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Chapter 10
The Normal Distribution
10.1 Introduction
The normal distribution is possibly the best known and most used continuous probability distribu- tion. It provides a good model for data in very many different applications, for example, the yields of crops, the heights of people, students’ marks. The outcomes of many production processes also follow normal distributions and hence it is used widely in industry.
The normal distribution has two parameters: the mean μ and the variance σ^2. The standard devia- tion σ =
σ^2 but we usually use the variance to specify the parameters. The probability density function of a normal distribution is often said to be “bell shaped” :
The formula for the pdf is
f (x) =
2 πσ^2
exp
(x − μ)^2 2 σ^2
There is no simple formula for calculating probabilities. However, they can be determined using tables or statistical packages such as Minitab.
There are four important characteristics of the normal distribution:
- It is symmetrical about its mean, μ.
- The mean, median and mode all coincide.
- The area under the curve is equal to 1.
- The curve extends for ever in both directions to infinity (i.e. to ±∞).
Below is a plot of the pdf of normal distributions for different values of μ and σ.
Note that the mean μ locates the distribution on the x–axis and the standard deviation σ affects the spread of the distribution, with larger values giving flatter and wider curves.
10.2 Notation
If a random variable X has a normal distribution with mean μ and variance σ^2 , then we write
X ∼ N
μ, σ^2
For example, a random variable X which follows a normal distribution with mean 10 and variance 25 is written as X ∼ N (10, 25) or X ∼ N (10, 52 ). It is important to note that the second parameter in this notation is the variance and not the standard deviation.
10.3 Some properties of normal variables
An important property concerns addition (or subtraction) of normal random variables. If X 1 and X 2 both have normal distributions and Y = X 1 + X 2 then Y also has a normal distribution. If Z = X 1 − X 2 then Z also has a normal distribution. If X 1 ∼ N (μ 1 , σ 12 ) and X 2 ∼ N (μ 2 , σ 22 ) then Y has mean μy = μ 1 + μ 2 and Z has mean μz = μ 1 − μ 2. If X 1 and X 2 are independent then Y and Z both have variance σ^2 y = σ z^2 = σ^21 + σ^22.
For example, if wagonloads of material have weights which are normally distributed with mean 3 tonnes and standard deviation 0.5 tonnes then the total weight from two wagonloads is normally distributed with mean 3 + 3 = 6 tonnes and standard deviation
- 52 + 0. 52 = 0. 7071 tonnes.
If X ∼ N (μ, σ^2 ) and a and b are fixed numbers and W = a + bX then W ∼ N (a + bμ, k^2 σ^2. For example, if the weight of a load of material has a normal distribution with mean 3 tonnes and
(e) The probability that the random variable Z lies between than − 1. 2 and 1. 5 is
P (− 1. 2 < Z < 1 .5) = P (Z < 1 .5) − P (Z ≤ − 1 .2) = 0. 9332 − 0. 1151 = 0. 8181.
We now consider how to determine probabilities for general normal distributions.
- Suppose we are interested in the IQ of 18-20 year olds and that IQs follow a normal dis- tribution with mean μ = 100 and standard deviation σ = 15. Mathematically we let the random variable X denote the IQ of a randomly chosen person from this age group and then X ∼ N (100, 152 ). Probability statements about IQs can be made as follows.
(a) The probability that an 18-20 year old has an IQ less than 85 is P (X < 85). Using the formula P (X ≤ x) = P
Z ≤
x − μ σ
we need to calculate z =
85 − μ σ
and from tables we obtain P (Z < −1) = 0. 1587. Therefore
P (X < 85) = 0. 1587.
(b) The probability that an 18-20 year old has an IQ greater than 142 is P (X > 142). Now P (X > 142) = 1 − P (X ≤ 142) and
z =
142 − μ σ
Using tables, we see that P (Z ≤ 2 .8) = 0. 9974 and so
P (X > 142) = 1 − 0 .9974 = 0. 0026.
- Suppose that the vitamin C content per 100 g tin of tomato juice is normally distributed with mean μ = 20mg and standard deviation σ = 4mg. Let X be the vitamin C content of a randomly chosen tin.
(a) The probability that the tin has less than 25 mg of vitamin C is P (X < 25). Now
z =
25 − μ σ
and from tables we obtain P (Z < 1 .25) = 0. 8944. Therefore
P (X < 25) = 0. 8944.
(b) The probability that the tin has more than 25 mg of vitamin C is P (X > 25). Now P (X > 25) = 1 − P (X ≤ 25) and so
P (X > 25) = 1 − 0 .8944 = 0. 1056.
