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 ΣX consisting of sums tends to be normally distributed and ΣΧ ~ N((n)(μΧ), ()(σΧ)).
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 ΣX has the following z-score associated with it:
- Σx is one sum.
-
- (n)(μX) = the mean of ΣX
- = standard deviation of
Using the TI-83, 83+, 84, 84+ Calculator
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), ()(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.
ΣX = the sum or total of 80 values. Since μX = 90, σX = 15, and n = 80, ~ N((80)(90),
()(15))
- mean of the sums = (n)(μX) = (80)(90) = 7,200
- standard deviation of the sums = (15)
- sum of 80 values = Σx = 7,500
a. Find P(Σx > 7,500)
P(Σx > 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)(μX, (σX))
normalcdf
(7500,1E99,(80)(90),(15)) = 0.0127
REMINDER
1E99 = 1099.
Press the EE
key for E.
b. Find Σx where z = 1.5.
Σx = (n)(μX) + (z)(σΧ) = (80)(90) + (1.5)()(15) = 7,401.2
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.
2nd DIStR
3:invNorm
k = invNorm (area to the left of k, (n)(mean), (standard deviation))
where:
- k is the kth 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 80th percentile for the sum of the 50 ages.
Solution
- μΣx = nμx = 50(34) = 1,700 and σΣx = σx = (15) = 106.07
The distribution is normal for sums by the central limit theorem. - P(1500 < Σx < 1800) =
normalcdf
(1,500, 1,800, (50)(34), (15)) = 0.7974 - Let k = the 80th percentile.
k =invNorm
(0.80,(50)(34),(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 90th 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 95th 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
- μΣx = nμx = 70(8.2) = 574 minutes and σΣx = = (1) = 8.37 minutes
- Let k = the 95th percentile.
k = invNorm (0.95,(70)(8.2),(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(Σx ≥ 600) =normalcdf
(600,E99,(70)(8.2),(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 84th and 16th percentiles for the sum of the sample. Interpret these values in context.