6: The Normal Probability Distribution

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 z = z0 will be indexed in Table 3, and the area that is obtained will be subtracted from 1. Denote the area obtained by indexing z = z0 in Table 3 by A(z0 ) and the desired area by A. Then, in the above example, A = 1- A(z0 ) .

3 To find the area under the standard normal curve between two values, z1 and z2, calculate the difference

in their cumulative areas, A = A( z2 ) - A( z1 ).

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, z = 0 , 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

Less than - 2 Greater than 1.16 Greater than 1.645 Between -2.33 and 2.33 Between 1.24 and 2.58 Less than or equal to 1.88

Write the probability

P(z < -2) P(z > 1.16) P(z > 1.645) P(-2.33 < z < 2.33) P(1.24 < z < 2.58) P(z 1.88)

Rewrite the Probability (if needed)

not needed

1- P(z 1.16) 1- P(z 1.645) P(z < 2.33) - P(z < -2.33) P(z < 2.58) - P(z < 1.24)

not needed

Find the probability

.0228 1- .8770 = .1230 1- .9500 = .0500 .9901- .0099 = .9802 .9951- .8925 = .1026

.9699

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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 Greater than 5

Write the probability

P(z > 5)

Rewrite the

Find the

Probability (if needed) probability

1- P(z 5)

1-1 = 0

Between -3 and 3

P(-3 < z < 3)

P(z < 3) - P(z < -3) .9987 - .0013 = .9974

Between -0.5 and 1.5

P(-0.5 < z < 1.5) P(z < 1.5) - P(z < -0.5) .9332 - .3085 = .6247

Less than or equal to -6.7 P(z -6.7)

not needed

0

Less than 2.81

P(z < 2.81)

not needed

.9975

Greater than 2.81

P(z > 2.81)

1- P(z 2.81)

1- .9975 = .0025

6.3 a It is necessary to find the area to the left of z = 1.6 . That is, A = A(1.6) = .9452. b The area to the left of z = 1.83 is A = A(1.83) = .9664. c A = A(.90) = .8159 d A = A(4.58) 1. Notice that the values in Table 3 approach 1 as the value of z increases. When the value of z is larger than z = 3.49 (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, A = A( z2 ) - A( z1 ). a A = A(1.4) - A(-1.4) = .9192 - .0808 = .8384 b A = A(3.0) - A(-3.0) = .9987 - .0013 = .9974

6.5 a P (-1.43 < z < .68) = A(.68) - A(-1.43) = .7517 - .0764 = .6753

b P (.58 < z < 1.74) = A(1.74) - A(.58) = .9591- .7190 = .2401

c P (-1.55 < z < -.44) = A(-.44) - A(-1.55) = .3300 - .0606 = .2694

d P ( z > 1.34) = 1- A(1.34) = 1- .9099 = .0901

e Since the value of z = -4.32 is not recorded in Table 3, you can assume that the area to the left of z = -4.32 is very close to 0. Then

P ( z < -4.32) 0

6.6 Similar to Exercise 6.5.

a P ( z < 2.33) = A(2.33) = .9901

b As in part a, P ( z < 1.645) = A(1.645). However, the value for z = 1.645 is not given in Table 3, but

falls halfway between two tabulated values, z = 1.64 and z = 1.65 . One solution is to choose an area

A(1.645) which lies halfway between the two tabulated areas, A(1.64) = .9495 and A(1.65) = .9505.

Then

A(1.645) = .9500 and P ( z < 1.645) = A(.9500).

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. P0 = A( z0 ) ) and let P1 and P2 be the two tabulated

probabilities with corresponding z values, z1 and z2.

Consider

z0 - z1 z2 - z1

which is the proportion of the

distance from z1 to z0.

134

3 Multiply

z0 z2

- -

z1 z1

(

P2

-

P1

)

to obtain a corresponding proportion for the probabilities and add this value to P1. This value is the

desired P0 = A( z0 ) . Thus, in this case,

z0 - z1 = 1.645 -1.64 = .005 = 1 z2 - z1 1.65 -1.64 .010 2

and

P0

=

A( z0 )

=

P1

+

z0 z2

- -

z1 z1

( P2

-

P1 )

=

.9495

+

1 2

[.9505 - .9495]

= .9495 + .0005

=

.9500

c P ( z > 1.96) = 1- A(1.96) = 1- .9750 = .0250

d P (-2.58 < z < 2.58) = A(2.58) - A(-2.58) = .9951- .0049 = .9902

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 P ( z > z0 ) = .025. This is equivalent to finding an indexed area of

1- .025 = .975 . Search the interior of Table 3 until you find the four-digit number .9750. The

corresponding z-value is 1.96; that is, A(1.96) = .9750 . Therefore, z0 = 1.96 is the desired z-value (see the

figure below).

b We need to find a z0 such that P ( z < z0 ) = .9251 (see below). Using Table 3, we find a value such that

the indexed area is .9251. The corresponding z-value is z0 = 1.44 .

135

6.8 We want to find a z-value such that P (-z0 < z < z0 ) = .8262 (see below).

Since A( z0 ) - A(-z0 ) = .8262 , the total area in the two tails of the distribution must be 1- .8262 = .1738 so that the lower tail area must be A(-z0 ) = .1738 2 = .0869 . From Table 3, -z0 = -1.36 and z0 = 1.36 .

6.9

a Similar to Exercise 6.7b. The value of z0 must be positive and A( z0 ) = .9505 . Hence, z0 = 1.65 .

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 z0 = -1.645 .

6.10 a Refer to the figure below. It is given that P (-z0 < z < z0 ) = .9000 . That is,

A( z0 ) + A(-z0 ) = .9000

A ( - z0

)

=

1 2

(1-

.9000)

A(-z0 ) = .0500

136

From Exercise 6.9b, z0 = 1.645 .

b Refer to the figure above and consider

P (-z0 < z < z0 ) = A1 + A2 = .9900

Then

A ( - z0

)

=

1 2

(1 -

.9900)

=

.0050

.

Linear interpolation must now be used to determine the value of

-z0 , which will lie between z1 = -2.57 and z2 = -2.58 . Hence, using a method similar to that in Exercise

6.6b, we find

z0

=

z1

+

P0 P2

- -

P1 P1

( z2

-

z1 )

=

-2.57

+

.4950 .4951

- -

.4949 .4949

( -2.58

+

2.57)

=

-2.57

-

1 2

(.01)

=

-2.575

If Table 3 were correct to more than 4 decimal places, you would find that the actual value of z0 is z0 = 2.576 ; many texts chose to round up and use the value z0 = 2.58 .

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 z = 0 , as shown for the 90th percentile in the figure below.

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 z = 1.28 with area .8997. Hence the 90th percentile is approximately z = 1.28 .

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 z = 1.645 . Hence the 95th percentile is z = 1.645 .

c The area to the left of the 98th percentile is .9800. From Table 3, the appropriate value of z is closest to z = 2.05 with area .9798. Hence the 98th percentile is approximately z = 2.05 .

d The area to the left of the 99th percentile is .9900. From Table 3, the appropriate value of z is closest to z = 2.33 with area .9901. Hence the 99th percentile is approximately z = 2.33 .

6.12 Since z = ( x - ? ) 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 z = x - ? = 13.5 -10 = 1.75 . Then

2

P ( x > 13.5) = P ( z > 1.75) = 1- .9599 = .0401

This probability is the shaded area in the right tail of the normal distribution on the next page.

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