The normal distribution

The Standard Normal Distribution

The standardized normal distribution is a type of normal distribution, with a mean of 0 and standard deviation of 1. It represents a distribution of standardized scores, called z-scores, as opposed to raw scores (the actual data values). A z-score indicates the number of standard deviation a score falls above or below the mean. Z-scores allow for comparison of scores, occurring in different data sets, with different means and standard deviations. It would not make sense to compare apples and oranges. Likewise, it does not make sense to compare scores from two different samples that have different means and standard deviations. Z-scores can be looked up in a Z-Table of Standard Normal Distribution, in order to find the area under the standard normal curve, between a score and the mean, between two scores, or above or below a score. The standard normal distribution allows us to interpret standardized scores and provides us with one table that we may use, in order to compute areas under the normal curve, for an infinite number of data sets, no matter what the mean or standard deviation.

A z-score is calculated as \(z=\frac{x-\mu }{\sigma }\). The score itself can be found by using algebra and solving for x. Multiplying both sides of the equation by σ gives: \((z)(\sigma )=x-\mu\). Adding μ to both sides of the equation gives \(\mu +(z)(\sigma )=x\).

Suppose we have a data set with a mean of 5 and standard deviation of 2. We want to determine the number of standard deviations the score of 11 falls above the mean. We can find this answer (or z-score) by writing

\[z=\frac{11-5}{2}=3\]

or

\[5+(z)(2)=11\text{,}\]

we can solve for z.

\[\begin{array}{l}2z=6 \\ z=3\end{array}\]

We have determined that the score of 11 falls 3 standard deviations above the mean of 5.

With a standard normal distribution, we indicate the distribution by writing Z ~ N(0, 1) which shows the normal distribution has a mean of 0 and standard deviation of 1. This notation simply indicates that a standard normal distribution is being used.

The Standard Normal Distribution

As described previously, if X is a normally distributed random variable and X ~ N(μ, σ), then the z-score is

\[z=\frac{x\ -\ \mu }{\sigma }\text{.}\]

The z-score tells you how many standard deviations the value x is above, to the right of, or below, to the left of, the mean, μ. Values of x that are larger than the mean have positive z-scores, and values of x that are smaller than the mean have negative z-scores. If x equals the mean, then x has a z-score of zero.

When determining the z-score for an x-value, for a normal distribution, with a given mean and standard deviation, the notation above for a normal distribution, will be given.

Example

Suppose X ~ N(5, 6). This equation says that X is a normally distributed random variable with mean μ = 5 and standard deviation σ = 6. Suppose x = 17. Then,

\[z=\frac{x-\mu }{\sigma }=\frac{17-5}{6}=2\text{.}\]

This means that x = 17 is two standard deviations (2σ) above, or to the right, of the mean μ = 5.

Notice that 5 + (2)(6) = 17. The pattern is μ + zσ = x.

Now suppose x = 1. Then, z = \(\frac{x-\mu }{\sigma }\) = \(\frac{1-5}{6}\) = –0.67, rounded to two decimal places.

This means that x = 1 is 0.67 standard deviations (–0.67σ) below or to the left of the mean μ = 5. This z-score shows that x = 1 is less than 1 standard deviation below the mean of 5. Therefore, the score doesn't fall very far below the mean.

Summarizing, when z is positive, x is above or to the right of μ, and when z is negative, x is to the left of or below μ. Or, when z is positive, x is greater than μ, and when z is negative, x is less than μ. The absolute value of z indicates how far the score is from the mean, in either direction.

If X is a random variable and has a normal distribution with mean µ and standard deviation σ, then the Empirical Rule states the following:

So, in other words, this is that about 68 percent of the values lie between z-scores of –1 and 1, about 95% of the values lie between z-scores of –2 and 2, and about 99.7 percent of the values lie between z-scores of -3 and 3. These facts can be checked, by looking up the mean to z area in a z-table for each positive z-score and multiplying by 2.

Condensed — the full section is in OpenStax Statistics.

The Standard Normal Distribution

Use the following information to answer the next two exercises: The life of Sunshine CD players is normally distributed with mean of 4.1 years and a standard deviation of 1.3 years. A CD player is guaranteed for three years. We are interested in the length of time a CD player lasts.

Using the Normal Distribution

The shaded area in the following graph indicates the area to the left of x. This area could represent the percentage of students scoring less than a particular grade on a final exam. This area is represented by the probability P(X < x). Normal tables, computers, and calculators are used to provide or calculate the probability P(X < x).

The area to the right is then P(X > x) = 1 – P(X < x). Remember, P(X < x) = Area to the left of the vertical line through x. P(X < x) = 1 – P(X < x) = Area to the right of the vertical line through x. P(X < x) is the same as P(X ≤ x) and P(X > x) is the same as P(X ≥ x) for continuous distributions.

