# Normal Calculations on TI-83/84 or TI-89

for Individuals and Samples

Copyright © 2002–2015 by Stan Brown, Oak Road Systems

for Individuals and Samples

Copyright © 2002–2015 by Stan Brown, Oak Road Systems

**Summary:**
Earlier, we used the empirical rule (68–95–99.7
rule) to find probabilities between certain values in a ND.
Now we extend that to calculate
**probabilities between any values**.
There are really only a few calculations,
but the variations can be hard to manage. This page
summarizes all the normal calculations, along with some important
related ideas.

Have boundary(ies), need area | Have area, need boundary | |
---|---|---|

working with z scores | `normalcdf(` left boundary,
right boundary`)` |
`invNorm(` area to left`)` |

working with raw (x) scores | `normalcdf(` left boundary,
right boundary, mean, standard deviation`)` |
`invNorm(` area to left, mean, standard deviation`)` |

TI-83/84 keystrokes | [`2nd` `VARS` makes `DISTR` ] [`2` ] |
[`2nd` `VARS` makes `DISTR` ] [`3` ] |

TI-89 keystrokes | [`CATALOG` ] [`F3` ] [plain `6` makes `N` ] [`ENTER` ] |
[`CATALOG` ] [`F3` ] [plain `9` makes `I` ] [`▼` 3 times ] [`ENTER` ] |

When you have only one boundary, and thus you’re computing area in
a tail, use ∞ or −∞ as the other
boundary. You get ∞ on your TI-83 as
`1`

[`2nd`

`,`

*makes* `EE`

] `99`

, or as `10^99`

.

**Caution:**
`invNorm`

works from **area to left**. If the
problem actually gives you an area to the right or an area in the
middle, you must convert it to an area to left by 1−area.

These are already in terms of area to left.
For instance, P_{40} is the x value such that 40% (0.40) of
the population scored lower. Therefore,
`invNorm(0.40,`

*mean, standard deviation*`)`

will give you the 40th percentile.

Suppose the mean score on the math SAT is 500 and the standard deviation is 100.

**Example 1**: What proportion of test
takers earn a score between 650 and 700?

**Solution**: This is a calculation for individuals:
μ = 500, σ = 100.
normalcdf(650, 700, 500, 100) = 0.0441. About 4.4%
of test takers earn a score between 650 and 700.

**Example 2**: What percentile is represented
by a score of 735?

**Solution**: This, too, is a calculation for individuals.
The percentile is
the percent of area to the left of x = 735.
normalcdf(−10^99, 735, 500, 100) =
0.9906, so a score of 735 is at the 99th percentile: 99% of
individual scores are lower than or equal to 735.

**Example 3**: What score is earned by the
best 15% of test takers?

**Solution**: The area of the right-hand tail is 0.15, so
the area to left is 1−0.15.
invNorm(1−.15, 500, 100) = 603.64. A score of
604 or above puts you in the top 15%.

As you know, the
mean of sample means is the same as the
population mean μ, but the standard deviation of sample means
(symbol σ_{x̅}, also known as the
**standard error of the mean, or SEM**)
is different,
equal to σ/√n, the population standard deviation
divided by the square root of sample size.

With just that one change, SEM instead of s.d., the calculations for a distribution of sample means are quite similar to the calculations for individuals.

Have boundary(ies), need area | Have area, need boundary | |
---|---|---|

working with raw (x) scores | `normalcdf(` left boundary,
right boundary, mean, st.dev.÷√n`)` |
`invNorm(` area to left, mean,
st.dev.÷√n`)` |

Suppose the mean score on the math SAT is 500 and the standard deviation is 100.

**Example 4**: You randomly select 48 test
takers, and compute their average score as 550. Does this surprise
you?

**Solution**: To ask whether that sample mean is
surprising, compute how likely it is to that random chance would
give a sample mean of 550 or further away from the supposed
population mean. Since 550>500, the question is the probability of
finding x̅≥550 for n=48.
You’re concerned with a sample mean, not an individual measurement, so the
relevant standard deviation is the SEM, σ/√n.
normalcdf(550, 10^99, 500, 100/√48) =
2.66×10^{-4}, or about 0.000 266.

Yes, this would be a surprising result. If μ = 500 and σ = 100, there’s less than a 3 in 10,000 chance that the mean of a random sample of size 48 will be ≥550.

This page is used in instruction at Tompkins Cortland Community College in Dryden, New York; it’s not an official statement of the College. Please visit www.tc3.edu/instruct/sbrown/ to report errors or ask to copy it.

For updates and new info, go to http://www.tc3.edu/instruct/sbrown/ti83/