Enter these data into your calculator!!! A researcher measured 30 newly hatched chicks and recorded their weights in grams as shown below. 79.5 87.5 88.5 89.2 91.6 84.5 82.1 82.3 85.7 89.8 84.0 84.8 88.2 88.2 82.9 89.8 89.2 94.1 88.0 91.1 91.8 87.0 87.7 88.0 85.4 94.4 91.3 86.4 85.7 86.0

Boxplots, Standard Deviation Section 9.7b  Last day of notes for PreCalculus!!

Boxplots Boxplot (Box-and-Whisker Plot) – a graphical representation of the five-number summary of a data set. Consists of a central rectangle (box) that extends from the first quartile to the third quartile, with a vertical segment marking the median. Line segments (whiskers) extend at the ends of the box to the minimum and maximum values.

Practice with Boxplots Let’s create a boxplot for the five-number summary for male life expectancies in South American nations from last class: 64.1 68.75 71.65 59.0 72.6 Min Max Q1 Med Q3 55 60 65 70 75 80 Are these data skewed in any way???  Skewed Left!!!

Practice with Boxplots Draw boxplots for the male and female data for life expectancies in South American nations and describe the information displayed. Males: Females: Males: Females: 55 60 65 70 75 80

Practice with Boxplots Draw boxplots for the male and female data for life expectancies in South American nations and describe the information displayed. Males: Females: 55 60 65 70 75 80 The middle half of female life expectancies are all greater than the median of male life expectancies. The median life expectancy for women is greater than the maximum for the men.

Practice with Boxplots Create a boxplot for Roger Maris’s annual home run totals: Five-Number Summary: 11 19.5 30.5 5 61 10 20 30 40 50 60 70 Is Maris’s 61 home run total an outlier ??? How can we tell???

Practice with Boxplots Our “Rule of Thumb”: A number in a data set can be considered an outlier if it is more than 1.5 x IQR below the first quartile or above the third quartile. Five-Number Summary for Maris’s home run totals: IQR = 30.5 – 11 = 19.5 Q3 + 1.5 x IQR = 30.5 + (1.5)(19.5) = 59.75 Since 61 > 59.75, our new rule identifies it as an outlier.

Practice with Boxplots When dealing with outliers, sometimes a modified boxplot is used, showing the outliers as isolated points… Two boxplots for Roger Maris’s annual home run totals Regular Boxplot: Modified Boxplot: 10 20 30 40 50 60 70

Variance and Standard Deviation These are measures of variability that are better indicators than the interquartile range… The standard deviation of the numbers is where x denotes the mean. The variance is , the square of the standard deviation. Note: The standard deviation is generally not very resistant…

Variance and Standard Deviation Most calculators actually give two standard deviations, the other denoted as s: The difference is that applies to the true parameter, which means only if the data is from the whole population. If the data comes from a sample, then the s formula actually gives a better estimate of the parameter…

Variance and Standard Deviation A researcher measured 30 newly hatched chicks and recorded their weights in grams as shown below. 79.5 87.5 88.5 89.2 91.6 84.5 82.1 82.3 85.7 89.8 84.0 84.8 88.2 88.2 82.9 89.8 89.2 94.1 88.0 91.1 91.8 87.0 87.7 88.0 85.4 94.4 91.3 86.4 85.7 86.0 Based on the sample, estimate the mean and standard deviation for the weights of newly hatched chicks. Are these measures useful in this case, or should we use the five-number summary? First, enter the data into your calculator  L 1 Then, choose STAT  CALC  1-Var Stats  ENTER

Variance and Standard Deviation A researcher measured 30 newly hatched chicks and recorded their weights in grams as shown below. Mean = x = 87.49 grams Standard Deviation = S = 3.510 grams x Because these data have no real outliers or skewness, the mean and standard deviation are appropriate measures (there is no need to include a five-number summary).

Our last new info in Chapter 9… These Distributions are Normal…

First, use your calculator to create a histogram of the data on weight of newly hatched chicks from last class (use Xscl = 2 and window [ 75, 98 ] by [ 0, 10 ] ): 79.5 87.5 88.5 89.2 91.6 84.5 82.1 82.3 85.7 89.8 84.0 84.8 88.2 88.2 82.9 89.8 89.2 94.1 88.0 91.1 91.8 87.0 87.7 88.0 85.4 94.4 91.3 86.4 85.7 86.0 Frequency Table: Histogram: 78.0-79.9 1 80.0-81.9 0 82.0-83.9 3 84.0-85.9 6 86.0-87.9 5 88.0-89.9 9 90.0-91.9 4 92.0-93.9 0 94.0-95.9 2

 The distribution is approximately NORMAL!!! First, use your calculator to create a histogram of the data on weight of newly hatched chicks from last class (use Xscl = 2 and window [ 75, 98 ] by [ 0, 10 ] ): Histogram: What do you notice about this histogram? The distribution is roughly symmetric, with no strong outliers or skewness. Most of the data cluster around a central point.  The distribution is approximately NORMAL!!!

Normal Distributions Examples??? In math-land, “normal” is actually a technical term… Graph the given function in the window [–3, 3] by [0, 1]: This curve, called the Gaussian curve or normal curve is a precise mathematical model for normal behavior. A great many naturally-occurring phenomena yield a normal distribution when displayed as a histogram. Examples???

The 68-95-99.7 Rule If the data for a population are normally distributed with mean and standard deviation , then Approximately 68% of the data lie between and Approximately 95% of the data lie between and Approximately 99.7% of the data lie between and

The 68-95-99.7 Rule About 68% of the data in any normal distribution lie within 1 standard deviation of the mean…

The 68-95-99.7 Rule About 95% of the data in any normal distribution lie within 2 standard deviations of the mean…

The 68-95-99.7 Rule About 99.7% of the data in any normal distribution lie within 3 standard deviations of the mean…

Standard Deviation = S = 3.510 grams Returning to the data for newly hatched chicks: 79.5 87.5 88.5 89.2 91.6 84.5 82.1 82.3 85.7 89.8 84.0 84.8 88.2 88.2 82.9 89.8 89.2 94.1 88.0 91.1 91.8 87.0 87.7 88.0 85.4 94.4 91.3 86.4 85.7 86.0 What are the mean and standard deviation??? Mean = x = 87.49 grams Standard Deviation = S = 3.510 grams x Based on these data, would a chick weighing 95 grams be in the top 2.5% of all newly hatched chicks? We assume that the weights of newly hatched chicks are normally distributed in the whole population. Since we do not know the mean and standard deviation for the whole population (the parameters and ), we use x and S as estimates. x

Standard Deviation = S = 3.510 grams 79.5 87.5 88.5 89.2 91.6 84.5 82.1 82.3 85.7 89.8 84.0 84.8 88.2 88.2 82.9 89.8 89.2 94.1 88.0 91.1 91.8 87.0 87.7 88.0 85.4 94.4 91.3 86.4 85.7 86.0 Mean = x = 87.49 grams Standard Deviation = S = 3.510 grams x Based on these data, would a chick weighing 95 grams be in the top 2.5% of all newly hatched chicks? Because 95% of the data must lie within 2 standard deviations, 2.5% of the data must be beyond this limit on either end. To be in the top 2.5%, a chick will have to weigh at least 2 standard deviations more than the mean: grams Since 95 > 94.51, a 95-gram chick is indeed in the top 2.5%!!!

Better than 97.5 percent of students Proctor measures the mean and standard deviation of his Ch. 9 test to be 44.3 points and 3.7 points, respectively. Assuming the test scores fall on a normal distribution, answer the following: 1. Approximately 68% of all students earned scores between what two numbers? Between 40.6 and 48 points 2. Yolanda earned a 33.2 on the test, meaning that she scored in the bottom _______ percent of students taking the test. Bottom 0.15 percent 3. Pip earned a 51.7 on the test, meaning that she scored better than what percentage of students taking the test? Better than 97.5 percent of students