Download 6.2 The Normal Distribution - MathFlight Learning Resources

6.2 - The Normal Distribution
Objectives:
1.
2.
3.
4.
Use continuous uniform distribution to find probabilities.
Use standard normal distribution to find probabilities.
Find z-scores from known areas.
Find probabilities when given z-scores.
Continuous Random Variables:
Continuous random variables have infinitely many values and those values can be associated with
measurements on a continuous scale without gaps or interruptions.
Uniform Distribution:
A continuous random variable has a uniform distribution if its values are spread evenly over the range of
probabilities. The graph of a uniform distribution results in a rectangular shape.
The uniform distribution allows us to see two very important properties:
1. The area under the graph of a probability distribution is equal to 1.
2. There is a correspondence between area and probability.
Density Curve:
The graph of the uniform distribution is called a density curve. A density curve is the graph of a
continuous probability distribution. It must satisfy the following properties:
1. The total area under the curve must equal 1.
2. Every point on the curve must have a vertical height that is 0 or greater. (That is, the curve
cannot fall below the x-axis.)
Area and Probability:
Because the total area under the density curve is equal to 1, there is a correspondence between area and
probability.
Example: Given the uniform distribution below, find the probability that a randomly selected voltage
level is greater than 124.5 volts.
Solution: The shaded area represents voltage levels greater than 124.5 volts. Because the total area
under the density curve is equal to 1, there is a correspondence between area and probability. The
shaded area is 25% of the total area and the probability is 0.25.
Example: Using the following uniform density curve, what is the probability that the random variable
has a value greater than 2?
Solution: Because the total area under the density curve is equal to 1, there is a correspondence
between area and probability. The area under the curve can be determined as follows:
A = LW
A = (6)(0.125)
A = 0.750
The probability is 0.750.
Example: Assume that the weight loss for the first month of a diet program varies between 6 pounds
and 12 pounds, and is spread evenly over the range of possibilities, so that there is a uniform
distribution. Find the probability that less than 11 pounds will be lost.
Solution: The horizontal axis ranges from 6 to 12. Therefore, in order for the area under the curve to be
equal to 1, the height of the rectangular region must be 1/6 (6*1/6=1). To find the probability that less
than 11 pounds will be lost we use the formula:
A = LW
1
A = (5) 
6
5
A=
6
Normal Distributions:
If a continuous random variable has a distribution with a graph that is symmetric and bell-shaped it can
be described by the equation below and we say that it has normal distribution.
f ( x) =
e
1  x−µ  2
− 

2 σ 
σ 2π
Standard Normal Distribution:
The standard normal distribution is a normal probability distribution with µ = 0 and σ = 1. The total area
under its density curve is equal to 1.
The standard normal distribution has three properties:
1. Its graph is bell-shaped.
2. Its mean is equal to 0 (µ = 0).
3. Its standard deviation is equal to 1 (σ = 1).
The density curve of a uniform distribution is a
horizontal line so we can find the area of any
rectangular region by applying the formula
A = LW.
Because the density curve of a normal distribution
has a complicated bell shape it is more difficult to
find the area. However the basic principle is the
same:
There is a correspondence between area and probability
Notation:
P(a < z < b) denotes the probability that the z score is between a and b.
P(z > a) denotes the probability that the z score is greater than a.
P(z < a) denotes the probability that the z score is less than a.
Finding Probabilities When Given Z-Scores:
It is not easy to find areas under the curve (requires calculus) so mathematicians have calculated many
different areas under the curve, and those areas are included in Table A-2 in Appendix A.
We will learn how to use Table A-2 to determine areas and probabilities. When using the table it is
essential to understand the following points:
1. It is designed only for the standard normal distribution, which has a mean of 0 and a standard
deviation of 1.
2. It is on two pages, with one page for negative z-scores and the other page for positive
z-scores.
3. Each value in the body of the table is a cumulative area from the left up to a vertical boundary
above a specific z-score.
4. When working with a graph, avoid confusion between z-scores and areas.
z Score - Distance along horizontal scale of the standard normal distribution;
refer to the leftmost column and top row of Table A-2.
Area - Region under the curve; refer to the values in the body of Table A-2.
5. The part of the z-score denoting hundredths is found across the top.
Table A-2:
Finding Probabilities with Excel:
The NORMDIST function in Excel can be used to find probabilities. When using this function you will
need to enter the following:
=NORMDIST (z-score, mean, standard deviation, true)
Example: The Precision Scientific Instrument Company manufactures thermometers that are supposed
to give readings of 0ºC at the freezing point of water. Tests on a large sample of these instruments reveal
that at the freezing point of water, some thermometers give readings below 0º (denoted by negative
numbers) and some give readings above 0º (denoted by positive numbers). Assume that the mean
reading is 0ºC and the standard deviation of the readings is 1.00ºC. Also assume that the readings are
normally distributed. If one thermometer is randomly selected, find the probability that, at the freezing
point of water, the reading is less than 1.27º.
Solution: The probability distribution of readings is a standard normal distribution, because the readings
are normally distributed with µ = 0 and σ = 1. We need to find the area below z = 1.27. The area below
z = 1.27 is equal to the probability of randomly selecting a thermometer with a reading of less than
1.27º.
From Table A-2 we find this area to be 0.8980.
