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Understanding the Normal Distribution: Means, Variances, and the Central Limit Theorem, Study notes of Statistics

National University of Ireland (NUI)Statistics

An in-depth explanation of the normal distribution, its significance in statistics, and how to use it to solve problems involving large datasets. It covers the concept of normal distributions with different means and variances, the importance of the mean and variance in determining the distribution, and the use of the standard normal distribution to find probabilities. The document also introduces the Central Limit Theorem and its implications for statistics.

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bg1
The Normal Distribution
Fall 2001 Professor Paul Glasserman
B6014: Managerial Statistics 403 Uris Hall
1. The normal distribution (the familiar b ell-shaped curve) is without question the most
important distribution in all of statistics. A broad range of problems, particularly those
involving large amounts of data, can be solved using the normal distribution. Although the
normal distribution is continuous, it is often used to approximate discrete distributions.
For example, we might say that the scores on an exam are (approximately) normally
distributed, even though the scores are discrete.
2. There are actually many different normal distributions. To fix a particular normal, we
must specify the mean µand the variance σ2.IfXhas this normal distribution, we write
X∼N(µ, σ2). This notation says “Xis normally distributed with mean µand variance
σ2.” We call µand σ2the parameters of the normal distribution.
3. Once we specify the parameters µand σ2, we have completely specified the distribution
of a normal random variable. This is not true for arbitrary random variables: ordinarily,
the mean and the variance do not completely determine the distribution, but for a normal
random variable they do.
4. Increasing µshifts the normal density to the right without changing its shape. Increasing
σ2flattens the density without shifting it. See Figure 1.
5. Some examples:
•A machine that fills bags of potato chips cannot put exactly the same weight of
chips into every bag. Suppose the quantity poured into an 8 ounce bag is normally
distributed with a mean of 8.3 ounces and a standard deviation of 0.2 ounces. We
might ask what proportion of bags contain less than 8 ounces of chips.
•An airplane manufacturer wants to build a smaller carrier with the same seating
capacity. If the height of men is normally distributed with a mean of 5 feet 9 inches
and a standard deviation of 2 inches, we might ask how low the ceiling can be so
that at most 2% of men will have to duck while walking down the aisle.
1
pf3
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pf5

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Download Understanding the Normal Distribution: Means, Variances, and the Central Limit Theorem and more Study notes Statistics in PDF only on Docsity!

The Normal Distribution

Fall 2001 Professor Paul Glasserman B6014: Managerial Statistics 403 Uris Hall

  1. The normal distribution (the familiar bell-shaped curve) is without question the most important distribution in all of statistics. A broad range of problems, particularly those involving large amounts of data, can be solved using the normal distribution. Although the normal distribution is continuous, it is often used to approximate discrete distributions. For example, we might say that the scores on an exam are (approximately) normally distributed, even though the scores are discrete.
  2. There are actually many different normal distributions. To fix a particular normal, we must specify the mean μ and the variance σ^2. If X has this normal distribution, we write X ∼ N (μ, σ^2 ). This notation says “X is normally distributed withmean μ and variance σ^2 .” We call μ and σ^2 the parameters of the normal distribution.
  3. Once we specify the parameters μ and σ^2 , we have completely specified the distribution of a normal random variable. This is not true for arbitrary random variables: ordinarily, the mean and the variance do not completely determine the distribution, but for a normal random variable they do.
  4. Increasing μ shifts the normal density to the right without changing its shape. Increasing σ^2 flattens the density without shifting it. See Figure 1.
  5. Some examples:
    • A machine that fills bags of potato chips cannot put exactly the same weight of chips into every bag. Suppose the quantity poured into an 8 ounce bag is normally distributed witha mean of 8.3 ounces and a standard deviation of 0.2 ounces. We might ask what proportion of bags contain less than 8 ounces of chips.
    • An airplane manufacturer wants to build a smaller carrier withthe same seating capacity. If the height of men is normally distributed with a mean of 5 feet 9 inches and a standard deviation of 2 inches, we might ask how low the ceiling can be so that at most 2% of men will have to duck while walking down the aisle.

−5^0 −4 −3 −2 −1 0 1 2 3 4 5

−^05 − 4 − 3 − 2 − 1 0 1 2 3 4 5

Figure 1: Left panel shows two normal distributions with different means (μ = 0 for solid, μ = 2 for dashed) and same standard deviation (σ = 1). Right panel shows two normal distributions withthe same mean ( μ = 0) and different standard deviations (σ = 1 for solid, σ = 2 for dashed).

