4: Probability and Probability Distributions - Murdochs Web

6: The Normal Probability Distribution
6.1 The Exercise Reps are designed to provide practice for the student in
evaluating areas under the normal curve. The following notes may be
of some assistance.
1 Table 3, Appendix I tabulates the cumulative area under a
standard normal curve to the left of a specified value of z.
2 Since the total area under the curve is one, the total area
lying to the right of a specified value of z and the total area to its
left must add to 1. Thus, in order to calculate a "tail area", such
as the one shown in Figure 6.1, the value of [pic]will be indexed in
Table 3, and the area that is obtained will be subtracted from 1.
Denote the area obtained by indexing [pic]in Table 3 by[pic] and the
desired area by A. Then, in the above example, [pic].
[pic] 3. To find the area under the standard normal curve between two
values, z1 and z2, calculate the difference in their cumulative
areas, [pic]
4. Note that z, similar to x, is actually a random variable which may
take on an infinite number of values, both positive and negative.
Negative values of z lie to the left of the mean, [pic], and
positive values lie to the right.
Reread the instructions in the My Personal Trainer section if
necessary. The answers are shown in the table.
|The Interval |Write the |Rewrite the |Find the |
| |probability |Probability (if |probability |
| | |needed) | |
| Less than ( 2 |[pic] |not needed |.0228 |
|Greater than 1.16|[pic] |[pic] |[pic] |
|Greater than |[pic] |[pic] |[pic] |
|1.645 | | | |
|Between (2.33 and|[pic] |[pic] |[pic] |
|2.33 | | | |
|Between 1.24 and |[pic] |[pic] |[pic] |
|2.58 | | | |
|Less than or |[pic] |not needed |.9699 |
|equal to 1.88 | | | | 6.2 Similar to Exercise 6.1. Reread the instructions in the My
Personal Trainer section if necessary. The answers are shown in the
table.
|The Interval |Write the |Rewrite the |Find the |
| |probability |Probability (if |probability |
| | |needed) | |
|Greater than 5 |[pic] |[pic] |[pic] |
|Between (3 and 3 |[pic] |[pic] |[pic] |
|Between (0.5 and |[pic] |[pic] |[pic] |
|1.5 | | | |
|Less than or |[pic] |not needed |[pic] |
|equal to (6.7 | | | |
| Less than 2.81 |[pic] |not needed |.9975 |
|Greater than 2.81|[pic] |[pic] |[pic] |
6.3 a It is necessary to find the area to the left of [pic]. That
is, [pic]
b The area to the left of [pic]is [pic]
c [pic]
d [pic] Notice that the values in Table 3 approach 1 as the value
of z increases. When the value of z is larger than [pic](the largest
value in the table), we can assume that the area to its left is
approximately 1. 6.4 To find the area under the standard normal curve between two values,
z1 and z2, calculate the difference in their cumulative areas, [pic]
a [pic]
b [pic] 6.5 a [pic]
b [pic]
c [pic]
d [pic]
e Since the value of [pic]is not recorded in Table 3, you can
assume that the area to the left of [pic]is very close to 0. Then
[pic]
6.6 Similar to Exercise 6.5.
a [pic]
b As in part a, [pic] However, the value for [pic]is not given
in Table 3, but falls halfway between two tabulated values, [pic]and
[pic]. One solution is to choose an area [pic] which lies halfway
between the two tabulated areas, [pic]and [pic] Then
[pic] and [pic]
This method of evaluation is called "linear interpolation" and is
performed as follows:
1. The difference between two entries in the table is called a "tabular
difference". Interpolation is accomplished by taking appropriate
portions of this difference.
2. Let P0 be the probability associated with z0 (i.e. [pic]) and let P1
and P2 be the two tabulated probabilities with corresponding z
values, z1 and z2. Consider [pic]which is the proportion of the
distance from z1 to z0.