(c) The probability that the tin has between 18 mg and 25 mg of vitamin C is
P r(18 < X < 25) = P (X < 25) − P (X ≤ 18).
We can determine P (X ≤ 18) from tables using
z =
18 − μ σ
giving P (X ≤ 18) = P (Z < − 0 .5) = 0. 3085. Therefore
P r(18 < X < 25) = 0. 8944 − 0 .3085 = 0. 5859.
We can also use the tables in reverse. For example, we might want to know below what value are 95% of the population. This is equivalent to determining the value of z that satisfies P (Z < z) = 0. 95. From tables, we can see that P (Z < 1 .64) = 0. 9495 and P (Z < 1 .65) = 0. 9505. Therefore the value we want for z lies between 1.64 and 1.65. If a more accurate value is needed we can interpolate between these values: 0.95 is half-way between 0.9495 and 0.9505 and so we take z = 1. 645. This is a more accurate answer and sufficient in most cases. However, the exact value for z can be found from more detailed tables or via a computer package such as Minitab. Here are some more examples.
- Below what value does 10% of the standard normal population fall? From tables we get
P (Z < − 1 .28) = 0. 1003 and P (Z < − 1 .29) = 0. 0985
and so we take
z = − 1 .29 +
× {− 1. 28 − (− 1 .29)}
× 0. 01
In other words P (Z < − 1 .2817) = 0. 1 and so 10% of the standard normal population falls below − 1. 2817.
- A similar calculation can be used to calculate the IQ that identifies the 10% of 18-20 year olds with the smallest IQ. We need the value of x, where P (X < x) = 0. 1. Now this population has μ = 100 and σ = 15. Also
P (X ≤ x) = P
Z ≤
x − μ σ
and so we need x so that P
Z ≤
x − 100 15
We know (from earlier) that P (Z < − 1 .2817) = 0. 1 and therefore we solve
x − 100 15
Minitab is very helpful with calculating normal probabilities. The following commands will cal- culate probabilities P (X < x) and also values of x that satisfy P (X < x) = p:
- Calc > Probability Distributions > Normal
opens up dialogue box
- Select Cumulative probability for P (X < x) or Inverse cumulative probability for the value of x satisfying P (X < x) = p
- Enter the Mean (μ) and the Standard Deviation (σ)
- Select Input Constant and enter the value for x or p (as appropriate)
- Click OK
- The answer is displayed in the Session Window:
10.4 The normal approximation to the binomial distribution
The normal distribution can be used as an approximation to the binomial distribution Bin(n, p) for large n and medium p. (Say if both np and n(1 − p) are greater than about 7). To approximate
a Bin(n, p) distribution we use a normal distribution with the same mean and variance. That is μ = np and σ^2 = np(1 − p).
Example. We plan to do a market research survey in which people will be asked whether or not they would buy a new product. A random sample of 600 people will be asked. Suppose that the true proportion of people in the population who would buy the product is 40%, i.e. p = 0. 4. Find the probability that, in our survey, between 220 and 260 (inclusive) answer “Yes.”
In this case n = 600 and p = 0. 4 so np = 240 and n(1 − p) = 360. Therefore we can use the approximation. We use the normal distribution with mean μ = np = 240 and variance σ^2 = np(1 − p) = 144. That is N (240, 144). The standard deviation is σ =
If X ∼ N (240, 144) then
Pr(X < 220) = Pr
X − 240
= Pr(Z < − 1 .67) = 0. 0475.
So Pr( fewer than 220 “Yes” ) ≈ 0. 0475.
We can sometimes obtain a better approximation by using a continuity correction. Since B(n, p) is a discrete distribution, Pr( number of “Yes” = 220) > 0. Using the continuity correction, we count as “220” everything between 219.5 and 220.5 so, for Pr( number of “Yes” < 220) we would use
Pr(X < 219 .5) = Pr
Z <
= Pr(Z < − 1 .71) ≈ 0. 0436.
Here the continuity correction makes a noticeable difference. Sometimes it does not.
In the same way
Pr(X < 260 .5) = Pr
Z <
= Pr(Z < 1 .71) ≈ 0 .9564 = 1 − 0. 0436.
So the probability that, in our survey, between 220 and 260 (inclusive) people say “Yes” is approx- imately 0. 9564 − 0 .0436 = 0. 9128 ≈ 0. 91.
- Bags of produce have a nominal weight of 1kg. In fact the weights have a normal distri- bution with mean 1064g and standard deviation 50g. A bag which weighs less than 1kg is considered to be underweight.
