AP Biology: Standard Deviation Carol Leibl National Math and Science Initiative Dallas, TX

Standard deviation A statistical computation that is about how far members of the population deviate from the mean is the standard deviation. It is a characteristic of the population, so it doesn't depend on how many members you sample (except that if you only sample a few members, you won't get a good estimate of SD).

Two Different Normal Distributions These two histograms have the same mean of five but they differ in their variation or range of values. Histogram 1 has a range of 8 units and has a range of 5 units. There is more variation in histogram 1. Both form a normal distribution.

Steps for Determining Standard Deviation Determine the mean for the set of data. Find the difference between each data point and the mean and square the difference. Sum all squared differences Divide by the number of data point less one. Finally, take the square root of this number.

Steps for Determining Standard Deviation

Your Turn Have the students determine the standard deviation for this example.

Determine the s for either the black-eyed peas or pinto beans. Your Turn Have the students determine the standard deviation for this example. Determine the s for either the black-eyed peas or pinto beans.

Two Different Standard Deviations The standard deviations, s, for the two diagrams below are different. s=2.053 for histogram #1 s=1.046 for histogram #2 This makes sense because histogram #1 has more deviation or variation from the mean of 5.

Interesting Facts About Standard Deviations and Normal Distributions The percentage of data points in a normal distribution that fall within plus or minus 1 standard deviation is 68.26% and the percentage of individuals falling within plus or minus 2 standard deviations is 95% and 3 standard deviations is 99%. This is often called the 68-95-99% rule.

Two Different Standard Deviations In histogram 1- 68% of all the data points fall between 2.947 and 7.053. 95% of all the data points fall between 9.106 and 0.894. In histogram 2- 68% of all the data points fall between 6.046 and 3.954. 95% of all the data points fall between 7.092 and 2.908

Your Turn A student was measuring the height of mung beans. The height of 200 plants was measured. The mean height of the mung beans was 4 cm and the standard deviation was .5 cm. What percentage of the plants had a height of 4 and 4.5 cm?   What percentage of the plants had a height between 3 and 4 cm?

Your Turn A student is germinating lettuce seeds in a petri dish and after three days, 95% of them are between 1.1 cm and 1.7 cm long. Assuming that the data is normally distributed determine the mean and the standard deviation.

Your Turn A student is measuring the mass of pill bugs and finds that 99.7% falls between .034 g and 0.18 g. Assuming that the data is normally distributed determine the mean and standard deviation.

Averaging or Determining What is Normal Shown are the histograms illustrating the frequency distribution of the mass of 300 black-eyed peas and 300 pinto beans. Mean S 2s Black-Eyed Peas 0.21 0.042 0.084 Pinto Beans .37 0.37 0.074 0.148

Averaging or Determining What is Normal It is possible to compare the means of the two populations using a simple bar graph. One bar representing the mean mass for each seed. The standard deviation or variation of the two populations can be represented by using vertical error bars.

Bar Graph with Illustrating the Means Seed Mean S Black-eyed Peas .21 .042 Pinto Beans .37 .074

Bar Graph with Illustrating the Means and + 1 s The represents the that variation that exist within 68% of the data points. It illustrates that the pinto beans has a greater variation in its sample population than the black-eyed peas.

Bar Graph with Illustrating the S and SE with error bars

AP Biology Content Specialist Carol Leibl AP Biology Content Specialist cleibl@nms.org