  • Suppose that weekly fluctuations in the price of XYZ stock are well approximated by a normal distribution witha mean of .3% and a standard deviation of .4%. What is the probability that the price will drop 1% or more in one week?
  1. We cannot answer these types of questions just by using a calculator because there is no formula to evaluate probabilities for the normal distribution. Instead, we use the table of cumulative probabilities for the standard normal distribution. The standard normal distribution is N (0, 1); i.e., the normal distribution with mean 0 and variance 1. Probabilities for any normal distribution N (μ, σ^2 ) can be found from a table for N (0, 1). (The table appears at the end of these notes.) To see this, we need a few properties of normal random variables.
  2. Converting a probability question concerning a general normal distribution N (μ, σ^2 ) into one concerning the standard normal N (0, 1) is called standardizing. Intuitively, switch- ing from N (μ, σ^2 ) involves thinking about the problem in terms of standard deviations away from the mean.
  3. The key to standardizing is the following property: If X ∼ N (μ, σ^2 ) th enZ defined by

Z =

X − μ σ is a standard normal random variable: Z ∼ N (0, 1). Notice that

X = μ + Zσ

so Z measures X in standard deviations away from μ.

  1. We need the following property: any linear transformation of a normal random variable is normal: if X is normal, then so is a + bX for any constants a and b.
  1. Keep in mind that the normal distribution is continuous, so P (X ≤ x) = P (X < x) for all x. In oth er words,P (X = x) = 0 for all x.
  2. Notice that the standard normal table only gives probabilities P (Z ≤ z) for positive values of z. To find P (Z ≤ −z) for negative values −z, we use the symmetry of the normal distribution. If z > 0, then

P (Z ≤ −z) = P (Z ≥ z) = 1 − P (Z ≤ z).

Thus, to find P (Z ≤ −z), look up P (Z ≤ z) in the table and subtract the tabulated value from 1.

  1. Examples: P (Z > − 1 .5) is the shaded area in Figure 3. By symmetry, this is the same as the shaded area in Figure 2. So, to find P (Z > − 1 .5) we look up P (Z < 1 .5) which is 0.9332. Now suppose we want P (Z < − 1 .5). This is the unshaded area in Figure 3. By symmetry, this is the same as the unshaded area in Figure 2. Since the total area under the curve is 1, this unshaded area is given by 1 − 0 .9332 = 0.0668.

−3.5^0 − 3 −2.5 − 2 −1.5 − 1 −0.5 0 0.5 1 1.5 2 2.5 3 3.

Figure 3: By symmetry of the normal distribution, P (Z > − 1 .5), the shaded area, is the same as P (Z < 1 .5), the shaded area in Figure 2. The unshaded areas in the two figures are also the same and are equal to P (Z < − 1 .5) and P (Z > 1 .5).

  1. How do we find P (− 1. 5 < Z < 1 .5) the shaded area in Figure 4? Write it as P (Z < 1 .5) − P (Z < − 1 .5): it’s th e area to th e left of 1.5 but with th e area to left of− 1 .5 taken away. Using the values we found previously, this is 0. 9332 − 0 .0668 = 0.8664.
  2. Example: Consider the potato chip problem above. We want to find P (X < 8) if X ∼ N (8. 3 , (0.2)^2 ). By standardizing, we find that

P (X < 8) = P (

X − 8. 3

) = P (Z < − 1 .5).

We now look up the value 1.5 in the table to get .9332. Now subtract from 1 to get 1-. = .0668. We conclude that P (X < 8) = P (Z < − 1 .5) = 1 − P (Z ≤ 1 .5) =. 0668.

  1. By inverting the steps that take us from X to Z, we get, once again,

X = μ + Zσ.

−3.5^0 − 3 −2.5 − 2 −1.5 − 1 −0.5 0 0.5 1 1.5 2 2.5 3 3.

Figure 4: The shaded area is P (− 1. 5 < Z < 1 .5). We find it from the normal table by expressing it as P (Z < 1 .5) − P (Z < − 1 .5).

As noted above, the standardized r.v. Z tells us how many standard deviations away from its mean the random variable X lands. In the potato-chip example, we should think of the 8-ounce minimum required as 1.5 standard deviations below the mean of 8.3 ounces.

  1. Sometimes, we use the table to go in the opposite direction. Example: Suppose the potato-chip dispenser can be adjusted to any mean while leaving the standard deviation unchanged. To what mean μ should we set the machine so that only 5% of bags contain less than 8 ounces? Answer: Let X be the weight of chips in a bag when the machine is set to a mean of μ; X ∼ N (μ, (0.2)^2 ). We want to set μ so that P (X < 8) = .05; i.e., . 05 = P (X < 8) = P ( X − μ 0. 2

8 − μ

  1. 2

= P (Z <

8 − μ

  1. 2

Clearly, μ will have to be bigger than 8, so (8 − μ)/ 0 .2 is negative. We should therefore write .05 = P (Z <

8 − μ

  1. 2

) = 1 − P (Z <

μ − 8

  1. 2

i.e., P (Z <

μ − 8

  1. 2

Now we look for .95 in the body of the table, since this is our target probability. We find that this corresponds to z = 1.65. Thus, we must choose μ so that μ − 8

  1. 2

In other words, μ = 8 + 1.65(0.2) = 8. 33. This tells us that we must set the machine 1.65 standard deviations above the threshhold of 8 to make sure that only 5% of bags fall below the threshhold.

