Chapter 6The Normal Distribution

Normal Distributions

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

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)

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.

Standard Normal Distribution

• Puts all normal distributions on same scale

– z has center (mean) at 0– z has spread (standard deviation) of 1

y

z

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)

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

y

z

Standardizing Variables

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

Standardizing

– Y ~ N(70,3). Standardize y = 68.

– y = 68 is 0.67 standard deviations below the mean

67.03

7068

y

z

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.

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

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

Finding Values

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

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

Finding values

• P(Z > 0.73)– Find row 0.7– Find column .03– Value from table

= 0.7673– P(Z > 0.73) = 0.2327

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

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

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

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

Solving Problems

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

Solving Problems

• What percent of men are shorter than 66 inches?

• P(Y < 66)

= P(Z< )

= P(Z<-1.33) = 0.0918

3

7066

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

3

7074

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

3

7071 3

7068

Solving Problems

• Scores on SAT verbal are known to be normally distributed with mean 500 and standard deviation 100.

• X ~ N(500,100)

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

100

500650

Solving Problems

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

zx

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?

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

Solving Problems

• Between what two scores would the middle 50% of people taking the SAT verbal be?– P(x1= –? < X < x2=?) = 0.50?

– P(-0.67 < Z < 0.67) = 0.50• x1 = (-0.67)(100)+500 = 433

• x2 = (0.67)(100)+500 = 567

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.

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

2.0

3.1616

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.

n

sxx

z

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.