Download Chapter 7: Normal Curve and Symmetrical Distribution

Chapter 7: Normal
Curve and Symmetrical Distribution
Normal Distribution is a mathematically theoretical concept or model that depicts a tendency that
the scores always tend to be distributed around the average. It is an arrangement of a data set in which most
values cluster in the middle of the range and the rest taper off symmetrically toward either extreme. It is
sometimes called the Gaussian Distribution, in honor of Carl Friedrich Gauss, who studied the curve. The
normal distribution is sometimes informally called the bell curve.
Normal Distribution in the Real-World
Height is one simple example of something that follows a normal distribution pattern: Most people
are of average height, the numbers of people that are taller and shorter than average are fairly equal and a
very small (and still roughly equivalent) number of people are either extremely tall or extremely short.
Properties of Normal Distribution
1.
2.
3.
4.
5.
6.
7.
It is bell-shaped.
The tails are asymptotic to the baseline.
The mean, median and mode, in a normal curve, coincide. (They have the same numerical values)
It is symmetrical about the mean.
The total area of the curve is 100% or 1.00.
The baseline of the curve is divided into 6 sigma distances from -3 sigma to +3 sigma.
Continuous for all values of X between -∞ and ∞ so that each conceivable interval of real numbers
has a probability other than zero.
8. -∞≤X≤∞
9. Two parameters, μ and σ. Note that the normal distribution is actually a family of distributions, since
μ and σ determine the shape of the distribution.
10. The rule for a normal density function is
11.
2
2
12. The notation N (μ, σ ) means normally distributed with mean μ and variance σ . If we say X ∼ N
2
2
(μ, σ ) we mean that X is distributed N (μ, σ ).
13. About 2/3 of all cases fall within one standard deviation of the mean, that is P (μ - σ ≤ X ≤ μ + σ) =
.6826.
14. About 95% of cases lie within 2 standard deviations of the mean, that is P (μ - 2σ ≤ X ≤ μ + 2σ) =
.9544
Why is the normal distribution useful?
1.
Many things actually are normally distributed, or very close to it. For example, height and intelligence
are approximately normally distributed; measurement errors also often have a normal distribution.
2. The normal distribution is easy to work with mathematically. In many practical cases, the methods
developed using normal theory work quite well even when the distribution is not normal.
3. There is a very strong connection between the size of a sample n and the extent to which a sampling
distribution approaches the normal form. Many sampling distributions based on large n can be
approximated by the normal distribution even though the population distribution itself is definitely
not normal.
Areas of Standard Normal Distribution
Let us look at an illustration to see how we read the table for the areas of standard normal
distribution. The z scores are in the margins: the left margin has unit digits and the tenths digits; the top has
the hundredths digits.
Example 1: Look up a z score of 1.28.
Solution:
We find 0.3997. It is the measure of area under the standard normal curve between z = 0 (which
locates the mean) and z = 1.28 (a number 1.28 standard deviation larger than the mean). This area is also the
measure of the probability associated with the same interval, that is, P (0 < z < 1.28) = 0.3997 read as: "the
probability that a value picked at random falls between the mean and 1.28 standard deviation above the
mean is 0.3997" or "the probability that z score picked at random will fall between 0 and 1.28 is 0.3997."
