Normal Calculations on TI-83/84 or TI-89
for Individuals and Samples
Copyright © 2002–2023 by Stan Brown, BrownMath.com
Copyright © 2002–2023 by Stan Brown, BrownMath.com
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, which means 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, P40 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 σx̅ , 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.
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