Normal Random Variables

1
Normal Random Variables

• In the class of continuous random variables, we
  are primarily interested in NORMAL random
  variables.
• These are a continuous random variable with a
  bell-shaped distribution. These normal or
  bell-shaped variables occur often in nature.
• Example Heights of Males
• are Normally Distributed
• Probability Density Function
• for Heights of Males ---gt

2
Properties of the Normal distribution

• There are infinitely many normal pdfs (curves).
  To fully describe a normal curve, we need the
  location (mean, µ) and the spread (standard
  deviation, s).
• Notation
• If X a normal random variable with mean E(X)µ
  and variance Var(X)s2 we write
• XN (µ, s2)
• For population mean and s.d. we use the Greek
  letters µ and s, for the sample mean and s.d. we
  use and s.
• The distribution is symmetrical around the mean
  µ.
• The median, and the mean are equal due to the
  symmetry of the distribution.
• The total area under the curve is equal to one.

3
The parameters µ and s
For all, s1
For all, µ 0
s1/2
µ0
µ2
µ-2
s1
s2
4
Probabilities with Normal RVs

• When we consider Normal Random Variables (or any
  continuous r.v.), we are interested in the
  probability that X falls into some INTERVAL.
• The probability a random variable XN(µ, s2) to
  take a value is equal to zero. In other words, if
  XN(µ, s2) then P(Xk)0, where k is some number.
• Example Suppose X is the height of a randomly
  chosen college woman. Further suppose that the
  heights of college women can be described as a
  normal, with µ 65inches (in), and s
  2.7 in.
• We might ask
• What is the proportion of women that are shorter
  than 62 in?
• What is the probability that X is between 65 and
  67in?

5
Graphical Representation of Probabilities
P(65ltXlt67)
P(Xlt62)

• The total area under the curve is equal to 1!
• The probability that X falls in an interval is
  equal to the area of the region below the curve
  and over the interval.

6
Probabilities with Normal RVs

• The total area under the curve is equal to 1!
• The probability that X falls in an interval is
  equal to the area of the region below the curve
  and over the interval.
• For example P(a ? X ? b) is equal to the area
  under the curve between a and b.
• Due to the continuity of the normal distribution
  we have that the probability a normal random
  variable to take a value is equal to zero, thus
• P(X ? a) P(X lt a)
• P(a ? X ? b) P(a lt X lt b)
• P(a ? X lt b) P(a lt X lt b) etc.

7
Standard Normal Distribution

• A normal r.v. with µ0 and
• s1 is called a standard
• normal random variable.
• We denote it with Z, so that
• ZN(0,1).

• We have tables for the probabilities of the form
• P(Z lt z) where z 0.
• e.g. P(Z 0.5), P(Z lt 2).
• Probabilities of the form P(Z lt - 0.4), P(Z gt
  1.2), P(Z gt - 0.25) have to be
  transformed into probabilities of the form P(Z lt
  z) where z 0.

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How to use the table

• The probability P(Zlt1.14) is the number in the
  table where the row of 1.1 and the column of .04
  are crossed. Thus, P(Zlt1.14)0.8729
• More examples P(Zlt0.57).7157 P(Zlt2).9972
  P(Zlt1.3).9032 P(Zlt0)0.5

9
Calculating Probabilities of ZN(0,1)

• For ZN(0,1), and a 0
• P(Z gt a) 1-P(Z lt a)
• P(Z lt -a) 1-P(Z lt a)
• P(Z gt -a) P(Z lt a)
• P(b ltZ lt c) P(Z lt c) - P(Z lt b), for any b and
  c.
• Draw a normal curve and shade the areas
  corresponding to the above probabilities.

10
Calculating Probabilities of any normal r.v.
XN( µ , s )
We can obtain any type of probabilities of
interest for any normal r.v XN(µ, s 2) by first
transforming X into Z using the following
standardization theorem
11
How to Calculate Probabilities

• If you want P(X lt x), first compute the z-score
• z (Value mean)/(Standard Deviation)
  (x-µ)/s,
• But P(X lt x) P(Z lt z) for which we have
  tables!!
• Example X height of a college woman, XN( 65,
  2.72)
• 1. P(X lt 62)
• z (62 65)/2.7 -3/2.7 -1.11
• P(X lt 62) P(Z lt -1.11) (now use Normal
  Table)
• 0.134 13.4
• 2. P(65 lt X lt 67)
• z1 (65 65)/2.7 0, z2 (67 65)/2.7
  1.11
• P(65 lt X lt 67) P(0 lt Z lt 1.11)
• P(Z lt 1.11) P(Z lt
  0)
• .867 .5 .367 or
  36.7
• 3. P(Xgt62) 1- P(Xlt62) 1-0.134 0.866

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Example Suppose verbal SAT scores of high-school
freshman are normally distributed with a mean of
500 and a standard deviation of 50.

