Density Curves and the Normal Distribution

Post on 31-Dec-2015

45 views 1 download

Tags:

description

Density Curves and the Normal Distribution. What is a Density Curve?. Consider the following histogram Remember that the shape of the histogram depends on the width of the classes we choose for the histogram, our data might look like this!. - PowerPoint PPT Presentation

Transcript of Density Curves and the Normal Distribution

Density Curves and theNormal Distribution

What is a Density Curve?

• Consider the following histogram

• Remember that the shape of the histogram depends on the width of the classes we choose for the histogram, our data might look like this!

Co

un

t

1

2

3

4

-4 -2 0 2 4Symmetric

Shape Histogram

Co

un

t

1

2

3

4

5

6

7

8

-4 -2 0 2 4Symmetric

Shape Histogram

• Using a smooth curve to describe the distribution will eliminate the effect of our choice of classes.

• Both the bars and the curve represent the proportion of observations on an interval

• Both the bars and the curve represent all the observations in the distribution

Co

un

t

1

2

3

4

5

Scores-0.2 -0.1 0.0 0.1 0.2

Count = x normalDensity

Shape Histogram

What is a Density Curve

• A Density Curve is a curve that:• Is always on or above the x-axis• Has an area of exactly 1 below it

Since the density curve represents the entire distribution, the area under the curve on any interval represents the proportion of observations in that interval.

Do all Density Curves look alike?

• We’ve already discovered that distributions take on different shapes.• Some are symmetric• Some are left-skewed• Some are right-skewed

• Since a density curve represents the distributions, they too take on different shapes.

Density Curvesy

0.04

0.08

0.12

0.16

0.20

x2 4 6 8 10 12

y = x chiSquareDensity

no data Function Plot

y

0.1

0.2

0.3

0.4

0.5

x1 2 3 4 5 6 7 8

y = x uniformDensity

no data Function Plot

y

0.1

0.2

0.3

0.4

0.5

x-3 -2 -1 0 1 2 3 4

y = x normalDensity

no data Function Plot

Measures of CenterC

ou

nt

1

2

3

4

symmetric-4 -2 0 2 4

Count = x normalDensitysymmetric mean = 0

symmetric median = 0

Collection 1 Histogram

In a symmetric distribution the mean, median and mode are located in the same place.

y0.04

0.08

0.12

0.16

0.20

x2 4 6 8 10 12

y = x chiSquareDensity

no data Function Plot

In a skewed distribution, the mean is pulled in the direction of the tail.

Normal DistributionThe Empirical Rule

• All normal distributions have particular characteristics with regard to the area under the curve on a given interval.

• We can use these characteristics to answer questions regarding the distributions.

• For example:• Within 1σ of the mean (-1σ < μ

< 1σ ), lies 68% of the area.y

0.10

0.20

0.30

0.40

x-3 -2 -1 0 1 2 3

y = x normalDensity = -1

= 1

no data Function Plot

68%

• Within 2σ of the mean, (-2σ < μ < 2σ ) lies 95% of the data

y

0.10

0.20

0.30

0.40

x-3 -2 -1 0 1 2 3

y = x normalDensity = -2

= 2

no data Function Plot

95%

• Within 2σ of the mean, (-2σ < μ < 2σ ) lies 95% of the data

y

0.10

0.20

0.30

0.40

x-3 -2 -1 0 1 2 3

y = x normalDensity = -3

= 3

no data Function Plot

99.7%

y

0.05

0.10

0.15

0.20

0.25

0.30

0.35

0.40

0.45

x-3 -2 -1 0 1 2 3

y = x normalDensity = -3

= 3 = -2 = -1

= 1 = 2

no data Function Plot

34%34%

13.5% 13.5%

2.35% 2.35%.15%

.15%

By symmetry, we can find the approximate area under the curve for each of the sections created in a normal curve…

Using the Empirical Rule to answer questions.

• There are three ways in which to ask the same question when dealing with a normal distribution.

• What is the area under the curve between and ?

• What is the proportion (percent) of observations between and ?

• What is the probability that an individual chosen at random will be between and ?

• Did you notice that all 3 questions asked about an interval?• That’s because (as we find out in

calculus) that there is NO area under the curve at a particular point. In a continuous distribution, as is the normal curve, we can only find area on an interval.

• The answers for all three of these questions are found in the same way! That’s because they are all asking the same question.

An Example:

• The army reports that the distribution of head circumference among male soldiers is approximately normal with mean 22.3 inches and standard deviation 1.1 inches. Use the empirical rule to answer the following:

Let’s start by drawing and labeling a normal distribution.

y

0.1

0.2

0.3

0.4

0.5

x-3 -2 -1 0 1 2 3

y = x normalDensity

no data Function Plot

19.5 20.6 21.7 22.8 23.9 25 26.1

What percent of soldiers have head circumference greater than 23.9 inches?

What percent of soldiers have head circumference greater than 23.9 inches?

.15% 2.35% 13.5% 34% 34% 13.5% 2.35% .15%

If we add the percentages from the shaded area we find that 13.5% + 2.35%+.15% = 16%

So, approximately 16% of soldiers have head circumferences greater than 23.9 inches

19.5 20.6 21.7 22.8 23.9 25 26.1

Finding percentiles• When we want to

know where an individual is in relation to the rest of the distribution we use percentiles.

• So, a head circumference of 23.0 inches would be what percentile?

y

0.05

0.10

0.15

0.20

0.25

0.30

0.35

0.40

0.45

x-3 -2 -1 0 1 2 3

y = x normalDensity = -3

= 3 = -2 = -1

= 1 = 2

no data Function Plot

.15% 2.5% 16% 50% 84% 87.5% 89.85%

.15% 2.35% 13.5% 34% 34% 13.5% 2.35% .15%

Percentiles

So, a head circumference of 23.9 inches would be at the 84th% percentile.

y

0.05

0.10

0.15

0.20

0.25

0.30

0.35

0.40

0.45

x-3 -2 -1 0 1 2 3

y = x normalDensity = -3

= 3 = -2 = -1

= 1 = 2

no data Function Plot

.15% 2.5% 16% 50% 84% 87.5% 89.85%

19.5 20.6 21.7 22.8 23.9 25 26.1

So, using the empirical percentages of the normal distribution is as simple as finding the corresponding area and adding the percentages for those areas.

Additional References

• YMM Text Pages 64-82• Homework Assignment:

Assignment 2.1• Please be sure to follow the four

steps outlined on the next slide when completing a normal distribution problem.

Sample Question: What percent of soldiers have head circumference greater than 23.9 inches?

• Write the question as a probability statement:

• P( X > 23.)• Draw a normal curve and label• Shade the appropriate area

− See slide 17.

• Find the appropriate area by adding corresponding areas

• Write the solution in context of the problem.