Suppose the graph above were to represent the percentage of students scoring less than 75 on a final exam, with this probability equal to 0.39. This would also indicate that the percentage of students scoring higher than 75 was equal to 1 minus 0.39 or 0.61.

Calculations of Probabilities

Probabilities are calculated using technology. There are instructions given as necessary for the TI-83+ and TI-84 calculators.

Example

If the area to the left is 0.0228, then the area to the right is 1 – 0.0228 = 0.9772.

Condensed — the full section is in OpenStax Statistics.

Isibonelo esisebenza: integrate e^(-x^2/2)/sqrt(2*pi) dx from -1 to 1

Practice (40)

Try each one on paper first. Reveal the answer to check; verified ones can be opened in the solver for every step.

1. What is the z-score of x, when x = 1 and X ~ N(12, 3)?

   Bonisa impendulo

   \(z=\frac{1-12}{3}\approx -3.67\)

2. a. Suppose a person lost 10 pounds in a month. The z-score when x = 10 pounds is z = 2.5 (verify). This z-score tells you that x = 10 is ________ standard deviations to the ________ (right or left) of the mean _____ (What is the mean?).

   Bonisa impendulo

   a. This z-score tells you that x = 10 is 2.5 standard deviations to the right of the mean five.
   

3. b. Suppose a person gained three pounds, a negative weight loss. Then z = __________. This z-score tells you that x = –3 is ________ standard deviations to the __________ (right or left) of the mean.

   Bonisa impendulo

   b. z = –4. This z-score tells you that x = –3 is four standard deviations to the left of the mean.
   

4. c. Suppose the random variables X and Y have the following normal distributions: X ~ N(5, 6) and Y ~ N(2, 1). If x = 17, then z = 2. This was previously shown. If y = 4, what is z?

   Bonisa impendulo

   c. z = \(\frac{y-\mu }{\sigma }\) = \(\frac{4-2}{1}\) = 2, where µ = 2 and σ = 1.

   The z-score for y = 4 is z = 2. This means that four is z = 2 standard deviations to the right of the mean. Therefore, x = 17 and y = 4 are both two of their own standard deviations to the right of their respective means.

   The z-score allows us to compare data that are scaled differently. To better understand the concept, suppose X ~ N(5, 6) represents weight gains for one group of people who are trying to gain weight in a six-week period and Y ~ N(2, 1) measures the same weight gain for a second group of people. A negative weight gain would be a weight loss. Since x = 17 and y = 4 are each two standard deviations to the right of their means, they represent the same, standardized weight gain relative to their means.

5. Fill in the blanks.

   Jerome averages 16 points a game with a standard deviation of four points. X ~ N(16, 4). Suppose Jerome scores 10 points in a game. The z-score when x = 10 is –1.5. This score tells you that x = 10 is _____ standard deviations to the ______ (right or left) of the mean______ (What is the mean?).

   Bonisa impendulo

   1.5, left, 16

6. a. Suppose a 15-to 18-year-old male from Chile was 168 cm tall in 2009–2010. The z-score when x = 168 cm is z = _______. This z-score tells you that x = 168 is ________ standard deviations to the ________ (right or left) of the mean _____ (What is the mean?).

   Bonisa impendulo

   a. –0.32, 0.32, left, 170
   

7. b. Suppose that the height of a 15-to 18-year-old male from Chile in 2009–2010 has a z-score of z = 1.27. What is the male’s height? The z-score (z = 1.27) tells you that the male’s height is ________ standard deviations to the __________ (right or left) of the mean.

   Bonisa impendulo

   b. 177.98 cm, 1.27, right

8. Use the information in to answer the following questions:

   1. Suppose a 15-to 18-year-old male from Chile was 176 cm tall from 2009–2010. The z-score when x = 176 cm is z = _______. This z-score tells you that x = 176 cm is ________ standard deviations to the ________ (right or left) of the mean _____ (What is the mean?).
   2. Suppose that the height of a 15-to 18-year-old male from Chile in 2009–2010 has a z-score of z = –2. What is the male’s height? The z-score (z = –2) tells you that the male’s height is ________ standard deviations to the __________ (right or left) of the mean.
   Bonisa impendulo

   Solve the equation z = \(\frac{x-\mu }{\sigma }\) for x. x = μ + (z)(σ)

   1. z = \(\frac{176-170}{6.28}\) ≈ 0.96, This z-score tells you that x = 176 cm is 0.96 standard deviations to the right of the mean 170 cm.
   2. X = 157.44 cm, The z-score(z = –2) tells you that the male’s height is two standard deviations to the left of the mean.

9. From 1984 to 1985, the mean height of 15-to 18-year-old males from Chile was 172.36 cm, and the standard deviation was 6.34 cm. Let Y = the height of 15-to 18-year-old males from 1984–1985, and y = the height of one male from this group. Then Y ~ N(172.36, 6.34).