(a) Find the probability that a bag is underweight. (b) Assuming that the weights of bags are independent, find an approximate value for the probability that, in a batch of 100 bags, more than 15 are underweight.
Probability Tables for the Standard Normal Distribution
- z -0.09 -0.08 -0.07 -0.06 -0.05 -0.04 -0.03 -0.02 -0.01 0. The table contains values of P r(Z < z), where Z ∼ N (0, 1).
- -2.9 0.0014 0.0014 0.0015 0.0015 0.0016 0.0016 0.0017 0.0018 0.0018 0.
- -2.8 0.0019 0.0020 0.0021 0.0021 0.0022 0.0023 0.0023 0.0024 0.0025 0.
- -2.7 0.0026 0.0027 0.0028 0.0029 0.0030 0.0031 0.0032 0.0033 0.0034 0.
- -2.6 0.0036 0.0037 0.0038 0.0039 0.0040 0.0041 0.0043 0.0044 0.0045 0.
- -2.5 0.0048 0.0049 0.0051 0.0052 0.0054 0.0055 0.0057 0.0059 0.0060 0.
- -2.4 0.0064 0.0066 0.0068 0.0069 0.0071 0.0073 0.0075 0.0078 0.0080 0.
- -2.3 0.0084 0.0087 0.0089 0.0091 0.0094 0.0096 0.0099 0.0102 0.0104 0.
- -2.2 0.0110 0.0113 0.0116 0.0119 0.0122 0.0125 0.0129 0.0132 0.0136 0.
- -2.1 0.0143 0.0146 0.0150 0.0154 0.0158 0.0162 0.0166 0.0170 0.0174 0.
- -2.0 0.0183 0.0188 0.0192 0.0197 0.0202 0.0207 0.0212 0.0217 0.0222 0.
- -1.9 0.0233 0.0239 0.0244 0.0250 0.0256 0.0262 0.0268 0.0274 0.0281 0.
- -1.8 0.0294 0.0301 0.0307 0.0314 0.0322 0.0329 0.0336 0.0344 0.0351 0.
- -1.7 0.0367 0.0375 0.0384 0.0392 0.0401 0.0409 0.0418 0.0427 0.0436 0.
- -1.6 0.0455 0.0465 0.0475 0.0485 0.0495 0.0505 0.0516 0.0526 0.0537 0.
- -1.5 0.0559 0.0571 0.0582 0.0594 0.0606 0.0618 0.0630 0.0643 0.0655 0.
- -1.4 0.0681 0.0694 0.0708 0.0721 0.0735 0.0749 0.0764 0.0778 0.0793 0.
- -1.3 0.0823 0.0838 0.0853 0.0869 0.0885 0.0901 0.0918 0.0934 0.0951 0.
- -1.2 0.0985 0.1003 0.1020 0.1038 0.1056 0.1075 0.1093 0.1112 0.1131 0.
- -1.1 0.1170 0.1190 0.1210 0.1230 0.1251 0.1271 0.1292 0.1314 0.1335 0.
- -1.0 0.1379 0.1401 0.1423 0.1446 0.1469 0.1492 0.1515 0.1539 0.1562 0.
- -0.9 0.1611 0.1635 0.1660 0.1685 0.1711 0.1736 0.1762 0.1788 0.1814 0.
- -0.8 0.1867 0.1894 0.1922 0.1949 0.1977 0.2005 0.2033 0.2061 0.2090 0.
- -0.7 0.2148 0.2177 0.2206 0.2236 0.2266 0.2296 0.2327 0.2358 0.2389 0.
- -0.6 0.2451 0.2483 0.2514 0.2546 0.2578 0.2611 0.2643 0.2676 0.2709 0.
- -0.5 0.2776 0.2810 0.2843 0.2877 0.2912 0.2946 0.2981 0.3015 0.3050 0.
- -0.4 0.3121 0.3156 0.3192 0.3228 0.3264 0.3300 0.3336 0.3372 0.3409 0.
- -0.3 0.3483 0.3520 0.3557 0.3594 0.3632 0.3669 0.3707 0.3745 0.3783 0.
- -0.2 0.3859 0.3897 0.3936 0.3974 0.4013 0.4052 0.4090 0.4129 0.4168 0.
- -0.1 0.4247 0.4286 0.4325 0.4364 0.4404 0.4443 0.4483 0.4522 0.4562 0. - 0.0 0.4641 0.4681 0.4721 0.4761 0.4801 0.4840 0.4880 0.4920 0.4960 0. - z 0.00 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.
- 0.0 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 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 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.
