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Chapter 6 The Normal Distribution

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Normal Distributions Bell Curve Area under entire curve = 1 or 100% Mean = Median – This means the curve is symmetric

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Normal Distributions Two parameters – Mean μ (pronounced “meeoo”) Locates center of curve Splits curve in half Shifts curve along x-axis – Standard deviation σ (pronounced “sigma”) Controls spread of curve Smaller σ makes graph tall and skinny Larger σ makes graph flat and wide Ruler of distribution Write as N(μ,σ)

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Standard Normal (Z) World Perfectly symmetric Centered at zero – Half numbers below the mean and half above. Total area under the curve is 1. – Can fill in as percentages across the curve.

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Standard Normal Distribution Puts all normal distributions on same scale – z has center (mean) at 0 – z has spread (standard deviation) of 1

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Standard Normal Distribution z = # of standard deviations away from mean μ – Negative z, number is below the mean – Positive z, number is above the mean Written as N(0,1)

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Portal from X-world to Z-world z has no units (just a number) Puts variables on same scale – Center (mean) at 0 – Spread (standard deviation) of 1 Does not change shape of distribution

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Standardizing Variables z = # of standard deviations away from mean – Negative z – number is below mean – Positive z – number is above mean

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Standardizing – Y ~ N(70,3). Standardize y = 68. – y = 68 is 0.67 standard deviations below the mean

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Your Table is Your Friend Get out your book and find your Z-table. Look for a legend at the top of the table. – Which way does it fill from? Find the Z values. Find the “middle” of the table. – These are the areas or probabilities as you move across the table. Notice they are 50% in the middle and 100% at the end.

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Areas under curve Another way to find probabilities when values are not exactly 1, 2, or 3 away from µ is by using the Normal Values Table – Gives amount of curve below a particular value of z – z values range from –3.99 to 3.99 – Row – ones and tenths place for z – Column – hundredths place for z

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Finding Values What percent of a standard Normal curve is found in the region Z < -1.50? P(Z < –1.50) – Find row –1.5 – Find column.00 – Value = 0.0668

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Finding Values P(Z < 1.98) – Find row 1.9 – Find column.08 – Value = 0.9761

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Finding values What percent of a std. Normal curve is found in the region Z >-1.65? P(Z > -1.65) – Find row –1.6 – Find column.05 – Value from table = 0.0495 – P(Z > -1.65) = 0.9505

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Finding values P(Z > 0.73) – Find row 0.7 – Find column.03 – Value from table = 0.7673 – P(Z > 0.73) = 0.2327

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Finding values What percent of a std. Normal curve is found in the region 0.5 < Z < 1.4? P(0.5 < Z < 1.4) – Table value 1.4 = 0.9192 – Table value 0.5 = 0.6915 – P(0.5 < Z < 1.4) = 0.9192 – 0.6915 = 0.2277

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Finding values P(–2.3 < Z < –0.05) – Table value –0.05 = 0.4801 – Table value –2.3 = 0.0107 – P(–2.3 < Z < –0.05) = 0.4801 – 0.0107 = 0.4694

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Finding values Above what z-value do the top 15% of all z-value lie, i.e. what value of z cuts offs the highest 15%? – P(Z > ?) = 0.15 – P(Z < ?) = 0.85 z = 1.04

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Finding values Between what two z-values do the middle 80% of the obs lie, i.e. what values cut off the middle 80%? – Find P(Z < ?) = 0.10 – Find P(Z < ?) = 0.90 – Must look inside the table P(Z<-1.28) = 0.10 P(Z<1.28) = 0.90

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Solving Problems The height of men is known to be normally distributed with mean 70 and standard deviation 3. – Y ~ N(70,3)

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Solving Problems What percent of men are shorter than 66 inches? P(Y < 66) = P(Z< ) = P(Z<-1.33) = 0.0918

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Solving Problems What percent of men are taller than 74 inches? P(Y > 74) = 1-P(Y<74) = 1 – P(Z< ) = 1 – P(Z<1.33) = 1 – 0.9082 = 0.0918

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Solving Problems What percent of men are between 68 and 71 inches tall? P(68 < Y < 71) = P(Y<71) – P(Y<68) =P(Z< )-P(Z< ) =P(Z<0.33) - P(Z<-0.67) = 0.6293 – 0.2514 = 0.3779

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Solving Problems Scores on SAT verbal are known to be normally distributed with mean 500 and standard deviation 100. X ~ N(500,100)

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Solving Problems Your score was 650 on the SAT verbal test. What percentage of people scored better? P(X > 650) = 1 – P(X<650) = 1 – P(Z< ) = 1 – P(Z<1.5) = 1 – 0.9332 = 0.0668

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Solving Problems To solve a problem where you are looking for y-values, you need to rearrange the standardizing formula:

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Solving Problems What would you have to score to be in the top 5% of people taking the SAT verbal? – P(X > ?) = 0.05? – P(X < ?) = 0.95?

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Solving Problems P(Z < ?) = 0.95? z = 1.645 x is 1.645 standard deviations above mean x is 1.645(100) = 164.5 points above mean x = 500 + 164.5 = 664.5 SAT verbal score: at least 670

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Solving Problems Between what two scores would the middle 50% of people taking the SAT verbal be? – P(x 1 = –? < X < x 2 =?) = 0.50? – P(-0.67 < Z < 0.67) = 0.50 x 1 = (-0.67)(100)+500 = 433 x 2 = (0.67)(100)+500 = 567

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Solving Problems Cereal boxes are labeled 16 oz. The boxes are filled by a machine. The amount the machine fills is normally distributed with mean 16.3 oz and standard deviation 0.2 oz.

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Solving Problems What is the probability a box of cereal is underfilled? Underfilling means having less than 16 oz. P(Y < 16) = P(Z< ) = P(Z< -1.5) = 0.0668

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Moving to Random Sample Even when we assume a variable is normally distributed if we take a random sample we need to adjust our formula slightly.

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Determining Normality If you need to determine normality I want you to use your calculators to make a box plot and to look for symmetry.

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