Enter these data Enter these data into your into your

calculator!!!calculator!!!A researcher measured 30 newly hatched chicks and recordedtheir weights in grams as shown below.

79.5 87.5 88.5 89.2 91.6 84.5 82.1 82.3 85.7 89.884.0 84.8 88.2 88.2 82.9 89.8 89.2 94.1 88.0 91.191.8 87.0 87.7 88.0 85.4 94.4 91.3 86.4 85.7 86.0

Boxplots,Boxplots,Standard Standard DeviationDeviation

Section 9.7b

Last day of notes for

PreCalculus!!

BoxplotsBoxplotsBoxplot (Box-and-Whisker Plot) – a graphicalrepresentation of the five-number summary of a data set.

Consists of a central rectangle (box) that extends from thefirst quartile to the third quartile, with a vertical segmentmarking the median. Line segments (whiskers) extend at theends of the box to the minimum and maximum values.

Practice with BoxplotsPractice with Boxplots

59.0,64.1,68.75,71.65,72.6

Let’s create a boxplot for the five-number summary for male lifeexpectancies in South American nations from last class:

55 60 65 70 75 80

59.064.1 68.75 71.65

72.6

MaxMinMedQ1 Q3

Are these data skewed in any way??? Skewed Left!!!Skewed Left!!!

Practice with BoxplotsPractice with Boxplots

59.0,64.1,68.75,71.65,72.6

Draw boxplots for the male and female data for life expectanciesin South American nations and describe the information displayed.

55 60 65 70 75 80

Males:

66.2,70.25,74.5,77.7,79.4Females:

Males:

Females:

Practice with BoxplotsPractice with BoxplotsDraw boxplots for the male and female data for life expectanciesin South American nations and describe the information displayed.

55 60 65 70 75 80

Males:

Females:

The middle half of female life expectancies are all greaterthan the median of male life expectancies. The medianlife expectancy for women is greater than the maximumfor the men.

Practice with BoxplotsPractice with Boxplots

14,28,16,39,61,33,23,26,8,13,9,5Create a boxplot for Roger Maris’s annual home run totals:

Five-Number Summary: 5,11,19.5,30.5,61

0 10 20 30 40 50 60 70

61511 19.5 30.5

Is Maris’s 61 home run total an outlier ??? How can we tell???

Practice with BoxplotsPractice with BoxplotsOur “Rule of Thumb”:

A number in a data set can be considered an outlier if it is morethan 1.5 x IQR below the first quartile or above the third quartile.

Five-Number Summaryfor Maris’s home run totals: 5,11,19.5,30.5,61

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 BoxplotsPractice with BoxplotsWhen dealing with outliers, sometimes a modified boxplot isused, showing the outliers as isolated points…

Two boxplots for Roger Maris’s annual home run totals

0 10 20 30 40 50 60 70

Regular Boxplot:

Modified Boxplot:

Variance andVariance andStandard DeviationStandard Deviation

1 2, , , nx x x

These are measures of variability that are better indicators thanthe interquartile range…

The standard deviation of the numbers is

21

1 n

ii

x xn

where x denotes the mean. The variance is , the squareof the standard deviation.

2

Note: The standard deviation is generally not very resistant…

Variance andVariance andStandard DeviationStandard Deviation

Most calculators actually give two standard deviations, theother denoted as s:

21

1

1

n

ii

s x xn

The difference is that applies to the true parameter, whichmeans only if the data is from the whole population. If the datacomes from a sample, then the s formula actually gives a betterestimate of the parameter…

Variance andVariance andStandard DeviationStandard Deviation

A researcher measured 30 newly hatched chicks and recordedtheir weights in grams as shown below.

79.5 87.5 88.5 89.2 91.6 84.5 82.1 82.3 85.7 89.884.0 84.8 88.2 88.2 82.9 89.8 89.2 94.1 88.0 91.191.8 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 deviationfor the weights of newly hatched chicks. Are these measuresuseful in this case, or should we use the five-number summary?

First, enter the data into your calculator L1

Then, choose STAT CALC 1-Var Stats ENTER

Variance andVariance andStandard DeviationStandard Deviation

A researcher measured 30 newly hatched chicks and recordedtheir weights in grams as shown below.