   The mean height of 15-to 18-year-old males from Chile in 2009–2010 was 170 cm with a standard deviation of 6.28 cm. Male heights are known to follow a normal distribution. Let X = the height of a 15-to 18-year-old male from Chile in 2009–2010, and x = the height of one male from this group. Then X ~ N(170, 6.28).

   Find the z-scores for x = 160.58 cm and y = 162.85 cm. Interpret each z-score. What can you say about x = 160.58 cm and y = 162.85 cm as they compare to their respective means and standard deviations?

   Bonisa impendulo

   The z-score for x = 160.58 cm is z = –1.5.
   The z-score for y = 162.85 cm is z = –1.5.
   Both x = 160.58 and y = 162.85 deviate the same number of standard deviations from their respective means and in the same direction.

10. In 2012, 1,664,479 students took the SAT exam. The distribution of scores in the verbal section of the SAT had a mean µ = 496 and a standard deviation σ = 114. Let X = a SAT exam verbal section score in 2012. Then, X ~ N(496, 114).

    Find the z-scores for x₁ = 325 and x₂ = 366.21. Interpret each z-score. What can you say about x₁ = 325 and x₂ = 366.21, as they compare to their respective means and standard deviations?

    Bonisa impendulo

    The z-score for x₁ = 325 is z₁ = –1.14.

    The z-score for x₂ = 366.21 is z₂ = –1.14.

    Student 2 scored closer to the mean than Student 1 and, since they both had negative z-scores, Student 2 had the better score.

11. Suppose X has a normal distribution with mean 25 and standard deviation five. Between what values of x do 68 percent of the values lie?

    Bonisa impendulo

    between 20 and 30.

12. From 1984–1985, the mean height of 15-to 18-year-old males from Chile was 172.36 cm, and the standard deviation was 6.34 cm. Let Y = the height of 15-to 18-year-old males in 1984–1985. Then Y ~ N(172.36, 6.34).

    1. About 68 percent of the y values lie between what two values? These values are ________________. The z-scores are ________________, respectively.
    2. About 95 percent of the y values lie between what two values? These values are ________________. The z-scores are ________________ respectively.
    3. About 99.7 percent of the y values lie between what two values? These values are ________________. The z-scores are ________________, respectively.

    Bonisa impendulo

    1. About 68 percent of the values lie between 166.02 cm and 178.7 cm. The z-scores are –1 and 1.
    2. About 95 percent of the values lie between 159.68 cm and 185.04 cm. The z-scores are –2 and 2.
    3. About 99.7 percent of the values lie between 153.34 cm and 191.38 cm. The z-scores are –3 and 3.

13. The scores on a college entrance exam have an approximate normal distribution with mean, µ = 52 points and a standard deviation, σ = 11 points.

    1. About 68 percent of the y values lie between what two values? These values are ________________. The z-scores are ________________, respectively.
    2. About 95 percent of the y values lie between what two values? These values are ________________. The z-scores are ________________, respectively.
    3. About 99.7 percent of the y values lie between what two values? These values are ________________. The z-scores are ________________, respectively.
    Bonisa impendulo

    1. About 68% of the values lie between the values 41 and 63. The z-scores are –1 and 1, respectively.
    2. About 95% of the values lie between the values 30 and 74. The z-scores are –2 and 2, respectively.
    3. About 99.7% of the values lie between the values 19 and 85. The z-scores are –3 and 3, respectively.

14. A bottle of water contains 12.05 fluid ounces with a standard deviation of 0.01 ounces. Define the random variable X in words. X = ____________.

    Bonisa impendulo

    ounces of water in a bottle

15. A normal distribution has a mean of 61 and a standard deviation of 15. What is the median?

16. X ~ N(1, 2)

    σ = _______

    Bonisa impendulo

    2

17. A company manufactures rubber balls. The mean diameter of a ball is 12 cm with a standard deviation of 0.2 cm. Define the random variable X in words. X = ______________.

18. X ~ N(–4, 1)

    What is the median?

    Bonisa impendulo

    –4

19. X ~ N(3, 5)

    σ = _______

20. X ~ N(–2, 1)

    μ = _______

    Bonisa impendulo

    –2

21. What does a z-score measure?

22. What does standardizing a normal distribution do to the mean?

    Bonisa impendulo

    The mean becomes zero.

