Normal Distribution PDF

# PROBABILITY DISTRIBUTIONS ## NORMAL DISTRIBUTION **(K.U.K. Dec. 2009; U.P.T.U. 2007)** The normal distribution is a continuous distribution. It can be derived from the Binomial distribution in the limiting case when _n_, the number of trials is very large and _p_, the probability of a success,...

# PROBABILITY DISTRIBUTIONS ## NORMAL DISTRIBUTION **(K.U.K. Dec. 2009; U.P.T.U. 2007)** The normal distribution is a continuous distribution. It can be derived from the Binomial distribution in the limiting case when _n_, the number of trials is very large and _p_, the probability of a success, is close to 1/2. The general equation of the normal distribution is given by: $f(x) = \frac{1}{\sigma \sqrt{2\pi}} e^{-\frac{1}{2}(\frac{x-\mu}{\sigma})^2}$ where $x$ can assume all values from -∞ to +∞, µ and σ, called the parameters of the distribution, are respectively the mean and the standard deviation of the distribution and -∞ < μ < ∞, σ > 0. _x_ is called the normal variate and _f(x)_ is called probability density function of the normal distribution. If a variable _x_ has the normal distribution with mean µ and standard deviation σ, we briefly write _x_: N(µ, σ²). The graph of the normal distribution is called the normal curve. It is bell-shaped and symmetrical about the mean µ. The two tails of the curve extend to +∞ and -∞ towards the positive and negative directions of the _x_-axis respectively and gradually approach the _x_-axis without ever meeting it. The curve is unimodal and the mode of the normal distribution coincides with its mean µ. The line _x_=µ divides the area under the normal curve above the _x_-axis into two equal parts. Thus, the median of the distribution also coincides with its mean and mode. The area under the normal curve between any two given ordinates _x_ = _x1_ and _x_ = _x2_ represents the probability of values falling into the given interval. The total area under the normal curve above the _x_-axis is 1. ### 5.28. STANDARD FORM OF THE NORMAL DISTRIBUTION If _X_ is a normal random variable with mean µ and standard deviation σ, then the random variable _Z_ = (X-µ)/σ has the normal distribution with mean 0 and standard deviation 1. The random variable _Z_ is called the standardized (or standard) normal random variable. The probability density function for the normal distribution in standard form is given by: _f(z) = 1/√2π e^(-1/2)(z^2)_ It is free from any parameter. This helps us to compute areas under the normal probability curve by making use of standard tables. **Note 1.** If _f(z)_ is the probability density function for the normal distribution, then: P(z₁ ≤ Z ≤ z₂) = ∫z₂z₁ f(z) dz = F(z₂) - F(z₁), where F(z) = ∫z(-∞) f(z) dz = P(Z ≤ z) The function _F(z)_ defined above is called the distribution function for the normal distribution. **Note 2.** The probabilities P(z₁ ≤ Z ≤ z₂), P(z₁ < Z ≤ z₂), P(z₁ ≤ Z < z₂) and P(z₁ < Z < z₂) are all regarded to be the same. ** Note 3.** F(-z₁) = 1 - F(z₁). ## ILLUSTRATIVE EXAMPLES **Example 1.** A sample of 100 dry battery cells tested to find the length of life produced the following results: _x_ =12 hours, σ = 3 hours. Assuming the data to be normally distributed, what percentage of battery cells are expected to have life: (i) more than 15 hours (ii) less than 6 hours (iii) between 10 and 14 hours length of life of dry battery cells. **(P.T.U. May 2005)** **Sol.** Here _x_ denotes the length of life of dry battery cells. z = (x - µ)/σ = (x - 12)/3 Also (i) When x = 15, z = 1 ∴ P(x > 15) = P(z > 1) =P(0 < z < ∞) – P(0 < z < 1) = .5 -0.3413 = 0.1587 = 15.87 %. (ii) When x = 6, z = − 2 ∴ P(x < 6) = P(z < - 2)=P(z > 2) = P(0 < z < ∞) − P(0 < z < 2) = .5-0.4772 = 0.0228 = 2.28 %. (iii) When x = 10, z = - 3/2 = - 0.67 When x = 14, z = 2/3 = 0.67 P(10 < x < 14) = P(-0.67 < z < 0.67) = 2P(0 < z < 0.67) = 2 × 0.2487 = 0.4974 = 49.74 %. **Example 2.** In a normal distribution, 31% of the items are under 45 and 8% are over 64. **(K.U.K., Dec. 2010)** **Sol.** Let μ and σ be the mean and S.D. respectively. 31% of the items are under 45. ⇒ Area to the left of the ordinate _x_ = 45 is 0.31 When _x_ = 45, let _z_ = _z1_ P(z₁ < z < 0) = .5 – .31 = .19 From the tables, the value of z corresponding to this area is 0.5 ∴ _z1_ = -0.5 [z₁ < 0] When _x_ = 64, let _z_ = _z2_ P(0 < z < z₂) = .5 − .08 = .42 From the tables, the value of z corresponding to this area is 1.4. ∴ _z2_ = 1.4 Since z = (x - µ)/σ -0.5 = (45 - µ)/σ and 1.4 = (64 - µ)/σ ⇒ 45 - µ = -0.5σ and 64 - µ = 1.4σ Subtracting 19 = -1.9σ .. σ = 10 From (1), 45 - µ = -0.5 × 10 = -5 .. µ = 50.