Contents

The Central Limit Theorem for Sums (Optional)

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

μ_X = the mean of Χ
σ_Χ = 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 and ΣΧ ~ N[(n)(μ_Χ), ( $\sqrt{n} \sqrt{n}$ )(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 ΣX has the following z-score associated with it:

Σx is one sum.
z = Σx–(n)( μ X ) ( n )( σ X ) z = Σx–(n)( μ X ) ( n )( σ X )
1. (n)(μ_X) = mean of ΣX
2. $(\sqrt{n}) (σ_{X}) (\sqrt{n}) (σ_{X})$ = standard deviation of $Σ X Σ X$

Using the TI-83, 83+, 84, 84+ Calculator

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), ( $\sqrt{n} \sqrt{n}$ )(standard deviation))

where,

mean is the mean of the original distribution,
standard deviation is the standard deviation of the original distribution, and
sample size = n.

Example 7.5

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

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.

Show solution

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.

ΣX = the sum or total of 80 values. Because μ_X = 90, σ_X = 15, and n = 80, $Σ X Σ X$ ~ N[(80)(90),
( $\sqrt{80} \sqrt{80}$ )(15)]

mean of the sums = (n)(μ_X) = (80)(90) = 7200
standard deviation of the sums = $(\sqrt{n})(σ_{X}) = (\sqrt{80}) (\sqrt{n})(σ_{X}) = (\sqrt{80})$ (15)
sum of 80 values = Σx = 7500

a. Find P(Σx > 7500)

P(Σx > 7500) = 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.

Figure 7.3

Using the TI-83, 83+, 84, 84+ Calculator

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

The parameter list is abbreviated(lower, upper, (n)(μ_X, $(\sqrt{n}) (\sqrt{n})$ (σ_X))

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

Reminder

1E99 = 10⁹⁹.

Press the EE key for E.

b. Find Σx where z = 1.5.

Σx = (n)(μ_X) + (z) $(\sqrt{n}) (\sqrt{n})$ (σ_Χ) = (80)(90) + (1.5)( $\sqrt{80} \sqrt{80}$ )(15) = 7401.2

Try It 7.5

An unknown distribution has a mean of 45 and a standard deviation of 8. 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.

Using the TI-83, 83+, 84, 84+ Calculator

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), $(\sqrt{n}) (\sqrt{n})$ (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, and
sample size = n.

Example 7.6

Problem

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

What are the mean and standard deviation for the sum of the ages of tablet users? What is the distribution?
Find the probability that the sum of the ages is between 1,500 and 1,800 years.
Find the 80^th percentile for the sum of the 50 ages.

Show solution

Solution

μ_Σx = nμ_x = 50(34) = 1,700 and σ_Σx = $\sqrt{n} \sqrt{n}$ σ_x = $(\sqrt{50}) (\sqrt{50})$ (15) = 106.01
The distribution is normal for sums by the central limit theorem.
P(1500 < Σx < 1800) = normalcdf (1500, 1800, (50)(34), $(\sqrt{50}) (\sqrt{50})$ (15)) = 0.7974
Let k = the 80^th percentile.
k = invNorm(0.80,(50)(34), $(\sqrt{50}) (\sqrt{50})$ (15)) = 1789.3

Try It 7.6

In a recent study reported Oct.29, 2012, the mean age of tablet users is 35 years. Suppose the standard deviation is 10 years. The sample size is 39.

What are the mean and standard deviation for the sum of the ages of tablet users? What is the distribution?
Find the probability that the sum of the ages is between 1,400 and 1,500 years.
Find the 90^th percentile for the sum of the 39 ages.

Example 7.7

Problem

The mean number of minutes for app engagement by a tablet user is 8.2 minutes. Suppose the standard deviation is one minute. Take a sample size of 70.

What are the mean and standard deviation for the sums?
Find the 95^th percentile for the sum of the sample. Interpret this value in a complete sentence.
Find the probability that the sum of the sample is at least 10 hours.

Show solution

Solution

μ_Σx = nμ_x = 70(8.2) = 574 minutes and σ_Σx = $(\sqrt{n}) (σ_{x}) (\sqrt{n}) (σ_{x})$ = $(\sqrt{70}) (\sqrt{70})$ (1) = 8.37 minutes
Let k = the 95^th percentile.
k = invNorm (0.95,(70)(8.2), $(\sqrt{70}) (\sqrt{70})$ (1)) = 587.76 minutes
Ninety-five percent of the app engagement times are at most 587.76 minutes.
10 hours = 600 minutes
P(Σx ≥ 600) = normalcdf(600,E99,(70)(8.2), $(\sqrt{70}) (\sqrt{70})$ (1)) = 0.0009

Try It 7.7

The mean number of minutes for app engagement by a tablet user is 8.2 minutes. Suppose the standard deviation is one minute. Take a sample size of 70.

What is the probability that the sum of the sample is between seven hours and 10 hours? What does this mean in context of the problem?
Find the 84^th and 16^th percentiles for the sum of the sample. Interpret these values in context.