In this lesson, we will use the TI-84
to determine the probability of a sum,
based upon the central limit theorem for sums
and also to determine a sum,
given the number of standard deviations,
the sum is a way from the mean of the sums.
To begin, let's review the central limit theorem for sums.
The central limit theorem for sums says
if you keep drawing more and more samples of size N
with a large enough N and taking their sums,
the sums form their own 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 and multiplied by the sample size
and a standard deviation equal
to the original standard deviation multiplied
by the square root of the sample size.
So going back to our example,
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.
Part A, we're asked to find the probability
that the sum of the 80 values
or the total of the 80 values is more than 7,500.
We can indicate this as the probability
that the sum of X is greater than 7,500.
Before we determine this probability,
let's find the mean or average of the sum of X
as well as the standard deviation of the sum of X.
So the mean of the sum of X is equal
to the sample size times the mean
of the original distribution which is, in this case,
80 times 90 is equal to 7,200
and the standard deviation of the sum of X is equal
to the square root of N,
which is the square root of the sample size
where the square root
of 80 times the original standard deviation
where the standard deviation of X, which is 15.
Let's go and leave this as 15 square root 80,
even though it could be simplified.
And now we will use the normal CDF function of a T-84
to determine the probability the sum
of X is greater than 7,500.
To begin, we press second, VARS, for the distribution menu.
Option two for normal CDF.
Now we need the lower bound upper bound mu and sigma.
And this information is for the sums,
not the original distribution.
So because we have the sum of X greater than 7,500,
the lower bound is 7,500,
enter.
Need to exaggerate the upper bound,
I'm going to use 999999.
Sometimes you'll also see scientific notation used,
but this should be good enough, enter.
Mu is the mean of the sums, which is 7,200.
And sigma is the standard deviation of the sums,
which is 15 square root 80.
[mouse clicking]
Enter, enter, enter.
And we have the probability to four decimal places,
we have 0.0127.
What is equal to 1.27%.
Next, we're asked to find the probability the sum
of 80 values is between 7,100 and 7,300.
Which is the probability the sum of X is greater than 7,100
and less than 7,300.
So going back to the calculator.
Second, VARS, option two for normal CDF.
Now the lower bound is 7,100,
enter.
The upper bound is 7,300,
enter.
The mean and standard deviation the same.
Enter, enter, enter.
And enter one more time.
To four decimal places, we have 0.5439.
[cursor scribbling]
Which is equal to 54.39%.
And for the last question,
we're asked to find the sum
that is 1.2 standard deviations below the mean of the sums.
We're told to round to the tenths place value.
So, the sum we are looking for, the sum of X,
is equal to the mean of the sum, which is 7,200.
This is the expected value.
Plus the Z score, which is negative 1.2
because the sum is 1.2 standard deviations below the mean,
times the standard deviation of the sums,
which is 15 square root 80.
And the same formula is given here below.
And now we'll go to the calculator
and get our decimal approximation.
We have 7,200 plus negative 1.2
times 15 square root 80.
[mouse clicking]
Which, to the tenths place value, is 7,039.0.
[cursor scribbling]
I hope you found this helpful.