# AP BIO EQUATIONS AND FORMULAS REVIEW SHEET #1 ```Formulas:
Mode = value that occurs most frequently in a data set
Median = middle value that separates the greater and lesser halves of a data set
Mean = sum of all data points divided by the number of data points
Range = value obtained by subtracting the smallest observation (sample minimum) from the greatest (sample
maximum)
Standard Deviation =
( xi  x ) 2
where x = mean and n = size of the sample
n 1
Example problem:
One of the lab groups collected the following data for the heights (in cm) of their Wisconsin Fast Plants:
5.4
7.2
4.9
9.3
7.2
8.1
8.5
5.4
7.8
10.2
Find the mode, median, mean, and range. Show your work where necessary.
4.9
5.4
5.4
7.2
7.2
7.8
8.1
8.5
9.3
10.2
Mode:__5.4, 7.2 ___
Median = average of middle 2 values = (7.2 + 7.8) / 2 = 7.5
Mean =
Median:____7.5_____
4.9  5.4  5.4  7.2  7.2  7.8  8.1  8.5  9.3  10.2 74

 7.4 Mean:______7.4____
10
10
Range = 10.2 – 4.9 = 5.3
Range:_____5.3_____
Find the standard deviation by filling in the following table.
Heights (x)
5.4
7.2
4.9
9.3
7.8
8.1
8.5
5.4
7.8
10.2
Mean ( x )
7.4
7.4
7.4
7.4
7.4
7.4
7.4
7.4
7.4
7.4
xx
(x  x)2
-2
-.2
-2.5
1.9
-.2
.7
1.1
-2
.4
2.8
4
.04
6.25
3.61
.04
.49
1.21
4
.16
7.84
27.64
 ( x  x ) 2
Standard deviation:
27.64
= 1.75
9
Interpret the standard deviation in the context of the problem.
The heights of Wisconsin Fast Plants in this sample are, on average, is about 1.75 cm away from the mean
height of 7.4 cm.
Formulas:
Chi Square  2  
(o  e) 2
e
o = observed individuals with observed genotype
e = expected individuals with observed genotype
Degrees of freedom equals the number of distinct possible outcomes
minus one
Degrees of Freedom
p
0.05
0.01
1
3.84
6.64
2
5.99
9.32
3
7.82
11.34
4
9.49
13.28
5
11.07
15.09
6
12.59
16.81
7
14.07
18.48
8
15.51
20.09
In peas, the allele for yellow seed color (Y) is dominant to the allele for green seed color (y). A student
crossed plants that came from yellow seeds with plants that came from green seeds. He believed that the
yellow seeds were heterozygous. The observed results are reported in the following table.
Seed color
Observed Number of Seeds in F1
Yellow seeds 62
Green seeds 38
Calculate the chi-squared (2) value for the null hypothesis that the yellow seed plants were heterozygous.
Null hypothesis – Yy x yy, expected ratio 50% Yellow, 50% Green
Phenotype
Yellow
Green
Obs
62
38
Exp
50
50
O-E
12
-12
O-E2
144
144
O-E2/E
144/50
144/50
Sum = 5.76
5.76 &gt; 3.84, therefore Null hypothesis is rejected
```