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09 Normal distribution

# 09 Normal distribution - University of California Los...

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University of California, Los Angeles Department of Statistics Statistics 100A Instructor: Nicolas Christou Normal distribution The normal distribution is the most important distribution. It describes well the distribution of random variables that arise in practice, such as the heights or weights of people, the total annual sales of a firm, exam scores etc. Also, it is important for the central limit theorem, the approximation of other distributions such as the binomial, etc. We say that a random variable X follows the normal distribution if the probability density function of X is given by f ( x ) = 1 σ 2 π e - 1 2 ( x - μ σ ) 2 , -∞ < x < This is a bell-shaped curve. We write X N ( μ, σ ). We read: X follows the normal distribution (or X is normally distributed) with mean μ , and standard deviation σ . The normal distribution can be described completely by the two parameters μ and σ . As always, the mean is the center of the distribution and the standard deviation is the measure of the variation around the mean. Shape of the normal distribution. Suppose X N (5 , 2). x f(x) -3 -1 1 3 5 7 9 11 13 0.00 0.05 0.10 0.15 0.20 X ~ N(5,2) 1

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The area under the normal curve is 1 (100%). Z -∞ 1 σ 2 π e - 1 2 ( x - μ σ ) 2 dx = 1 The normal distribution is symmetric about μ . Therefore, the area to the left of μ is equal to the area to the right of μ (50% each). Useful rule (see figure above): The interval μ ± 1 σ covers the middle 68% of the distribution. The interval μ ± 2 σ covers the middle 95% of the distribution. The interval μ ± 3 σ covers the middle 100% of the distribution. Because the normal distribution is symmetric it follows that P ( X > μ + α ) = P ( X < μ - α ) The normal distribution is a continuous distribution. Therefore, P ( X a ) = P ( X > a ) , because P ( X = a ) = 0 . Why? How do we compute probabilities? Because the following integral has no closed form solution P ( X > α ) = Z α 1 σ 2 π e - 1 2 ( x - μ σ ) 2 dx = . . . the computation of normal distribution probabilities can be done through the standard normal distribution Z : Z = X - μ σ Theorem: Let X N ( μ, σ ). Then Y = αX + β follows also the normal distribution as follows: Y N ( αμ + β, ασ ) Therefore, using this theorem we find that Z N (0 , 1) It is said that the random variable Z follows the standard normal distribution and we can find probabilities for the Z distribution from tables (see next pages). 2
The standard normal distribution table: 3

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4
5

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Example: Suppose the diameter of a certain car component follows the normal distribution with X N (10 , 3). Find the proportion of these components that have diameter larger than 13.4 mm . Or, if we randomly select one of these components, find the probability that its diameter will be larger than 13.4 mm . Answer: P ( X > 13 . 4) = P ( X - 10 > 13 . 4 - 10) = P X - 10 3 > 13 . 4 - 10 3 = P ( Z > 1 . 13) = 1 - 0 . 8708 = 0 . 1292 .
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