1. Unit 13 Normal Distribution
Reference: (Material source and pages)
2018/6/8
Unit 13 Normal Distribution
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2. ITD1111 Discrete Mathematics & Statistics STDTLP
13.1 Normal distribution
A random variable X is said to have a Normal distribution, with parameters  and 2 if it can take any real value and has p.d.f.
In this case, we write X ~ N(, 2). It can be shown that  is the expected value of X and 2 is the variance.
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P(a < X < b) = =
This integral cannot be done algebraically and its value has to be found by numerical methods of integration. The value of P(a < X < b) can also be viewed as the area under the curve of f(x) from
x = a to x = b.
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4. 13.2 Characteristics of Normal Distribution
1. Bell shape
2. Symmetric
3. Mean = Mode = Median
4. The 2 tails extend indefinitely
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5. 13.3 Standard Normal Distribution
A standard normal distribution is the normal distribution with  = 0 and 2 = 1, i.e. the N(0,1) distribution.
A standard normal random variable is often denoted by Z.
If Z ~ N(0,1), its c.d.f. is usually written as
(z) = P(Z  z) =
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6. ITD1111 Discrete Mathematics & Statistics STDTLP
Note:
(1) (z) may be interpreted as the area to the left of z under the standard normal curve.
(2) P(Z < -z) = P(Z > z), since the standard normal curve is symmetrical about the line Z = 0.
(3) The area between Z= -1 and +1 is 68%
Z= -2 and +2 is 95%
Z= -3 and +3 is 99%
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7. ITD1111 Discrete Mathematics & Statistics STDTLP
Example (1/2)
Given Z ~ N(0,1), find the following probabilities using the standard normal table.
(a) p( Z  1.25)
(b) P( Z > 2.33)
(c) P(0.5 < Z < 1.5)
(d) P( Z < -1.25)
(e) p (-1.5< Z <-0.5)
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Example (2/2)
(a) P(Z  1.25) =
P(Z > 2.33) = 1 - (2.33)
= 1 – =
(c) P(0.5 < Z < 1.5) = (1.5) - (0.5)
= – =0.2417
(d) P( Z < -1.25) = P(Z > 1.25)
= 1 - (1.25)
= 1 – =
(e) P(-1.5 < Z < -0.5) =
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Example (1/4)
Given Z ~ N(0,1), find the value of c if
(a) P(Z  c) =
(b) P( Z > c) = 0.37
(c) P(Z < c) = 0.025
(d) P(0  Z  c) =
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10. ITD1111 Discrete Mathematics & Statistics STDTLP
Example (2/4)
(a) (c) =
c = 1.22
(b) 1 - (c) = 0.37
(c) = 0.63
c = 0.332
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11. Example (3/4)
(c) Obviously c is negative and the standard normal table cannot be used directly.
Recall that P(Z < -z) = P(Z > z)
0.025 = P(Z < c) = P(Z > -c) = 1 –(-c)
(-c) = 0.975
-c = 1.96
c = -1.96
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Example (4/4)
(d) P(0  Z  c) = (c) – 0.5 =
(c) =
c = 2.43
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13. 13.4 Standardization of normal random variables
If X ~ N(, 2), then it can be shown that
Z = ~ N(0,1) .
In other words is the standard normal random variable.
The process of transforming X to Z is called standardization of the random variable X.
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14. ITD1111 Discrete Mathematics & Statistics STDTLP
Example (1/4)
Suppose X ~ N(10, 6.25). Find the following probabilities
(a) P(X < 13)
(b) P(X > 5)
(c) P(8 < X < 15)
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15. ITD1111 Discrete Mathematics & Statistics STDTLP
Example (2/4)
Let Z =
(a) P(X < 13) = P(Z < )
= P(Z < 1.2) =
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16. ITD1111 Discrete Mathematics & Statistics STDTLP
Example (3/4)
(b) P(X > 5) = P(Z > )
= P(Z > -2)
= P(Z < 2) =
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17. ITD1111 Discrete Mathematics & Statistics STDTLP
Example (4/4)
(c) P(8 < X < 15) = P( )
= P(-0.8 < Z < 2)
= (2) - (-0.8)
= (2) – (1 - (0.8))
= – (1 – ) =
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18. ITD1111 Discrete Mathematics & Statistics STDTLP
Example (1/2)
The lifetime of light bulbs produced in a certain factory is normally distributed with mean 3000 hours and standard deviation 50 hours. Find the probability that a randomly chosen light bulb has a lifetime less than 2900 hours.
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19. ITD1111 Discrete Mathematics & Statistics STDTLP
Example (2/2)
Let X be the lifetime of a randomly chosen light bulb.
Then X ~ N(3000, 502).
Let Z =
P(X < 2900) = P(Z < -2)
= 1 – =
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20. ITD1111 Discrete Mathematics & Statistics STDTLP
Example (1/3)
If a random variable has the normal distribution with  = 40 and  = 2.4, find the probabilities that it will take a value
(a) less than 43.6
(b) greater than 38.2
(c) between 40.6 and 43.0
(d) between 35.8 and 44.2.
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21. ITD1111 Discrete Mathematics & Statistics STDTLP
Example (2/3)
Let X be the random variable.
(a) P(X < 43.6) = P(Z < )
= P(Z < 1.5) =
(b) P(X > 38.2) = P(Z > )
= P(Z > -0.75)
= P(Z < 0.75) =
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22. ITD1111 Discrete Mathematics & Statistics STDTLP
Example (3/3)
(c) P(40.6 < X < 43.0)
= P( )
= P(0.25 < Z < 1.25) =
(d) P(35.8 < X < 44.2)
= P( )
= P(-1.75 < Z < 1.75) =
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