Skip to document
This is a Premium Document. Some documents on Studocu are Premium. Upgrade to Premium to unlock it.

MATH1342 Hawkes Lesson 8.3 The Standard Normal Distribution

MATH1342 Hawkes Lesson 8.3 The Standard Normal Distribution Professor:...
Course

Probability and Statistics (MATH1342)

34 Documents
Students shared 34 documents in this course
Academic year: 2021/2022
Uploaded by:
Anonymous Student
This document has been uploaded by a student, just like you, who decided to remain anonymous.
Stephen F. Austin State University

Comments

Please sign in or register to post comments.

Preview text

Hawkes Learning Discovering Statistics and Data 3rd Edition Lesson 8 – The Standard Normal Distribution

Standard Normal Distribution – is a normal distribution with a mean of zero and a standard deviation of 1.

Example 8.3. – Compute the probability that a standard normal random variable is less than 1.

Solution - Drawing a picture, even when the problem is rather simple, is a good idea. Remember that the probability is represented by the area under the standard normal curve. Determining the area under the standard normal curve to the left of a particular value requires little effort since the tables are constructed to give the cumulative probabilities. That is, the table gives probabilities that the random variable z is less than (or less than or equal to) some value (i., P(z<z0) where z0 is the number of standard deviations above or below the mean). In this case, the construction of the Standard Normal Table: Area −∞ to z exactly matches the kind of interval we are examining. Thus, merely looking up the value corresponding to 1 in the table is sufficient to obtain the probability.

Example 8.3 – Compute the probability that a standard normal random variable is between 0 and 1.

Solution – First, draw a picture.

To determine the area under the standard normal curve requires little effort. Table C is constructed to give the probability between 0 and some value of z. In this case, the table’s construction exactly matches the kind of interval you are examining. Thus, merely looking up the value corresponding to 1 in the table is sufficient to obtain the probability.

P(0 < z < 1) = 0.

Because the standard normal distribution is symmetric and the total area under the curve is equal to 1, we know that the probability that z is less than 0 is equal to 0 (since 0 is the mean or the central value). We can find P(z<−1) to be 0 from looking up −1 in the Standard Normal Table: Area −∞ to −z. Thus, to find the area between −1 and 0 we subtract the area to the left of −1 from 0 as follows.

Example 8.3 – What is the probability that a standard normal random variable will be between 1 and 2?

Solution – Again, first draw a picture.

Because the Standard Normal Table: Area −∞ to z gives us cumulative probabilities, we can find the probability that z is less than 2 and then subtract the probability that z is less than 1, yielding the probability that z is between 1 and 2. Thus, we have the following. The figure below illustrates the areas involved in this calculation.

Example 8.3 - Given that z is a standard normal random variable, find the value of z for each situation.

a. The area to the left of z is 0. b. The area between 0 and z is 0. c. The area to the left of z is 0. d. The area to the right of z is 0.

Solution – first draw a picture. Note that this problem is slightly different from the previous one. In a previous example, you were asked to find a probability, given that you know the value of z. In this example, you are given a probability and asked to find the corresponding value of z. Recall that Table A and Table B in Appendix A give you the cumulative probability of the area less than some value of z. Since the value of z shown in the figure above is positive (greater than 0), look in the body of Table B and find the probability value 0. Once you’ve found the value (the probability), determine the corresponding value of z. In this case, the value of z is 1.

P( z < 1) = 0.

And the value of z is 1 with the area to the left of it being 0.

Note that from the picture, we have the area to the right of z. However, we know that the total area under the curve is 1. Thus, if the area to the right of z is 0, then the area to the left of z is 1−0.7967=0. From the picture, it is clear that if we find 0 in the body of Appendix A, Table A, the corresponding value of z is the value we are interested in. This value of z is −0. Therefore, the value of z with the area 0 to the right is −0.

Was this document helpful?
This is a Premium Document. Some documents on Studocu are Premium. Upgrade to Premium to unlock it.

MATH1342 Hawkes Lesson 8.3 The Standard Normal Distribution

Course: Probability and Statistics (MATH1342)

34 Documents
Students shared 34 documents in this course
Was this document helpful?

This is a preview

Do you want full access? Go Premium and unlock all 7 pages
  • Access to all documents

  • Get Unlimited Downloads

  • Improve your grades

Upload

Share your documents to unlock

Already Premium?
Hawkes Learning Discovering Statistics and Data 3rd Edition
Lesson 8.3 – The Standard Normal Distribution
Standard Normal Distribution – is a normal distribution with a mean of zero and a standard
deviation of 1.
Example 8.3.1. – Compute the probability that a standard normal random variable is less than
1.27.
Solution - Drawing a picture, even when the problem is rather simple, is a good idea. Remember
that the probability is represented by the area under the standard normal curve. Determining the
area under the standard normal curve to the left of a particular value requires little effort since the
tables are constructed to give the cumulative probabilities. That is, the table gives probabilities
that the random variable z is less than (or less than or equal to) some value (i.e., P(z<z0) where
z0 is the number of standard deviations above or below the mean). In this case, the construction
of the Standard Normal Table: Area −∞ to z exactly matches the kind of interval we are
examining. Thus, merely looking up the value corresponding to 1.27 in the table is sufficient to
obtain the probability.

Why is this page out of focus?

This is a Premium document. Become Premium to read the whole document.

Why is this page out of focus?

This is a Premium document. Become Premium to read the whole document.