Chapter 3

Essential Statistics Chapter 3 1

Chapter 3

The Normal Distributions

Z-Score Explained http://www.youtube.com/watch?v=AT-HH0W_swA&feature=related

Basics of Using the Std Normal Table http://www.youtube.com/watch?v=y6sbghmHwQA&feature=related

Normal Distribution & Z-score http://www.youtube.com/watch?v=mai23vW8uFM&feature=related


We’ll Learn The Topics

Review Histogram Density Curve Normal Distribution 68 – 95 – 99.7 Rule Z-score Standard Normal Distribution



Density CurvesExample: here is a histogram of vocabulary scores of 947 seventh graders.

- We can describe the histogram with a smooth curve, a bell- shaped curve.

- It corresponding to a normal distribution Model.


Density Curves

Example: the areas of the shaded bars in this histogram represent the proportion of scores that are less than or equal to 6.0. This proportion in the observed data is equal to 0.303.


Density Curves■ now the area under the smooth curve to the left of 6.0 is shaded.

■ The scale is adjusted, the total area under the curve is exactly 1, this curve is called a density curve.

■ The proportion of the area to the left of 6.0 is now equal to 0.293.


Density Curves

Always on or above the horizontal axis

Have area exactly 1 underneath curve

Display the bell-shaped pattern of a distribution

A histogram becomes a density curve if the scale is adjusted so that the total area of the bars is 1.


Mean & Standard Deviation

The mean and standard deviation computed from actual observations (data) are denoted by and s, respectively

x

The mean and standard deviation of the distribution represented by the density curve are denoted by µ (“mu”) and (“sigma”), respectively.

The mean of a density curve is the "balance point" of the curve.


Bell-Shaped Curve:The Normal Distribution

standard deviation

mean


The Normal Distribution

■ Knowing the mean (µ) and standard deviation () allows us to make various conclusions about Normal distributions.

■ Notation: N(µ,).


68-95-99.7 Rule forAny Normal Curve

68% of the observations fall within one standard deviation of the mean

95% of the observations fall within two standard deviations of the mean

99.7% of the observations fall within three standard deviations of the mean



68%

+- µ

+3-3

99.7%

µ

+2-2

95%

µ




Health and Nutrition Examination Study of 1976-1980

Heights of adult men, aged 18-24

– mean: 70.0 inches

– standard deviation: 2.8 inches

– heights follow a normal distribution, so we

have that heights of men are N(70, 2.8).



68-95-99.7 Rule for men’s heights68% are between 67.2 and 72.8 inches

[ µ = 70.0 2.8 ]

95% are between 64.4 and 75.6 inches[ µ 2 = 70.0 2(2.8) = 70.0 5.6 ]

99.7% are between 61.6 and 78.4 inches[ µ 3 = 70.0 3(2.8) = 70.0 8.4 ]



What proportion of men are less than 72.8 inches tall?

?

70 72.8 (height values)

+1

? = 84%

68% (by 68-95-99.7 Rule)

16%

-1


Standard Normal Distribution Z – Score

The standard Normal distribution N(0,1) is the Normal distribution has a mean of zero and a standard deviation of one

Normal distributions can be transformed to standard normal distributions by Z-score

zx



What proportion of men are less than 68 inches tall?

?

68 70 (height values)

How many standard deviations is 68 from 70?


Standardized Scores

standardized score (Z-score) =(observed value minus mean) / (std dev)

[ = (68 70) / 2.8 = 0.71 ] The value 68 is 0.71 standard

deviations below the mean 70.




-0.71 0 (standardized values)68 70 (height values)

?


Table A:Standard Normal Probabilities

See pages 464-465 in text for Table A.(the “Standard Normal Table”)

Look up the closest standardized score (z) in the table.

Find the probability (area) to the left of the standardized score.





z .00 .02

0.8 .2119 .2090 .2061

.2420 .2358

0.6 .2743 .2709 .2676

0.7

.01

.2389





.2389



What proportion of men are greater than 68 inches tall?

.2389


1.2389 = .7611



How tall must a man be to place in the lower 10% for men aged 18 to 24?

.10 ? 70 (height values)


See pages 464-465 in text for Table A.

Look up the closest probability (to .10 here) in the table.

Find the corresponding standardized score.

The value you seek is that many standard deviations from the mean.




z .07 .09

1.3 .0853 .0838 .0823

.1020 .0985

1.1 .1210 .1190 .1170

1.2

.08

.1003



How tall must a man be to place in the lower 10% for men aged 18 to 24?

-1.28 0 (standardized values)

.10 ? 70 (height values)


Observed Value for a Standardized Score

Need to “reverse” the z-score to find the observed value (x) :

zx

x z observed value =

mean plus [(standardized score) (std dev)]


Observed Value for a Standardized Score

observed value =mean plus [(standardized score) (std dev)]

= 70 + [(1.28 ) (2.8)]

= 70 + (3.58) = 66.42

A man would have to be approximately 66.42 inches tall or less to place in the lower 10% of all men in the population.



The Entry in Table A

Using random variable z to get the entrance in Table A. Variable z is z-score which follows the standard normal

distribution N(0, 1)

Z-score:

When search entry for a z value

♫ look up the most left column first, locate the most close value to z value

♫ look up the top row to locate the 2th decimal place for a z value


zx

The Entry in Table A

Table A’s entry is an area underneath the curve, to the left of z

Table A’s entry is a percent of the whole area, to the left of z-score

Table A’s entry is a probability, corresponding to the z-score value.

Math formula:

P (z ≤ z0) = 0.xxxx

P (z ≤ -0.71) = 0.2389


Problem type I

If z ≈ N(0, 1), P (z ≤ z0) = ? By checking the Table A, find out the

answer. For type I problem, check the table and

get the answer directly.For example, P (z ≤ -0.71) = 0. 2389


Problem type II

If z ≈ N(0, 1), P (z ≥ z0) = ? This type’s problem, cannot check the

table directly. Using the following operation.

P (z ≥ z0) = 1 - (z ≤ z0) For example, p (z ≥ -0.71) = ?

◙ P (z ≥ -0.71) = 1 - (z ≤ - 0.71) = 1 – 0.2389 = 0.7611


Problem Type II



0.2389 0.7611

Problem Type III

If z ≈ N(0, 1), P ( z2 ≤ z ≤ z1) = ?

random variable z is between two numbers Look up z1 → P1

Look up z2 → P2

The result is: P ( z2 ≤ z ≤ z1) = P1 - P2

For example, P ( -1.4 ≤ z ≤ 1.3) = ?look up 1.3 P1 = 0.9032

look up -1.4 P2 = 0.0808

P ( -1.4 ≤ z ≤ 1.3) = 0.9032 – 0.0808 = 0.8224


Steps Summary Write down the normal distribution N(µ,) for observation

data set Locate the specific observation value X0

Transform X0 to be Z0 by z-score formula

Check table A using random variable Z0 to find out table entry P(z ≤ z0)

If is problem type I, the result is P(z ≤ z0) If is problem type II, the result is:

P (z ≥ z0) =1- P(z ≤ z0) If is problem type III, the result is:

P ( z2 ≤ z ≤ z1) = P1 - P2

P1 = P(z ≤ z1), P2 = P(z ≤ z2)


Chapter 3

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