Example 2: From previous example, find the area under the normal curve to the right of z = 1.28
Solution: The area to the right of the mean is 0.5000. But here, 0.3997 is not included. Therefore:
Area = 0.5000 - 0.3997
Area = 0.1003 or 10.03% or P (z > 1.28) = 0.1003
Example 3: From previous examples, find the area to the left of z = 1.28
Solution:
The area to the left of the mean is 0.5000. Therefore:
Area = 0.5000 + 0.3997
Area = 0.8997 or 89.9% or P (z < 1.28) = 0.8997
Example 4: Find the area between the mean (z = 0) and z = -2.4
Solution:
Area= 0.4927 or 49.27% or P (-2.4 < z > 0) = 0.4297
Example 5: Find the area to the left of z = -1.30
Solution:
The area to the left of the mean is 0.5000. But here, 0.4032, which is the area from the mean to 1.30 is
not included. Therefore:
Area = 0.5000 - 0.4032
Area = 0.0968 or 9.8% or P (z < -1.3) = 0.0968
Example 6: Find the area between z = -1.93 and z = 1.28
Solution:
P (-1.93 < z > 0) + P (0 < z > 1.28)
=0.4732 + 0.3997
= 0.8729 or 87.29%
Example 7: Find the area between z = 0.9 and z = 2.3
Solution:
The area of all the shadings in the figure is 0.4893. But here, 0.3159, which is the are from 0 to 0.9 is
not included. Therefore:
Area = P (0 < z < 2.3) - P (0 < z < 0.9)
Area = 0.4893 - 0.3159
Area = 0.1734 or 17.34% or P (0.9 < z < 2.3) = 0.1734
Note: The normal distribution table can also be used to determine a z score if the area is given.
Example 8: Find the z score if the area to the right of the mean is 0.3790.
Solution:
From the table, z = 1.17
Example 9: Find the z score if the area to the left of z is 0.8665
Solution:
The area to the left of the mean is 0.5000. Then, 0.8665 - 0.5000 = 0.3665. 0.3665 is the area from
the mean to z, hence, z = 1.11
Example 10: Find z if the area to the right of z is 0.0735
Solution:
The area to the right of the mean is 0.5000, 0.5000 - 0.735 = 0.4265. Then, 0.4265 is the area from
the mean to z, therefore, z =1.45
Application of the Normal Distribution
Example 11: Given a normal distribution of 700 scores with a mean of 100 and a standard deviation
of 10, what is the probability that a score, randomly picked, will lie:
a) between the scores 100 and 115?
b) above a score of 124?
c) below a score of 110?
d) above a score of 85?
Example 12: Past experience has shown that the standard deviation of the weights of the bubble
gum balls produced by a certain machine is 0.5 grams. If 90% of the population weighs more than 8
grams each, what is the mean of the population, assuming that the weights are approximately
normally distributed?
Example 13: The test grades of a group of students are approximately normally distributed with a
mean of 60 and a standard deviation of 10. If we wish to assign passing grades of 90% of the group,
what point should divide the failing grades from the passing grades?
Let's Practice 1:
Direction: Answer the following correctly.
1. Find the area under the normal curve that lies between the following pairs of z values:
a) z = 0 to z = 1.7
b) z = 0 to z = -3.9
c) z = 0 to z = -2.68
d) z = 0 to z = 2.45
2. Find the area under the normal curve that lies between the following pairs of z values.
a) z = -1.90 to z = 1.92
b) z = -3.3 to z = -0.65
c) z = -2.10 to z = 3.28
d) z = -2.3 to z = -1.28
e) z = 2.45 to z = 3.24
f) z = 1.64 to z = 2.83
Let's Practice 2:
Direction: Answer the following correctly.
1. The salaries of teachers in a certain school in Pasig have a mean of P8,500.00 and a standard
deviation of P1,500.00. What is the probability that a teacher picked at random has a salary of:
a) more than P9,000.00
b) more than P7,000.00 but less than P10,000.00
c) more than P11,000.00
d) more than P12,000.00
2. The mean weight of the tokens in an arcade has been established as 12 grams. If the weights are
approximately normally distributed and if we know that 80% of the tokens weight between 10 and
16 grams, what is the standard deviation of the weights?
Assignment:
Direction: In one whole yellow paper, answer the following correctly.
1. Find the probability that a piece of data picked at random from a normally distributed population
will have a standard score (z) that is as follows:
a) less than 2.50
b) less than -3.0
c) greater than 3.47
d) greater than -1.45
e) greater than 2.35
f) less than -3.54
2. The waiting time x at a certain restaurant is approximately normally distributed with a mean of
12.3 minutes and a standard deviation of 4.8 minutes.
a) Find the probability that a randomly selected customer has to wait less than 8 minutes.
b) Find the probability that a randomly selected customer has to wait more than 15 minutes.
c) Find the value of the 75th percentile for x.