Chapter 3 The Normal DistributionsDensity Curves vs. Histograms:11023456789246810121416Histogram of 66 Observations- Histogram displays count of obs in a given category16/66 = .242440/66 = .6061- Density curves describe what proportion of the observations fall into each category (not the count of the obs)=116 out of 66 obs fall between 5-6

A Density Curve is a curve that:- Is always on or above the horizontal axis, and- has area exactly 1 underneath it. Describes the overall pattern of the distributionAreas under the curve and above any range of values on the x-axis represent the proportion of total observations taking those valuesMean and Median of a Density Curve The Median of a density curve is the = areas point of the area on one side and on the other.. The Mean of a density curve is the point at which it would balance if made of solid material

Symmetric Density CurveM x.5.5

M

- Density Curves are idealized descriptions of the data distribution so we need to distinguish between the actually computed values of the mean and standard deviation and the mean and standard deviation of a density curvethey MAY be differentm (mu)s (sigma)** English = Computed Values / Greek = Density Curve **

Stats 1.3 ContinuedNormal Curves:

Symmetric Single-Peaked Bell shaped Describes a Normal Distribution All normal distributions have the same overall shape Described by giving the mean () and std. deviation () as N() ex: N(25, 4.7) Mean = Median

Locating

68 - 95 - 99.7 Rule**Applies to all normal distributions**

ex: Heights of women 18-24: N(64.5, 2.5) (inches)2 = 2 x (2.5) = 5 inches 3 = 3 x (2.5) = 7.5 inches 69.572259.557

Why is normal important?Represents some distributions of real data (ex: test scores, biological populations, etc)

Provides good approximations to chance outcomes (ex: coin tosses)

Statistical Inference procedures based on normal distributions work well for roughly symmetric distributions.

WARNING! WARNING! MANY DISTRIBUTIONSARE NOT NORMAL!!

1.3 contd - The Standard Normal Distribution Standardized Observations / Standardizing Theory: All normal distributions are the same if we measure in units of size from as the center = std. deviations / common scale Changing to these units is called standardizing Notation / Formula for Standardizing an observation:(Z = z-score or z-number) Z can be negative - tells us how many std. devs. away from the mean we are AND in which direction (+/-)

ex: if x > , Z = (+) / if x < , Z = (-) ex: Heights of women 18-24 = N(64.5, 2.5) ex: z-score for height of 68 inchesex: z-score for height of 60 inches

ex: Weights of different speciesSouth American vs. African CrocodilesSouth American: N(1200, 200)African: N(900, 100)For a South American croc at 1400 lbs and an African croc at 1100 lbs, which is the greater anomaly?South American z-score:= 1.0African z-score:= 2.0** The African croc at +2 is more of an anomaly **

1.3 contd - Normal Distribution Calculations Calculating area under a density curve Area = proportion of the observations in a distribution Because all normal distributions are the same after we standardize, we can use one normal curve to compute areas for ANY normal distribution. The one curve we use is called the Standard Normal Distribution Can use the calculator OR Table-A inside front cover of book to find areas under the std. normal curve.

ex: What proportion of all women 18-24 are less than 68 in. tall?Area = .9192

Finding Normal Curve Areas - Calculator Steps2)

Finding Normal Curve Areas - Using Table -A1) Find z-value to tenths place2) Find correct hundreths place column3) Find intersection of z-value and hundreths place

Finding Normal Proportions1) State the problem in terms of the desired variable.2) Standardize / Draw pic 3) Use Calculator / Tbl-A ex: Blood cholesterol levels distribute normally in people of the same age and sex. For 14 yr old males the distribution is N(170, 30) (mg/dl). Levels above 240 will require medical attention. Find the proportion of 14 yr old males that have a blood cholesterol level above 240.1) x = level of cholesterol

= .0098 --> about 1% Finding a value given a proportionex: SAT scores distribute normally N(430, 100). What score would be necessary to be in the top 10% of all scores?

Finding value given a proportion - Calculator2ndVARSinvNorm(area to the left of value, mean, std.dev)= 558 or greater to be in top 10%(DISTR)

Finding a value given a proportion - Reverse Table-A lookup 1) Scan table for area value closest to what is needed2) Trace left and up to obtain z-number to two placesFor area = .9 to the left, z = 1.28Plug z-value into z-formula solved for x:

x - z = For N(430, 100) and z = 1.28, x = (1.28)(100) + 430 x = 558Solving z formula for x: