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_{X}*Χ**σ*= the standard deviation of_{Χ}*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}*), ($\sqrt{n}$)(

*σ*)).

_{Χ}**The** central limit theorem for sums says that if you repeatedly draw samples of a given size (such as repeatedly rolling ten dice) and calculate the sum of each sample, these sums tend to follow a normal distribution. As sample sizes increase, the distribution of means more closely follows the normal distribution. **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:

- Σ
*x*is one sum. - $z\text{=}\frac{\Sigma x\u2013(n)({\mu}_{X})}{(\sqrt{n})({\sigma}_{X})}$
- (
*n*)(*μ*) = the mean of Σ_{X}*X* - $(\sqrt{n})({\sigma}_{X})$ = standard deviation of $\Sigma X$

- (

## 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), ($\sqrt{n}$)(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 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.

### 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.**

Σ*X* = the sum or total of 80 values. Since *μ _{X}* = 90,

*σ*= 15, and

_{X}*n*= 80, $\Sigma X$ ~

*N*((80)(90),

($\sqrt{\text{80}}$)(15))

- mean of the sums = (
*n*)(*μ*) = (80)(90) = 7,200_{X} - standard deviation of the sums = $\text{(}\sqrt{n}\text{)(}{\sigma}_{X}\text{) = (}\sqrt{\text{80}}\text{)}$(15)
- sum of 80 values =
*Σx*= 7,500

a. Find *P*(Σ*x* > 7,500)

*P*(Σ*x* > 7,500) = 0.0127

## 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}*, $\left(\sqrt{n}\right)$(

*σ*))

_{X}`normalcdf`

(7500,1E99,(80)(90),$\left(\sqrt{80}\right)$(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*)$\left(\sqrt{n}\right)$(

*σ*) = (80)(90) + (1.5)($\sqrt{80}$)(15) = 7,401.2

_{Χ}## Try It 7.5

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.

## 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), $\text{(}\sqrt{n})$(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*sample size*=*n*

## Example 7.6

### Problem

In a recent study, it was reported that the mean age of iPad users is 34 years. Suppose the standard deviation is 15 years. The sample of size is 50.

- What are the mean and standard deviation for the sum of the ages of iPad 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.

### Solution

*μ*=_{Σx}*nμ*= 50(34) = 1,700 and_{x}*σ*= $\sqrt{n}$_{Σx}*σ*= $(\sqrt{\text{50}}\text{)}$(15) = 106.07_{x}

The distribution is normal for sums by the central limit theorem.*P*(1500 < Σ*x*< 1800) =`normalcdf`

(1,500, 1,800, (50)(34), $(\sqrt{\text{50}}\text{)}$(15)) = 0.7974- Let
*k*= the 80^{th}percentile.*k*=`invNorm`

(0.80,(50)(34),$(\sqrt{\text{50}}\text{)}$(15)) = 1,789.3

## Try It 7.6

In a recent study, it was reported that the mean age of iPad users is 35 years. Suppose the standard deviation is ten years. The sample size is 39.

- What are the mean and standard deviation for the sum of the ages of iPad 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 of size 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 ten hours.

### Solution

*μ*=_{Σx}*nμ*= 70(8.2) = 574 minutes and_{x}*σ*= $\left(\sqrt{n}\right)\left({\sigma}_{x}\right)$ = $(\sqrt{\text{70}}\text{)}$(1) = 8.37 minutes_{Σx}- Let
*k*= the 95^{th}percentile.*k*= invNorm (0.95,(70)(8.2),$(\sqrt{\text{70}}\text{)}$(1)) = 587.76 minutes

Ninety five percent of the sums of app engagement times are at most 587.76 minutes. - ten hours = 600 minutes
*P*(Σ*x*≥ 600) =`normalcdf`

(600,E99,(70)(8.2),$(\sqrt{\text{70}}\text{)}$(1)) = 0.0009

## Try It 7.7

The mean number of minutes for app engagement by a table use 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 ten 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.