8. THE NORMAL DISTRIBUTION, Summaries of Statistics

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8. THE NORMAL DISTRIBUTION

The normal distribution with mean μ and variance σ^2 has the following density function:

The normal distribution is sometimes called a Gaussian Distribution, after its inventor, C.F. Gauss (1777-1855).

We won't need the formula for the normal f ( x ) , just tables of areas under the curve.

• f ( x ) has a bell shape, is symmetrical about μ, and reaches its maximum at μ.
• μ and σ determine the center and spread of the distribution.
• The empirical rule holds for all normal distributions: 68% of the area under the curve lies between (μ − σ, μ + σ) 95% of the area under the curve lies between (μ − 2 σ, μ + 2σ) 99.7% of the area under the curve lies between (μ − 3 σ, μ + 3σ)

• There are many different normal distributions, one for each choice of the parameters μ and σ.

The normal distribution plays an extremely important role in statistics because

1. It is easy to work with mathematically

2. Many things in the world have nearly normal distributions:

Heights of ocean waves (but not Tsunamis!) IQ scores (by design). Stock Returns, according to Black-Scholes Theory. Weights of “4 ounce” bags of M&Ms. The high temperature in Central Park on January 1. The distance from the darts to the bulls-eye on a dartboard.

A normal random variable with μ = 0 and σ^2 = 1 is said to have the standard normal distribution.

Although there are infinitely many normal distributions, there is only one standard normal distribution.

All normal distributions are bell-shaped, but the bell for the standard normal distribution has been standardized so that its center is at zero, and its spread (the distance from the center to the inflection points) is 1. This is the same standardization used in computing z-scores, so we will often denote a standard normal random variable by Z. We will use φ( z ) to denote the density function for a standard normal.

To calculate probabilities for standard normal random variables, we need areas under the curve φ( z ). These are tabulated in Table 5.

To avoid confusion, replace z in Table 5 by z 0. In the diagram, change z to z 0 , and then label the horizontal axis as the z-axis.

Table 5 gives the areas under φ( z ) between z = 0 and z = z 0. This is the probability that a standard normal random variable will take on a value between 0 and z 0.

For example, the probability that a standard normal is between 0 and 2 is 0.4772.

The table includes only positive values z 0. To get areas for more general intervals, use the symmetry property, and the fact that the total area under φ( z ) must be 1.

Eg: If Z is standard normal, compute P(− 1 ≤ Z ≤ 1), P(− 2 ≤ Z ≤ 2) and P(− 3 ≤ Z ≤ 3).

Solution : P(− 1 ≤ Z ≤ 1) = 2(0.3413) = 0. P(− 2 ≤ Z ≤ 2) = 2(0.4772) = 0. P(− 3 ≤ Z ≤ 3) = 2(0.4987) = 0.9974.

Note: This shows that the empirical rule holds for standard normal distributions. We still need to prove it for general normal distributions.

Eg: Compute the probability that a standard normal RV will be a) Between 1 and 3 b) Greater than −0. c) Less than −1.35.

Eg : What's the 95th^ percentile of a standard normal distribution?

Solution: Since we've been given the probability and need to figure out the z-value, we have to use Table 5 in reverse. Since the 50 th^ percentile of a standard normal distribution is zero, the 95 th^ percentile is clearly greater than zero.

So we need to find the entry inside of Table 5 which is as close as possible to 0.45. In this case, there are two numbers which are equally close to 0.45. They are 0.4495 ( z =1.64) and 0. ( z =1.65).

So the 95th^ percentile is 1.645. In other words, there is a 95% probability that a standard normal will be less than 1.645.

Eg: z-scores on an IQ test have a standard normal distribution. If your z -score is 2.7, what is your percentile score?

Solution: To figure out what percentile this score is in, we need to find the probability of getting a lower score, and then multiply by 100.

We have Pr(Z<2.7) = 0.5 + 0.4965 = 0.9965. So the percentile score is 99.65.

Important Property: If X is normal, then Z = (X−μ)/σ is standard normal, that is, E(Z) = 0, Var(Z) = 1.

Therefore, P(a<X<b) can be computed by finding the probability that a standard normal is between the two corresponding z -scores, (a−μ)/σ and (b−μ)/σ.

Fortunately, we only need one normal table: the one for the standard normal. This makes sense, since all normal distributions have the same shape. Things would be much more complicated if we needed a different table for each value of μ and σ!

• For any normal random variable, we can compute the probability that it will be within 1,2,3 standard deviations of its mean.

This is the same as the probability that the z-score will be within 1,2,3 units from zero. (Why?) Since Z is standard normal, the corresponding probabilities are 0.6826, 0.9544, 0.9974, as computed earlier.

• Thus, the “empirical rule” is exactly correct for any normal random variable.

Eg 1: Suppose the current price of gold is $930/Ounce. Suppose also that the price 1 month from today has a normal distribution with mean μ=930 and standard deviation σ= (obtained from recent estimates of volatility). a) Compute the probability that the price in 1 month will be at or below $900/Ounce. b) If you own 1 Ounce of gold what is the 5% VaR?

Eg 2: Suppose that GMAT scores are normally distributed with a mean of 530 and a variance of 10,000. a) What is the probability that a randomly selected student's score is at most 780? b) At least 400? c) Between 500 and 600? d) Below what score do 95% of the scores lie?

Eg 3: A company manufactures 1/8” rivets for use in an airplane wing. Due to imperfections in the manufacturing process, the diameters of the rivets are actually normally distributed with mean μ =1/8” and standard deviation σ. In order for the rivets to fit properly into the wing, their diameters must meet the 1/8” target to within a tolerance of ±0.01". To what extent must the company control the variation in the manufacturing process to ensure that at least 95% of all rivets will fit properly?