23. Is X ~ N(0, 1) a standardized normal distribution? Why or why not?

24. What is the z-score of x = 12, if it is two standard deviations to the right of the mean?

    Bonisa impendulo

    z = 2

25. What is the z-score of x = 9, if it is 1.5 standard deviations to the left of the mean?

26. What is the z-score of x = –2, if it is 2.78 standard deviations to the right of the mean?

    Bonisa impendulo

    z = 2.78

27. What is the z-score of x = 7, if it is 0.133 standard deviations to the left of the mean?

28. Suppose X ~ N(2, 6). What value of x has a z-score of three?

    Bonisa impendulo

    x = 20

29. Suppose X ~ N(8, 1). What value of x has a z-score of –2.25?

30. Suppose X ~ N(9, 5). What value of x has a z-score of –0.5?

    Bonisa impendulo

    x = 6.5

31. Suppose X ~ N(2, 3). What value of x has a z-score of –0.67?

32. Suppose X ~ N(4, 2). What value of x is 1.5 standard deviations to the left of the mean?

    Bonisa impendulo

    x = 1

33. Suppose X ~ N(4, 2). What value of x is two standard deviations to the right of the mean?

34. Suppose X ~ N(8, 9). What value of x is 0.67 standard deviations to the left of the mean?

    Bonisa impendulo

    x = 1.97

35. Suppose X ~ N(–1, 2). What is the z-score of x = 2?

36. Suppose X ~ N(12, 6). What is the z-score of x = 2?

    Bonisa impendulo

    z = –1.67

37. Suppose X ~ N(9, 3). What is the z-score of x = 9?

38. Suppose a normal distribution has a mean of six and a standard deviation of 1.5. What is the z-score of x = 5.5?

    Bonisa impendulo

    z ≈ –0.33

39. In a normal distribution, x = 5 and z = –1.25. This tells you that x = 5 is ____ standard deviations to the ____ (right or left) of the mean.

40. In a normal distribution, x = 3 and z = 0.67. This tells you that x = 3 is ____ standard deviations to the ____ (right or left) of the mean.

    Bonisa impendulo

    0.67, right

Symbols used here

\int f(x)\,dx,\ \int_a^b

integral
Antiderivative (indefinite) or signed area from a to b (definite).

\sqrt{x},\ \sqrt[n]{x}

square root, n-th root
The non-negative number whose square (n-th power) is x.

\pi

pi
Ratio of a circle's circumference to its diameter, 3.14159…

\sigma,\ s,\ \sigma^2

standard deviation, sample s.d., variance
Typical distance from the mean; its square.

\bar{x},\ \mu

sample mean, population mean
Average of the data; average of the whole population.

e

Euler's number
2.71828…, the base whose exponential is its own derivative.

\approx

approximately equal
Equal to the precision shown, not exactly.

\leq,\ \geq

less/greater than or equal
Inequalities that allow equality; < and > exclude it.

\pm

plus or minus
Both signs at once: x = 3 ± 2 means 5 and 1.

n!

factorial
n × (n−1) × … × 1; the number of orderings of n things. 0! = 1.

\binom{n}{k}

binomial coefficient, "n choose k"
Number of k-element subsets of n things: n!/(k!(n−k)!).

\sum_{k=1}^{n} a_k

summation
Add a_k for k = 1 up to n.

A \cup B,\ A \cap B,\ A \setminus B

union, intersection, difference
In either; in both; in A but not B.

P(A),\ P(A \mid B)

probability, conditional probability
Chance of A; chance of A given that B happened.

E[X],\ \operatorname{Var}(X)

expected value, variance
Probability-weighted average of X; its spread.

N(\mu, \sigma^2),\ z

normal distribution, z-score
The bell curve with mean μ and variance σ²; (x − μ)/σ.

How to: The normal distribution

1. First find an antiderivative F, then evaluate F(b) − F(a).
2. Pull the constant \frac{\sqrt{2}}{2 \sqrt{\pi}} out of the integral.
3. Erf rule.
4. Fundamental theorem of calculus: plug in the limits.
5. Simplify.

Questions people ask

Mean or median — which should I use?

Median when the data have outliers or a long tail (incomes, house prices); mean when the data are roughly symmetric and you want every value to count. Report both if they disagree — the gap is itself information.

What does a p-value actually say?

The probability of seeing data at least this extreme if the null hypothesis were true. It is not the probability that the null hypothesis is true.

Why divide by n − 1 for the sample variance?

The sample mean sits closer to the sample than the true mean does, so squared deviations from it are slightly too small on average; dividing by n − 1 instead of n corrects the bias.

Zama wena

⌨ Type ✎ Draw

Inkinga yakho ye-matematics

Isinqumo

x²√÷×π( )=∫d/dxlim[ ]sinlogeˣ

(72 - 65)/3.5integrate e^(-x^2/2)/sqrt(2*pi) dx from -2 to 2integrate e^(-x^2/2)/sqrt(2*pi) dx from -oo to oo(50 - 60)/8

Parts of this page are adapted from OpenStax Statistics (CC BY 4.0). Condensed and re-explained here; errors are ours.

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