Transcript 12.5 – The Normal Distribution Discrete and Continuous Random Variables Discrete random variable: A random variable that can take on only certain fixed.

12.5 – The Normal Distribution
Discrete and Continuous Random Variables
Discrete random variable: A random variable that can take
on only certain fixed values.
The number of even values of a single die.
The number of heads in three tosses of a fair coin.
Continuous random variable: A variable whose values are
not restricted.
The diameter of a growing tree.
The height of third graders.
12.5 – The Normal Distribution
Definition and Properties of a Normal Curve
A normal curve is a symmetric, bell-shaped curve.
Any random continuous variable whose graph has this
characteristic shape is said to have a normal distribution.
On a normal curve the horizontal axis is labeled with the
mean and the specific data values of the standard deviations.
If the horizontal axis is labeled using the number of standard
deviations from the mean, rather than the specific data values,
then the curve the standard normal curve
12.5 – The Normal Distribution
Sample Statistics
Normal Curve
– 2.8
– 1.4
5.5
1.4
Standard Normal Curve
2.8
–2
–1
0
or
5.5
1
2
12.5 – The Normal Distribution
Normal Curves
B
A
S
C
0
S is standard, with mean = 0, standard deviation = 1
A has mean < 0, standard deviation = 1
B has mean = 0, standard deviation < 1
C has mean > 0, standard deviation > 1
12.5 – The Normal Distribution
Properties of Normal Curves
The graph of a normal curve is bell-shaped and symmetric
about a vertical line through its center.
The mean, median, and mode of a normal curve are all equal
and occur at the center of the distribution.
Empirical Rule: the approximate percentage of all data lying
within 1, 2, and 3 standard deviations of the mean.
within 1 standard deviation
68%
within 2 standard deviations
95%
within 3 standard deviations.
99.7%
12.5 – The Normal Distribution
Empirical Rule
68%
95%
99.7%
12.5 – The Normal Distribution
Example: Applying the Empirical Rule
A sociology class of 280 students takes an exam. The
distribution of their scores can be treated as normal. Find the
number of scores falling within 2 standard deviations of the
mean.
A total of 95% of all scores lie within 2 standard deviations of
the mean.
(.95)(280) =
266 scores
12.5 – The Normal Distribution
Normal Curve Areas
In a normal curve and a standard normal curve, the total area
under the curve is equal to 1.
The area under the curve is presented as one of the following:
Percentage (of total items that lie in an interval),
Probability (of a randomly chosen item lying in an
interval),
Area (under the normal curve along an interval).
12.5 – The Normal Distribution
A Table of Standard Normal Curve Areas
To answer questions that involve regions other than 1, 2, or 3
standard deviations, a Table of Standard Normal Curve Areas
is necessary.
The table shows the area under the curve for all values in a
normal distribution that lie between the mean and z standard
deviations from the mean.
The percentage of values within a certain range of z-scores, or
the probability of a value occurring within that range are the
more common uses of the table.
Because of the symmetry of the normal curve, the table can be
used for values above the mean or below the mean.
12.5 – The Normal Distribution
Example: Applying the Normal Curve Table
Use the table to find the percent of all scores that lie between the mean
and 1.5 standard deviations above the mean.
x
z = 1.50
Find 1.50 in the z column.
z = 1.5
The table entry is .4332
Therefore, 43.32% of all values lie between the mean and 1.5 standard
deviations above the mean.
or
There is a .4332 probability that a randomly selected value will lie
between the mean and 1.5 standard deviations above the mean.
12.5 – The Normal Distribution
Example: Applying the Normal Curve Table
Use the table to find the percent of all scores that lie between the mean
and 2.62 standard deviations below the mean.
z = –2.62
z = – 2.62
x
Find 2.62 in the z column.
The table entry is 0.4956
Therefore, 49.56% of all values lie between the mean and 2.62 standard
deviations below the mean.
or
There is a 0.4956 probability that a randomly selected value will lie
between the mean and 2.62 standard deviations below the mean.
12.5 – The Normal Distribution
Example: Applying the Normal Curve Table
Find the percent of all scores that lie between the given z-scores.
z = –1.7
x
z = – 1.7
The table entry is 0.4554
z = 2.55
The table entry is 0.4946
z = 2.55
0.4554 + 0.4946 = 0.95
Therefore, 95% of all values lie between – 1.7 and 2.55 standard
deviations.
12.5 – The Normal Distribution
Example: Applying the Normal Curve Table
Find the probability that a randomly selected value will lie between the
given z-scores.
x
z = 0.61
z = 0.61
The table entry is 0.2291
z = 2.63
The table entry is 0.4957
z = 2.63
0.4957 – 0.2291 = 0.2666
There is a 0.2666 probability that a randomly selected value will lie
between 0.61 and 2.63 standard deviations.
12.5 – The Normal Distribution
Example: Applying the Normal Curve Table
Find the probability that a randomly selected value will lie above the
given z-score.
x
z = 2.14
z = 2.14
The table entry is 0.4838
Half of the area under the curve is 0.5000
0.5000 – 0.4838 = 0.0162
There is a 0.0162 probability that a randomly selected value will lie 2.14
standard deviations.
12.5 – The Normal Distribution
Example: Applying the Normal Curve Table
The volumes of soda in bottles from a small company are distributed
normally with a mean of 12 ounces and a standard deviation .15 ounces.
If 1 bottle is randomly selected, what is the probability that it will have
more than 12.33 ounces?
z = 2.2
x
12.33
The table entry is 0.4861
Half of the area under the curve is 0.5000
0.5000 – 0.4861 = 0.0139
There is a 0.0139 probability that a
randomly selected bottle will contain more
than 12.33 ounces.
12.5 – The Normal Distribution
Example: Finding z-scores for Given Areas
Assuming a normal distribution, find the z-score meeting the condition
that 39% of the area is to the right of z.
11% = 0.11
39% = 0.39
50% of the area lies to the
right of the mean.
The areas from the Normal
Curve Table are based on the
area between the mean and
the z-score.
area between the mean and the z-score = 0.50 – 0.39 = 0.11
From the table, find the area of 0.1100 or the closest value and read the
z-score.
z-score = 0.28
12.5 – The Normal Distribution
Example: Finding z-scores for Given Areas
Assuming a normal distribution, find the z-score meeting the condition
that 76% of the area is to the left of z.
26% = 0.26
50%
0.5000
50% of the area lies to the left
of the mean.
The areas from the Normal
Curve Table are based on the
area between the mean and
the z-score.
area between the mean and the z-score = 0.76 – 0.50 = 0.26
From the table, find the area of 0.2600 or the closest value and read the
z-score.
z-score = 0.71