unit 4: normal distributions part 1 statistics mr. evans
DESCRIPTION
Normal Distribution The normal distribution of a set of data is a very important class of statistical distributions.TRANSCRIPT
Unit 4: Normal Distributions
Part 1StatisticsMr. Evans
Focus Points
• Graph a normal curve and summarize its important properties
• Apply the empirical rule to solve real-world problems.
Normal Distribution
• The normal distribution of a set of data is a very important class of statistical distributions.
Properties of Normal Curve
• The curve is bell-shaped with the highest point over the mean μ• It is symmetrical about the vertical line through the mean μ• The curve approaches the horizontal axis but never touches or
crosses it• The transition points between cupping upward and downward
occur above μ + σ and μ – σ
Empirical Rule
• For a distribution that is symmetrical and bell-shaped• Approximately 68% of the data values will lie within one
standard deviation on each side of the mean• Approximately 95% of the data values will lie within two
standard deviation on each side of the mean• Approximately 99.7% of the data values will lie within three standard deviations on each side of the mean
Guided Exercise # 1
• Sketch the graph with mean μ = 35 and standard deviation σ = 8.
11 19 27 35 43 51 59
Guided Exercise # 1
• Sketch the graph with mean μ = 19.3 and standard deviation σ = 4.7.
5.2 9.9 14.6 19.3 24.0 28.7 33.4
Empirical Rule Graphs
Empirical Rule Graphs
Empirical Rule Graphs
Empirical Rule Graphs
Empirical Rule Graphs
Empirical Rule Graphs
Guided Exercise # 2
The yearly wheat yield per acre on a particular farm is normally distributed with mean μ = 35 bushels and standard deviation σ = 8.a) Shade the area under the curve that represents between
19 and 35 bushels. b) Find the percentage of area over the interval between
19 and 35.
11 19 27 35 43 51 59
If we don’t match any specific graph we can use the 4th graph
x x 13.5+34=47.5%
Guided Exercise # 2
Assuming the heights of college women are normally distributed with mean μ = 65 in and standard deviation σ = 2.5 in.a) Shade the area under the curve that represents shorter
than 67.5 in.b) Find the percentage of area under 67.5 in.x x
xxx
57.5 60 62.5 65 67.5 70 72.5
Matches the right area of the to right graph on the back
84%