3. Multiply [pic]
to obtain a corresponding proportion for the probabilities and add
this value to P1. This value is the desired [pic]. Thus, in this
case,
[pic]
and
[pic]
c [pic]
d [pic] 6.7 Now we are asked to find the z-value corresponding to a particular
area.
a We need to find a z0 such that [pic] This is equivalent to
finding an indexed area of [pic]. Search the interior of Table 3
until you find the four-digit number .9750. The corresponding z-value
is 1.96; that is, [pic]. Therefore, [pic]is the desired z-value (see
the figure below).
[pic]
b We need to find a z0 such that [pic](see below). Using Table 3,
we find a value such that the indexed area is .9251. The
corresponding z-value is [pic].
[pic]
6.8 We want to find a z-value such that [pic](see below).
[pic] Since [pic], the total area in the two tails of the distribution must
be [pic]so that the lower tail area must be [pic]. From Table 3,
[pic]and [pic]. 6.9 a Similar to Exercise 6.7b. The value of z0 must be positive and
[pic]. Hence, [pic].
b It is given that the area to the left of z0 is .0505, shown as
A1 in the figure below. The desired value is not tabulated in Table 3
but falls between two tabulated values, .0505 and .0495. Hence, using
linear interpolation (as you did in Exercise 6.6b) z0 will lie halfway
between -1.64 and -1.65, or [pic]. [pic] 6.10 a Refer to the figure below. It is given that [pic]. That is,
[pic]
[pic] From Exercise 6.9b, [pic].
b Refer to the figure above and consider
[pic]
Then [pic]. Linear interpolation must now be used to determine the
value of [pic], which will lie between [pic]and [pic]. Hence, using a
method similar to that in Exercise 6.6b, we find
[pic]
If Table 3 were correct to more than 4 decimal places, you would find
that the actual value of z0 is [pic]; many texts chose to round up and
use the value [pic]. 6.11 The pth percentile of the standard normal distribution is a value of
z which has area p/100 to its left. Since all four percentiles in
this exercise are greater than the 50th percentile, the value of z
will all lie to the right of [pic], as shown for the 90th percentile
in the figure below. [pic] a From the figure, the area to the left of the 90th percentile is
.9000. From Table 3, the appropriate value of z is closest to
[pic]with area .8997. Hence the 90th percentile is approximately
[pic].
b As in part a, the area to the left of the 95th percentile is
.9500. From Table 3, the appropriate value of z is found using linear
interpolation (see Exercise 6.9b) as [pic]. Hence the 95th percentile
is [pic].
c The area to the left of the 98th percentile is .9800. From
Table 3, the appropriate value of z is closest to [pic]with area
.9798. Hence the 98th percentile is approximately [pic].
d The area to the left of the 99th percentile is .9900. From
Table 3, the appropriate value of z is closest to [pic]with area
.9901. Hence the 99th percentile is approximately [pic]. 6.12 Since [pic]measures the number of standard deviations an observation
lies from its mean, it can be used to standardize any normal random
variable x so that Table 3 can be used.
a Calculate [pic]. Then
[pic]
This probability is the shaded area in the right tail of the normal
distribution on the next page.
[pic] b Calculate [pic]. Then
[pic]
This probability is the shaded area in the left tail of the normal
distribution above.
c Refer to the figure below.
[pic]
Calculate [pic] and [pic]. Then
[pic] 6.13 Similar to Exercise 6.12.
a Calculate [pic] and [pic]. Then
[pic]
b Calculate [pic]. Then
[pic]
c Calculate [pic] and [pic]. Then
[pic] 6.14 It is given that x is normally distributed with [pic]but with unknown
mean[pic], and that [pic]. This normal distribution is shown below.
[pic] Consider the probability [pic]. In terms of the standard normal
random variable z, we can write the z-value corresponding to [pic]as
[pic]
And [pic]. From Table 3, the value [pic] must be negative, with [pic]
or [pic]. Solving for[pic],
[pic]. 6.15 The 99th percentile of the standard normal distribution was found in
Exerci