The annual rainfall in a certain region
is approximately normally distributed
with a mean of 42.2 inches
and a standard deviation of 5.4 inches.
We're told to round answers
to the nearest 10th of a percent.
To save some time, I've already set up
the normal distribution curve below
where we have the mean at 42.2 at the center
of the distribution and the horizontal axis
is scaled by the standard deviation of 5.4.
For part A, what percentage of years will have
an annual rainfall of less than 46 inches?
We can indicate this as the probability
or percentage that x is less than 46.
Looking at the graph, notice 46 is approximately here.
To determine the percentage of years
that will have an annual rainfall less than 46 inches,
we will use the TI84 to determine the area under the curve
to the left of 46.
This will give us a decimal which will
then convert to a percentage.
Remember the total area under the normal curve
is equal to one.
And now we'll find this are a using the normal cumulative
density function on the TI84.
So going to the calculator, we press second bars
number two for the normal cumulative density function.
Now from here because we're looking for the area
to the left of 46, we'll have to exaggerate
the left bounder or lower bound.
Notice right now it says negative one to E99
which is the default, which is negative one times 10
raised to power of 99 which is a great exaggeration
for the lower bound or left bound.
But we can also just use something like negative 999999.
And then we press enter.
The upper bound or right bound is the given value of 46.
Enter, mew is 42.2 the mean.
Enter and sigma, the standard deviation is 5.4.
Enter and we press ENTER again on paste
which brings the information to the home screen.
If you have an older TI84 you may have to enter
the information as it looks here from the home screen.
And now we press enter,
which are three decimal places approximately 0.759.
Which is equal to 75.9%.
Part B, what percentage of years
will have an annual rainfall of more than 32 inches?
We can indicate this as the probability or percentage
that x is greater than 32.
Looking at the graph 32 is approximately here.
Because we're looking for the percentage of years
that will have an annual rainfall more than 32 inches,
we will now find the area to the right of 32
which is this area here.
Notice we don't have to exaggerate the right bound
or upper bound.
Press second bars, number two.
The lower bound is now 32 because you want
the area to the right.
So we enter 32 enter.
The upper bound needs to be exaggerated.
For example, 999999 works well, enter.
Mew and sigma stay the same.
So we go down to paste, press enter, and enter again.
To three decimal places, we have approximately 0.971,
which is 97.1%.
And then for part C what percentage of years will
have an annual rainfall of between 38 inches and 43 inches?
Which we could indicate as the probability
that x is greater than 38 and less than 43.
Where 38 is approximately here
and 43 is approximately here.
So now we're looking for the area between these two values.
So going back to the calculator, second bars number two,
the lower bound and left bound is 38, enter.
The upper bound or right bound is 43 enter.
Everything else stays the same.
Go down to paste, press enter, and enter again,
which gives us approximately 0.341,
which is 34.1%.
I hope you found this helpful.