Section 3: Standard Deviation

Transcription

1 Section 3: Standard Deviation The most important measure of the spread of any distribution is the standard deviation. The standard deviation of a set of data is a measure of the average distance from the mean. The larger the standard deviation the more spread out the data, the smaller the standard deviation the less spread out the data. To calculate the standard deviation of a data set, we do the following: 1. Find the mean ( x) of the data. x i x x = n i = Next, we calculate the deviation from the mean ( x i x) for each score. x ( x i x) i For example: x i x = = Next, we find the squared deviation ( x x) 2 i for each score. x ( x i x) ( ) 2 i x i x

2 4. The variance is the average squared deviation: s = xi x n Variance = ( ) For this data set, we have s = ( ) = (34.86) = Since we squared the deviations to make them positive, we now have a measurement in square units. To undo the process and get the units back to their original values, we take the square root. The standard deviation ( s x ) is the square root of the variance: s 1 ( ) 2 x = xi x n 1 s = 5.81 = 2.41 x The sample standard deviation is designated s or s x. The population standard deviation will be designated with the Greek letter sigma - σ. Notice that we divide by n 1 to find the standard deviation instead of n. This ensures that the sample standard deviation is the best possible approximation to the population standard deviation. Note: You can find the standard deviation using the 1-Var Stats command on the TI-83. Example 1: Finding the standard deviation of data Find the standard deviation of the data: Example 2: Finding the standard deviation of data Find the standard deviation of the data in the table below. Compare this to your answer for Example 1 and explain

3 How do we use the standard deviation? A density curve is a smooth curve used to describe the shape of the distribution of a variable. It has two key features: it always lies above or on the x-axis, and it has total area of 1 underneath the curve. This is a mathematical model of the distribution. The most useful density curve for our purposes will be the normal distribution. The normal distribution is single-peaked (unimodal), symmetric, and bell shaped as shown in the figure below. A normal distribution is completely determined by its mean and standard deviation. normal distribution For a normal distribution, the mean is equal to the median rule If X is a normally distributed variable with mean µ and standard deviation σ, we write X N ( µ, σ ) and the following general approximation holds:, Number of standard deviations from the mean Percent of scores within this range µ ± σ 68% of the scores µ ± 2σ 95% of the scores µ ± 3σ 99.7% of the scores

4 Example 1: Using the Rule If the distribution of heights of women ages 18 to 24 is normal with mean 64.5 standard deviationσ = 2.5, find the following a) Between what two heights do 68% of the heights lie? µ = inches and b) What percent of the heights are below 62 inches? c) What percent of the heights are above 62 inches? d) What percent of the heights are between 57 inches and 72 inches? e) What percent of the heights are above 72 inches? f) What percent of the heights are below 72 inches?

5 Example 2: Using the Rule Birth weights at a local hospital have a normal distribution with a mean of 110 oz and a standard deviation of 15 oz. a) What proportion of infants have birth weights under 95 oz.? b) What proportion of infants have birth weights above 95 oz.? c) What proportion of infants have birth weights under 80 oz.? d) What proportion of infants have birth weights between 80 oz. and 95 oz.? e) What proportion of infants have birth weights above 125 oz.?

6 z-scores The rule is useful but what happens if the data value we are considering is not exactly 1, 2, or 3 standard deviations from the mean. We will use z-scores to find areas under normal density curves for these values. Definition: The z-score of a data value data value lays from the mean. z = x µ gives the number of standard deviations that the σ Example 3: Finding z-scores The scores on a standardized test have a mean of 21 a) Find the z-score of x = 24. µ = and a standard deviation of σ = 3. b) Find the z-score of x = 22. c) Find the z-score of x = 18. d) Find the z-score of x = 19. µ = and standard deviation 1 Fact: z has a normal distribution with mean 0 σ =. This can be abbreviated using the notation z N(0,1). This distribution is called the standard normal distribution.

7 Finding percents using z scores The z-score can be used to find the percent of scores above or below a given value by using the table in the front of your book (Table A). The table gives the proportion of scores to the left of the given z score. The table gives Px ( z). To find the proportion of scores greater than a given value subtract from 1: Px ( z) = 1 Px ( z). Example 4: Finding percents Find the percent of z values that satisfy each of the following conditions: a) z < 1.5 b) z < 0.5 c) z > 0.5 d) 1.2 < z < 0.5 Example 5: Using z -scores If the distribution of heights of women ages 18 to 24 is normal with mean 64.5 standard deviation σ = 2.5, find the following a) What percent of the heights are above 70 inches? µ = inches and b) What percent of the heights are below 60 inches? c) What percent of the heights are between 60 inches and 70 inches?

8 Example 6: Using z -scores If the distribution of heights of adult men is normal with mean 69 deviation σ = 2.5, find the following a) Between what two heights do 95% of the heights lie? µ = inches and standard b) What percent of the heights are below 61.5 inches? c) What percent of the heights are above 61.5 inches? d) What percent of the heights are between 60 inches and 72 inches? e) What percent of the heights are above 72 inches? f) What percent of the heights are below 60 inches?

9 Sometimes it is useful to find the score that corresponds to a certain percent. We can solve the formula x µ z = for x to getx = µ + σz. To use the formula, we find the z-score that corresponds to the σ given percent and then substitute into the formula to find x. Example 7: Using z scores If the distribution of heights of adult men is normal with mean 69 σ = 2.5, find the following: a) Above what height do the top 10% of heights lie? µ = inches and standard deviation b) Below what height does the bottom 10% of heights lie? Example 8: Using z scores If the distribution of scores on a test is normal with mean 75 σ = 8, and the top 10% will get an A s and the bottom 10% will get F s. Find the following: a) The score you need to get an A. µ = inches and standard deviation b) The score below which you will get an F.