Download Using the TI-83 to find areas under the standard normal curve and

Using the TI-83 to find areas under the standard normal curve and any normal
curve
Remember, the standard normal is a special normal distribution whose mean is 0
and standard deviation is 1. The TI-83 will calculate the area under th curve
between two values.
According to the Empirical Rule, approximately 68% of the data for a normal
distribution falls within 1 standard deviation of the mean. In other words,
P(-1<z<1)=.68. To verify this using the TI clear all "y=" and turn off all stat plots.
To see a drawing of the curve:
2nd Distr…DRAW… ShadeNorm( -1 , 1 , 0 , 1 ) enter. You should see
the normal curve shaded in from -1 to 1 showing an area of .682689.
If your graph doesn't look quite right, adjust the window and try again. Xmin = -3,
xmax= 3, xscl = .5, ymin= 0, ymax= .5, yscl= 1, xres = 1. Since we're doing the
standard normal curve, the x axis should go from -3 to 3 and the y axis from 0 to
about .4 or .5. Generally though, most people don’t do the drawing since it
requires some finesse with the window settings and instead, merely compute the
area under the curve.
To compute the area and not have a drawing of the curve:
2nd Distr…#2 for Normalcdf(-1,1,0,1) enter.
When using Normalcdf, you must input:
Normalcdf (lower boundary, upper boundary, mean, standard deviation)
The lower boundary of negative infinity is -EE99 and the upper boundary of
positive infinity in EE99
To compute P(z>1.3): normalcdf(1.3,EE99,0,1)
To compute P(z<1.3): normalcdf(-EE99,1.3,0,1)
Finding probabilities for nonstandard normal distributions
Suppose a test of coordination for first graders is scored and the mean is 50 with
a standard deviation of 15. Assume the distribution is normal.
What percent of the frist graders will have the following scores?
1) below 30 normalcdf(-EE99,30,50,15) = .0912 or 9.12%
2) between 40 and 75 normalcdf(40,75,50,15) = .6997 or 69.97%
3) above 75 normalcdf(75,EE99,50,15) = .0478 or 4.78%
Notice the normalpdf is not used!