Mean = x = 87.49 grams

xStandard Deviation = S = 3.510 grams

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 Our last new info in Chapter 9…in Chapter 9…

These Distributions are Normal…

First, use your calculator to create a histogram of the data onweight of newly hatched chicks from last class (use Xscl = 2and window [ 75, 98 ] by [ 0, 10 ] ):

79.5 87.5 88.5 89.2 91.6 84.5 82.1 82.3 85.7 89.884.0 84.8 88.2 88.2 82.9 89.8 89.2 94.1 88.0 91.191.8 87.0 87.7 88.0 85.4 94.4 91.3 86.4 85.7 86.0

Frequency Table:78.0-79.9 180.0-81.9 082.0-83.9 384.0-85.9 686.0-87.9 588.0-89.9 990.0-91.9 492.0-93.9 094.0-95.9 2

Histogram:

First, use your calculator to create a histogram of the data onweight of newly hatched chicks from last class (use Xscl = 2and window [ 75, 98 ] by [ 0, 10 ] ):

Histogram:

What do you notice about this histogram?

The distribution is roughlyThe distribution is roughlysymmetric, with no strong outlierssymmetric, with no strong outliersor skewness. Most of the dataor skewness. Most of the datacluster around a central point.cluster around a central point.

The distribution is approximately NORMAL!!!

Normal DistributionsIn math-land, “normal” is actually a technical term…

Graph the given function in the window [–3, 3] by [0, 1]:

2 2xf x eThis curve, called the Gaussian curve or normal curve is aprecise mathematical model for normal behavior.

A great many naturally-occurring phenomena yield anormal distribution when displayed as a histogram.

Examples???Examples???

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

and1 1

• Approximately 95% of the data lie between

and2 2

• Approximately 99.7% of the data lie between

and3 3

The 68-95-99.7 Rule

About 68% of the data in any normal distributionAbout 68% of the data in any normal distributionlie within 1 standard deviation of the mean…lie within 1 standard deviation of the mean…

The 68-95-99.7 Rule

About 95% of the data in any normal distributionAbout 95% of the data in any normal distributionlie within 2 standard deviations of the mean…lie within 2 standard deviations of the mean…

The 68-95-99.7 Rule

About 99.7% of the data in any normal distributionAbout 99.7% of the data in any normal distributionlie within 3 standard deviations of the mean…lie within 3 standard deviations of the mean…

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 89.884.0 84.8 88.2 88.2 82.9 89.8 89.2 94.1 88.0 91.191.8 87.0 87.7 88.0 85.4 94.4 91.3 86.4 85.7 86.0

Mean = x = 87.49 gramsxStandard Deviation = S = 3.510 grams

What are the mean and standard deviation???

Based on these data, would a chick weighing 95 grams be in thetop 2.5% of all newly hatched chicks?

We assume that the weights of newly hatched chicks areWe assume that the weights of newly hatched chicks arenormally distributed in the whole population. Since we donormally distributed in the whole population. Since we donot know the mean and standard deviation for the wholenot know the mean and standard deviation for the wholepopulation (the parameters and ), we use population (the parameters and ), we use xx and S and Sas estimates.as estimates.

x

79.5 87.5 88.5 89.2 91.6 84.5 82.1 82.3 85.7 89.884.0 84.8 88.2 88.2 82.9 89.8 89.2 94.1 88.0 91.191.8 87.0 87.7 88.0 85.4 94.4 91.3 86.4 85.7 86.0

Mean = x = 87.49 gramsxStandard Deviation = S = 3.510 grams

Based on these data, would a chick weighing 95 grams be in thetop 2.5% of all newly hatched chicks?

Because 95% of the data must lie within 2 standardBecause 95% of the data must lie within 2 standarddeviations, 2.5% of the data must be beyond this limit ondeviations, 2.5% of the data must be beyond this limit oneither end. To be in the top 2.5%, a chick will have to weigheither end. To be in the top 2.5%, a chick will have to weighat least 2 standard deviations more than the mean:at least 2 standard deviations more than the mean:

2 87.49 2 3.51 94.51x Sx grams

Since 95 > 94.51, a 95-gram chick is indeed in the top 2.5%!!!Since 95 > 94.51, a 95-gram chick is indeed in the top 2.5%!!!

Proctor measures the mean and standard deviation of his Ch. 9test to be 44.3 points and 3.7 points, respectively. Assumingthe test scores fall on a normal distribution, answer the following:

1. Approximately 68% of all students earned scores betweenwhat two numbers?

2. Yolanda earned a 33.2 on the test, meaning that she scoredin the bottom _______ percent of students taking the test.

3. Pip earned a 51.7 on the test, meaning that she scored betterthan what percentage of students taking the test?

Bottom 0.15 percentBottom 0.15 percent

Between 40.6 and 48 points

Better than 97.5 percent of studentsBetter than 97.5 percent of students