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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.