Cumulative Probabilities for the Standard Normal Distribution.

This table gives probabilities to the left of given z values for the

  • z 0.00 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0. standard normal distribution - 0 0.5000 0.5040 0.5080 0.5120 0.5160 0.5199 0.5239 0.5279 0.5319 0.
    • 0.1 0.5398 0.5438 0.5478 0.5517 0.5557 0.5596 0.5636 0.5675 0.5714 0.
    • 0.2 0.5793 0.5832 0.5871 0.5910 0.5948 0.5987 0.6026 0.6064 0.6103 0.
    • 0.3 0.6179 0.6217 0.6255 0.6293 0.6331 0.6368 0.6406 0.6443 0.6480 0.
    • 0.4 0.6554 0.6591 0.6628 0.6664 0.6700 0.6736 0.6772 0.6808 0.6844 0.
    • 0.5 0.6915 0.6950 0.6985 0.7019 0.7054 0.7088 0.7123 0.7157 0.7190 0.
    • 0.6 0.7257 0.7291 0.7324 0.7357 0.7389 0.7422 0.7454 0.7486 0.7517 0.
    • 0.7 0.7580 0.7611 0.7642 0.7673 0.7704 0.7734 0.7764 0.7794 0.7823 0.
    • 0.8 0.7881 0.7910 0.7939 0.7967 0.7995 0.8023 0.8051 0.8078 0.8106 0.
    • 0.9 0.8159 0.8186 0.8212 0.8238 0.8264 0.8289 0.8315 0.8340 0.8365 0.
      • 1 0.8413 0.8438 0.8461 0.8485 0.8508 0.8531 0.8554 0.8577 0.8599 0.
    • 1.1 0.8643 0.8665 0.8686 0.8708 0.8729 0.8749 0.8770 0.8790 0.8810 0.
    • 1.2 0.8849 0.8869 0.8888 0.8907 0.8925 0.8944 0.8962 0.8980 0.8997 0.
    • 1.3 0.9032 0.9049 0.9066 0.9082 0.9099 0.9115 0.9131 0.9147 0.9162 0.
    • 1.4 0.9192 0.9207 0.9222 0.9236 0.9251 0.9265 0.9279 0.9292 0.9306 0.
    • 1.5 0.9332 0.9345 0.9357 0.9370 0.9382 0.9394 0.9406 0.9418 0.9429 0.
    • 1.6 0.9452 0.9463 0.9474 0.9484 0.9495 0.9505 0.9515 0.9525 0.9535 0.
    • 1.7 0.9554 0.9564 0.9573 0.9582 0.9591 0.9599 0.9608 0.9616 0.9625 0.
    • 1.8 0.9641 0.9649 0.9656 0.9664 0.9671 0.9678 0.9686 0.9693 0.9699 0.
    • 1.9 0.9713 0.9719 0.9726 0.9732 0.9738 0.9744 0.9750 0.9756 0.9761 0.
      • 2 0.9772 0.9778 0.9783 0.9788 0.9793 0.9798 0.9803 0.9808 0.9812 0.
    • 2.1 0.9821 0.9826 0.9830 0.9834 0.9838 0.9842 0.9846 0.9850 0.9854 0.
    • 2.2 0.9861 0.9864 0.9868 0.9871 0.9875 0.9878 0.9881 0.9884 0.9887 0.
    • 2.3 0.9893 0.9896 0.9898 0.9901 0.9904 0.9906 0.9909 0.9911 0.9913 0.
    • 2.4 0.9918 0.9920 0.9922 0.9925 0.9927 0.9929 0.9931 0.9932 0.9934 0.
    • 2.5 0.9938 0.9940 0.9941 0.9943 0.9945 0.9946 0.9948 0.9949 0.9951 0.
    • 2.6 0.9953 0.9955 0.9956 0.9957 0.9959 0.9960 0.9961 0.9962 0.9963 0.
    • 2.7 0.9965 0.9966 0.9967 0.9968 0.9969 0.9970 0.9971 0.9972 0.9973 0.
    • 2.8 0.9974 0.9975 0.9976 0.9977 0.9977 0.9978 0.9979 0.9979 0.9980 0.
    • 2.9 0.9981 0.9982 0.9982 0.9983 0.9984 0.9984 0.9985 0.9985 0.9986 0.
      • 3 0.9987 0.9987 0.9987 0.9988 0.9988 0.9989 0.9989 0.9989 0.9990 0.
    • 3.1 0.9990 0.9991 0.9991 0.9991 0.9992 0.9992 0.9992 0.9992 0.9993 0.
    • 3.2 0.9993 0.9993 0.9994 0.9994 0.9994 0.9994 0.9994 0.9995 0.9995 0.
    • 3.3 0.9995 0.9995 0.9995 0.9996 0.9996 0.9996 0.9996 0.9996 0.9996 0.
    • 3.4 0.9997 0.9997 0.9997 0.9997 0.9997 0.9997 0.9997 0.9997 0.9997 0.