2 nd DIStR
3:invNorm
k = invNorm (area to the left of k , ( n )(mean), ( n ) (standard deviation))

where:

  • k is the k th percentile
  • mean is the mean of the original distribution
  • standard deviation is the standard deviation of the original distribution
  • sample size = n

In a recent study reported Oct. 29, 2012 on the Flurry Blog, the mean age of tablet users is 34 years. Suppose the standard deviation is 15 years. The sample of size is 50.

  1. What are the mean and standard deviation for the sum of the ages of tablet users? What is the distribution?
  2. Find the probability that the sum of the ages is between 1,500 and 1,800 years.
  3. Find the 80 th percentile for the sum of the 50 ages.
  1. μ Σx = x = 50(34) = 1,700 and σ Σx = n σ x = ( 50  ) (15) = 106.01
    The distribution is normal for sums by the central limit theorem.
  2. P (1500<Σ x <1800) = normalcdf (1,500, 1,800, (50)(34), ( 50  ) (15)) = 0.7974
  3. Let k = the 80 th percentile.
    k = invNorm (0.80,(50)(34), ( 50  ) (15)) = 1,789.3

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Source:  OpenStax, Introduction to statistics i - stat 213 - university of calgary - ver2015revb. OpenStax CNX. Oct 21, 2015 Download for free at http://legacy.cnx.org/content/col11874/1.3
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