Download Normal Distributions

Normal Distributions
Density Curve


A density curve is a smooth function meant to
approximate a histogram.
The area under a density curve is one.
Density Curve
Density Curves: Properties
Density Curves

Mean of density curve is point at which the
curve would balance.

For symmetric density curves, balance point
(mean) and the median are the same.
Characterization



A normal distribution is
bell-shaped and
symmetric.
The distribution is
determined by the mean
mu, m, and the standard
deviation sigma, s.
The mean mu controls
the center and sigma
controls the spread.
Definitions


Mean is located in center,
or mode of normal
curve.
The standard deviation is
the distance from the
mean to the inflection
point of the normal
curve, the place where
the curve changes from
concave down to
concave up.
Construction




A normal curve is drawn by first drawing a
normal curve.
Next, place the mean, mu on the curve.
Then place sigma on curve by placing the
segment from the mean to the upper (or lower)
inflection point on your curve.
From this information, the scale on the
horizontal axis can be placed on the graph.
Examples

Draw normal curve with mean=mu=100, and
standard deviation = sigma = 10.

Draw normal curve with mean = 20, sigma=2.
68-95-99.7 Rule


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
For any normal curve with
mean mu and standard
deviation sigma:
68 percent of the
observations fall within one
standard deviation sigma of
the mean.
95 percent of observation fall
within 2 standard deviations.
99.7 percent of observations
fall within 3 standard
deviations of the mean.
Example Questions

If mu=30 and sigma=4, what are the values (a, b)
around 30 such that 95 percent of the observations
fall between these values?

If mu=40 and sigma=5, what are the bounds (a, b)
such that 99.7 percent of the values fall between
these values?
Standard Normal Distribution

The standard normal distribution has mean = 0 and
standard deviation sigma=1.
Normal Table Usage


What proportion of standard normal distribution values
Z are less than 1.40? That is, P(Z < 1.40) = ?
Ans:.9192 or 91.92 percent of values.
Standard Normal


P( 0 < Z < 1.40) = ?
Ans: P(Z < 1.40) – P(Z<0) = .9192 - .5 = .4192
Example

P( Z < - 2.15) = ?
Normal Table Usage
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
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P( .64 < Z < 1.23) = ?
Ans: P(Z<1.23) – P(Z < .64) = .8907 - .7389 =
.1518
P(Z > 2.24) = CAREFUL !!!!!
Ans: Either = 1 – P(Z < 2.24) = 1 - .9875 =
or by symmetry = P(Z < - 2.24) = .0125. In this
approach you are using the fact that both tails of a
standard normal are the same and so P(Z>2.24) =
P(Z< -2.24) = .0125.
Z-Score Formula

Any normal distribution with mean=mu and standard
deviation= sigma, can be converted into a standard
normal Z distribution by the following transformation:
Example

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Consider a distribution with mean=mu=100 and
standard deviation = sigma = 10. Draw density
curve with number line provided.
Now re-draw the curve and number line on
horizontal axis after subtracting 100 from each
value. Notice this centers the curve at zero.
Then draw the resulting number line after
dividing the previous number line values by 10.
Voila ! We are now back to Z scale !
Example
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Example 1.26 in Page 75.
X=The SAT score of a randomly chosen
student. X has N(m=1019, s=209).
What percent of all students had SAT scores of
at least 820? That is, P( X > 820) = ?
X  820
Solution
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
P( X > 820 ) =
Solution = .8289
Example
Problem 1.86 (Moore&Mc)

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Eleanor gets 680 on SAT
math exam. Mean on
this exam is 500 and sd is
100.
Eleanor’s standardized
score is:
1.86 Continued

Gerald got 27 on ACT
math. Mean is 18 with
sd of 6.
Gerald’s Z-Score is:

Eleanor did better !

Human Pregnancies

What proportion of births are premature? That is, what
proportion is below 240 days? P(X<240)= ?
London Bus Drivers
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Calorie intake for drivers averages 2821 cals per day with
sd=sigma=436.
What proportion of drivers have calorie intakes, X, less
than 2000 calories per day? P(X < 2000)?
P( X  2000) = P(
X m
s
P( Z  1.88) = .0301
2000  2821

)
436
London Bus Drivers

What proportion of drivers consume between 2000 and
2500 cals per day? P(2000<X<2500)?
2000  2821 X  m 2500  2821
P(2000  X  2500) = P(


)
436
s
436
P(1.88  Z  .74) = .2296  .0301 = .1995
Finding a Percentile
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Backwards problem. We are now given a fraction and
need to find the X-value.
In past, we were provided X and found a proportion.
Use Formula:
X =mZ s
*
London Bus Drivers


Find the calorie intake at the 90th percentile of the calorie
distribution.
Insert mean and sd into backward formula, then
determine correct Z-star value.
X = m  Z s = 2821  Z (436)
X = 2821  1.28(436) = 3379.08
*
*
Finding a Percentile

Plugging in the mean and sd are not hard. The difficulty is
finding Z-star. It is simply the same percentile you are
trying to find, except for the standard normal distribution.
This requires you to use an inverse lookup in your z-table.
X =mZ s
*
TV Viewing


Neilsen ratings service found that tv viewing for children
aged 2-11 had a normal distribution with mean 23.02
hours and sigma=6.23 hours.
What proportion of children watch more than 24 hours
of tv per week?
P( X  24) = P(
X m
s
P( Z  .16) = .4364
24  23.02

)
6.23
TV Viewing

How many hours of tv does a child watch that is at the 95th
percentile of the tv viewing distribution?
X = m  Z s = 23.02  1.64(6.23) = 33.24
*
The Central Limit Theorem
(for the sample mean x)
If a random sample of n observations is selected
from a population (any population), then when
n is sufficiently large, the sampling distribution
of x will be approximately normal.
(The larger the sample size, the better will be the
normal approximation to the sampling
distribution of x.)

How Large Should n Be?

For the purpose of applying the central limit
theorem, we will consider a sample size to be
large when n > 30.