• What is the probability of a randomly chosen
  individual having a score greater than 600?
• z-score 600-500/50 2
• P(Xgt600) P(Zgt2) 1- P(Z ? 2) 1-P(Zlt2)
  1-.9772 0.228
• Note that the only difference in the two
  graphs below is the scale on the tow axes.
  However, the shaded areas are equalsince the
  total area under any of this curves is one.

P(Xgt600)
P(Zgt2)
0
2
-2
-4
4
13

• What is the probability of a randomly chosen
  individual having a score between 400 and 500?
• We want P(400ltXlt500).
• z-score1 z1 400-500/50 -2
• z-score2 z2 500-500/50 0
• P(400ltXlt500) P(-2 lt Z lt 0)
• P(Zlt0) P(Zlt-2)
  .5-.228 .4772 (from Table)

P(-2ltZlt0)
That is the probability of a randomly chosen
student having a score between 400 and 500 is
about .48 or 48.
14

• What is the probability of a randomly chosen
  individual having a score between 350 and 450?
• z-score1 z1 350-500/50 -3
• z-score2 z2 450-500/50 -1
• P(350ltXlt450) P(-3 lt Z lt -1)
• P(Zlt-1) P(Zlt-3)
• .1587-.0013 .1574
  (from normal Table)

P(-3ltZlt-1)
That is the probability of a randomly chosen
student having a score between 350 and 450 is
about .16 or 16.
15
The Empirical Rule and Normal Distrib.

• The Empirical Rule states that for any
  bell-shaped distribution, approximately
• 68 of the values fall within 1 standard
  deviation of the mean in either direction. (in
  the interval µ s )
• 95 of the values fall within 2 standard
  deviations of the mean in either direction. (in
  the interval µ 2s )
• 99.7 of the values fall within 3 standard
  deviations of the mean in either direction. (in
  the interval µ 3s )
• This empirical rule is valid for all bell-shaped
  distributions but it is exactly right in the case
  of the normal distribution.
• Check the following probabilities
• P(-1ltZlt1) ___ , P(-2ltZlt2) ___ , P(-3ltZlt3) ___

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How can we find percentiles?

• Question For a normal r.v. X with mean µ and
  standard deviation s , how can we find x (a
  value of X ), such that P( X x) a, where a
  is a known probability.
• e.g. if a 95 , the 95th percentile of X is
  the value of X such that 95 of its possible
  values are less than that.
• Solution First we get the a-th percentile for Z,
  
• P(Z z) 0.95 gives z 1.64.
•   and we get x using x s z µ.
• Example What is the 90th percentile of the
  height of college women? Recall that XN( 65,
  2.72)
• P(Z z) 0.90, then z 1.38 since
  P(Zlt1.38)0.8997 ?the closest value to 0.90 in
  the table.
• x2.71.386568.726,
• Thus the 90th percentile of the height of college
  women is 68.726in.

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SummaryDefinitions and theory for Normal r.vs.

• Knowing µ and s, specifies the particular normal
  distribution out of the class of all normal
  distributions.
• The pdf of any normal r.v XN(µ , s 2), also
  called normal curve, is symmetric, bell shaped
  and centered at µ.
• The standard normal random variable, Z, has µ 0
  and s 2 s 1.
• We have the tables for all the probabilities of
  the form P(Z z) P(Z lt z).
• For any normal r.v X N(µ , s 2 ), we can obtain
  any probabilities of interest using the
  standardization theorem   
• P(X x) P (X- µ)/ s (x- µ)/ s P(Z
  z),
• Where z (x- µ)/ s, is called the z-score of x.

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SummaryFinding Probabilities of XN(µ , s 2 )

• First find the z-score of x (or xs if more than
  one) to be able to use the tables.
• Write the probability in terms of Z.
• Think what is the area under the curve that
  corresponds to this probability
• Having in mind that the normal curve is symmetric
  and that the total area under the curve is equal
  to 1 figure out how to transform this probability
  into the form P(Zltz) rules on slide .
• Finally, obtain the probability using the table.
•  
• To find the a-th percentile of X
• We want to find x (a value of X), such that P( X
  x) a
• First we get the a-th percentile for Z, for
  example if a95)
• P(Z z) 0.95, then z 1.64
• We get x using x s z µ