P(z < 1.27) = 0.8980
=NORMDIST (1.27, 0, 1, true) = 0.8980
Interpretation: the probability of randomly selecting a thermometer with a reading less than 1.27º is
equal to the area of 0.8980 as shown in the diagram above. Another way to interpret this result is to
conclude that 89.90% of the thermometers will have readings below 1.27º.
Caution - It is important to remember that the area/probability given in the table represents the
area/probability below that z-score.
Example: Find the area of the shaded region. The graph depicts the standard normal distribution with
mean 0 and standard deviation 1.
Solution: From Table A-2 or NORMDIST (1.13, 0, 1, true) we find this area to be 0.8708.
Example: If thermometers have an average (mean) reading of 0 degrees and a standard deviation of 1
degree for freezing water, and if one thermometer is randomly selected, find the probability that it reads
(at the freezing point of water) above –1.23 degrees.
Solution: From table A-2 we determine that a z-score of -1.23 has a probability of 0.1093. This
represents the cumulative area from the left to the z-score of -1.23. Because we want to find the area
that is greater than -1.23, we need to subtract this probability from 1 to obtain the area above the z-score
of 0.8907.
P (z > –1.23) = 1 - NORMDIST (-1.23, 0, 1, true) = 0.8907
Interpretation: Probability of randomly selecting a thermometer with a reading above –1.23º is 0.8907.
Example: Find the area of the shaded region. The graph depicts the standard normal distribution with
mean 0 and standard deviation 1.
Solution: From Table A-2 or 1- NORMDIST (0.59, 0, 1, true) we find this area to be 0.2776
Example: A thermometer is randomly selected. Find the probability that it reads (at the freezing point of
water) between –2.00 and 1.50 degrees.
Solution: The probability of selecting a thermometer that reads between –2.00 and 1.50 degrees
corresponds to the shaded area of the diagram below. To find this area, use table A-2 or NORMDIST
to:
1. Find the area corresponding to a z-score of 1.50.
2. Find the area corresponding to a z-score of -2.00
3. Subtract the areas to obtain the shaded area.
NORMDIST (1.50, 0, 1, true) - NORMDIST (-2.00, 0, 1, true)
Interpretation: The probability that the chosen thermometer has a reading between – 2.00 and 1.50
degrees is 0.9104. In other words, if many thermometers are selected and tested at the freezing point of
water, then 91.04% of them will read between –2.00 and 1.50 degrees.
Example: Find the area of the shaded region. The graph depicts the standard normal distribution with
mean 0 and standard deviation 1.
Solution: From Table A-2 or NORMDIST we find this area to be 0.6424
NORMDIST (0.92, 0, 1, true) - NORMDIST (-0.92, 0, 1, true)
Finding a Z-Score When Given a Probability Using Table A-2:
If we already know a probability and want to find the corresponding z-score, we find it as follows.
1. Draw a bell-shaped curve and identify the region under the curve that corresponds to the given
probability. If that region is not a cumulative region from the left, work instead with a known
region that is a cumulative region from the left.
2. Using the cumulative area from the left, locate the closest probability in the body of Table A-2
and identify the corresponding z score.
In Excel we can use the NORMINV (probability, mean, standard deviation) function in the same
manner as the NORMDIST function.
Example: Find the z-score that corresponds to an area (probability) of 0.95.
Solution: Referring to table A-2, we search for the area of 0.95 in the body of the table and then find the
corresponding z-score. In table A-2 we find the areas of 0.9495 and 0.9505, but there’s an asterisk with
a special note indicating that 0.9500 corresponds to a z-score of 1.645.
NORMINV (0.95, 0, 1) = 1.645
Example: Find the indicated z score. The graph depicts the standard normal distribution with mean 0
and standard deviation of 1. Shaded area is 0.0901.
Solution: From Table A-2 we find the z-score to be – 1.34.
NORMINV (0.0901, 0, 1) = -1.34
Example: Find the indicated z score. The graph depicts the standard normal distribution with mean 0
and standard deviation of 1. Shaded area is 0.8599.
Solution: From Table A-2 we find the z-score to be – 1.08.
NORMINV (0.8599, 0, 1) = -1.08
Example: Find the z-score separating the bottom 2.5% and the top 2.5%.
Solution: To find the z-score located to the left, we search the body of Table A-2 for an area of 0.025.
The result is z = -1.96. To find the z-score located to the right, we search the body of table A-2 for an
area of 0.975. The result is z = 1.96.
Note: In this type of situation, one z score will always be negative and the other positive
NORMINV (0.025, 0, 1) = -1.96
NORMINV (0.975, 0, 1) = 1.96
Critical Values:
For a normal distribution, a critical value is a z-score on the borderline separating z-scores that are likely
to occur from those that are unlikely Common critical values are z = -1.96 and z = 1.96.
Notation:
The following notation is used for critical z values found by using the standard normal distribution.
Z α denotes the z-score with an area of α to its right.
Example: Find Z 0.005
Solution: Table A-2 will give cumulative area from the left; however we are looking for area to the
right. Therefore, locate the area 1 - 0.005 = 0.995 in table A-2 to find a z-score of 2.575.
Example: Find Z 0.02
Solution: Table A-2 will give cumulative area from the left; however we are looking for area to the
right. Therefore, locate the area 1 - 0.02 = 0.98 in table A-2 to find a z-score of 2.05.