Standard Deviation by hand
The formula for standard deviation is
= − −1
Here’s what you do:
1.
2.
3.
4.
5.
6.
Calculate the mean of the data. This is denoted “x-bar”: .
Calculate the difference between each data point and the mean.
Square each answer in step 2.
Add up all the numbers from step 3.
Divide by − 1, where n is the number of data points.
Square root the result from step 5.
Example: The data is 45, 54, 46, 48, 44, 50, 52, 52, 51, 49, 45, 42.
There are 12 data points, so n = 12.
= 48.17
Step 1: The mean is = Steps 2 and 3: I suggest a table:
45
54
46
48
44
50
52
52
51
49
45
42
− 45 – 48.17 = -3.17
54 – 48.17 = 5.83
46 – 48.17 = -2.17
48 – 48.17 = -0.17
44 – 48.17 = -4.17
50 – 48.17 = 1.83
52 – 48.17 = 3.83
52 – 48.17 = 3.83
51 – 48.17 = 2.83
49 – 48.17 = 0.83
45 – 48.17 = -3.17
42 – 48.17 = -6.17
− (-3.17)2 = 10.0489
5.822 = 33.9889
(-2.17)2 = 4.7089
(-0.17)2 = 0.0289
(-4.17)2 = 17.3889
1.832 = 3.3489
3.832 = 14.6689
3.832 = 14.6689
2.832 = 8.0089
0.832 = 0.6889
(-3.17)2 = 10.0489
(-6.17)2 = 38.0689
Step 4: Sum the last column: 155.6668.
Step 5: Divide by 11, which is the number of data points, minus 1:
.
!
= 14.1515 …
Step 6: Square root it: $14.1515 … = 3.76. This is your standard deviation. It is denoted on the TI83/84 calculators.
Normal distributions on the TI calculator:
If you need to calculate an area under a normal curve, you can use a table, or your TI calculator.
Select 2nd-DISTR. Then select 2:normalcdf(.
Type in the lower bound, upper bound, mean and standard deviation, then hit ENTER.
Example: Bags of candy weigh on average (mean) 5 lbs with a standard deviation of 0.6 lbs, normally
distributed. What is the probability a bag of candy weighs between 4.6 and 5.2 lbs?
Answer:
normalcdf(4.6, 5.2, 5, 0.6)
[ENTER] 0.378, or 37.8%.
(On the standard normal, the mean is 0 and the standard deviation is 1.)
For open-ended intervals, use a “far away” number for the open end.
Example: The average kitty cat weighs 11 lbs with a standard deviation of 3 lbs, normally distributed. What’s
the probability a kitty cat weighs above 9 lbs?
Answer. The lower bound is 9, but the upper bound is not specified. Pick any large number for upper bound,
for example 50. normalcdf(9, 50, 11, 3) [ENTER] 0.588, or 58.8%.
Working backwards: We know an area, now we want a “z-score” that matches it.
Select 2nd-DISTR. Then select 3:InvNorm(
Type in the area, the mean and the standard deviation. Area is always to the left of the z-score.
Example: A class took a test with a mean of 82 and a standard deviation of 5. What score would put you at
the 85th percentile (that is, you scored better than 85% of other students?
Answer: the area is 0.85: InvNorm(0.85, 82, 5) [ENTER] 87.2. A min score would be 87.
Example: The professor gives an A to anyone in the top 5% of the class. What’s the score needed to get an
A?
Answer: The top 5% is the same as the 95th percentile, so we use 0.95 as our area:
InvNorm(0.95, 82, 5) [ENTER] 90.22. The min score needed is 90.