1. NORMAL
DISTRIBUTION
(A Continuous Probability Distribution)
2. Normal Distribution
In probability theory, the normal (or Gaussian) distribution is a
continuous probability distribution. The graph of the
associated probability density function is "bell"-shaped, and is known as
the Gaussian function or bell curve
Definition: A continuous random variable X is said to follow Normal
Distribution if its probability density function is given as
-∞ < x < ∞
where parameter μ is the mean (location of the peak) and
σ 2 is the variance (the measure of the width of the distribution).
X ~ N (μ, σ 2 )
4. The Standard Normal Distribution:
If X is a normal variable with mean μ and variance σ 2 then
the variable z = (x- μ )/σ is called standard normal
variable A standard normal distribution is a normal
distribution with zero mean ( ) and unit variance (
), given by the probability density function
P(Z=z) =
The distribution of with μ = 0 and σ 2 = 1 is called the
standard normal.
• If X ~ N (μ, σ 2 )
~ N ( 0, 1 )
5. For normal variable infinite values of mean and variance
are possible that’s why it is to be standardize…
6. Properties of Normal Curve:
The normal curve with mean µ and standard deviation σ has following
properties
1. The equation of the curve is
2. The curve is symmetric about the line x = µ and x ranges from -∞ to
+∞.
3.Mean, mode and median coincide at x = µ as distribution is symmetrical.
4.The points of inflection of the curve are x = µ + σ and x = µ - σ.
5. In normal distribution Q.D.: M.D. : S.D. = 10 : 12 : 15
6.The M.D. about mean = 4/5 * S.D.
7.The normal curve has single peak i.e. unimodal.
11. Normal Probability:
1] As normal distribution is continuous distribution,
probability at a point is zero i.e. P ( z = a ) = 0
2] As the standard normal distribution is symmetric about
mean (=0). P (z < 0 ) = P ( z > 0 ) = ½
3] P ( 0 < z < a) then table value for a gives the required
probability.
4] P (-a < z < 0) = P ( 0 < z < a) as the curve is symmetric
about 0.
5] P( z < -a) = P ( z > a) = P ( z > 0 ) - P ( 0 < z < a)
12. • 1]Question: What is the relative frequency of observations
below 1.18?
• That is, find the relative frequency of the event Z < 1.18. (Here
small z is 1.18.) (Or P(Z < 1.18 )
• P( z < 1.18) = P( z < 0 ) + P ( 0 < z < 1.18)
• = 0.5 + 0.3810
• = 0.8810
14. 4] P( -1.65 < z < 1.65 ) = 2 P ( 0< z < 1.65 )
5] P(z<0.84) = 0.8000
15. • Normal tables can be of this type P ( z < a )
• z. 00 .01.... 08. 09
• 0.0 .5000 .5040.... .5319 .5359
• 0.1 .5398 .5438.... .5714 .5753
• |
• |
• 1.0 .8413 .8438... .8599 .8621
• 1.1 .8643 .8665... .8810 .8830
• 1.2 .8849 .8869... .8997 .9015
16. • Example 1
An average light bulb manufactured by the Acme
Corporation lasts 300 days with a standard deviation of
50 days. Assuming that bulb life is normally distributed,
what is the probability that an Acme light bulb will last
at most 365 days?
• Example 2
Suppose scores on an IQ test are normally distributed.
If the test has a mean of 100 and a standard deviation
of 10, what is the probability that a person who takes
the test will score between 90 and 110?
[P( 90 < X < 110 ) = P( X < 110 ) - P( X < 90 )
P( 90 < X < 110 ) = 0.84 - 0.16
P( 90 < X < 110 ) = 0.68
• Thus, about 68% of the test scores will fall between 90
and 110.]
17. • Problem 3
• Molly earned a score of 940 on a national achievement
test. The mean test score was 850 with a standard
deviation of 100. What proportion of students had a
higher score than Molly? (Assume that test scores are
normally distributed.)
• [z = (X - μ) / σ = (940 - 850) / 100 = 0.90
• we find P(Z < 0.90) = 0.8159.
Therefore, the P(Z > 0.90) = 1 - P(Z < 0.90) = 1 - 0.8159
= 0.1841. Thus, we estimate that 18.41 percent of the
students tested had a higher score than Molly. ]
18. 4.The Acme Light Bulb Company has found that an average light bulb
lasts 1000 hours with a standard deviation of 100 hours. Assume
that bulb life is normally distributed. What is the probability that
a randomly selected light bulb will burn out in 1200 hours or less?
5. Bill claims that he can do more push-ups than 90% of the boys in
his school. Last year, the average boy did 50 push-ups, with a
standard deviation of 10 pushups. Assume push-up performance is
normally distributed. How many pushups would Bill have to do to
beat 90% of the other boys?
***************
19. • Example
• You and your friends have just measured the heights of your dogs
(in millimeters):
• The heights (at the shoulders) are: 600mm, 470mm, 170mm,
430mm and 300mm.
• Find out the Mean, the Variance, and the Standard Deviation.
• Your first step is to find the Mean:
20. • Answer:
• Mean = [600 + 470 + 170 + 430 + 300]/5 = 1970/5 = 394
•
• so the mean (average) height is 394 mm. Let's plot this on the
chart:
• Now, we calculate each dogs difference from the Mean:
21. • To calculate the Variance, take each difference, square it, and then average the
result:
• Variance: σ2 = [2062 + 762 + (-224)2 + 362 + (-94)2 ]/5 = 108,520/5 = 21,704
• So, the Variance is 21,704.
• And the Standard Deviation is just the square root of Variance, so:
• Standard Deviation: σ = √21,704 = 147.32... = 147 (to the nearest mm)
•
• And the good thing about the Standard Deviation is that it is useful. Now we can
show which heights are within one Standard Deviation (147mm) of the Mean:
• So, using the Standard Deviation we have a "standard" way of knowing what is
normal, and what is extra large or extra small.
