Topic:Probabilities and the Standard Normal Curve

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Contents 1 Topic Highlights 2 Introduction and Motivation 3 Learning Activities 4 Learning Objectives 5 Lecture Notes: The Standard Normal Curve 5.1 The Standard Normal Curve 5.1.1 Example 1 5.1.2 Example 2 5.1.3 Note 5.2 Probabilities from the Standard Normal Curve 5.2.1 Example 3 5.2.2 Example 4 5.3 Finding any Area Under the Curve 5.3.1 Example 5 5.3.2 Example 6 5.3.3 Example 7 5.4 A Few More ... 5.4.1 Example 8 5.4.2 Example 9 5.4.3 Example 10 5.4.4 Example 11 6 Lecture Notes: General Approach to Solving Problems 7 Practice Problems 7.1 Practice Problem 1 7.2 Practice Problem 2 (Updated) 7.3 Practice Problem 3 7.4 Practice Problem 4 (Updated)

Topic Highlights

(What you will learn)

• The basics of the standard normal curve - what it is and what it means
  
• How to solve probability problems using the standard normal curve, including:
  
  • The many different ways of looking at areas under the curve
    
  • How to use the appropriate statistical table
    
• How to make any normal problem look like the standard normal problem

Introduction and Motivation

(Why learn it)

In the last topic on Continuous and Normal Random Variables, we looked at the concepts of continuous random variables and continuous probabilities distributions. We also introduced the concept of the normal curve, a term used to refer to the probability distribution for normal random variables.

Then, we saw how probability for a continuous probability curve is given by the area under the curve.

We take things a step further in this topic by looking deeper at the concept of a normal curve and at how you can obtain numerical probabilities using a table of "areas of the standard normal distribution".

Learning Activities

(How the levels of understanding will be gained)

Learning activities for this topic

Type	Name	Direction
Reading	Section 5.3 of Bowerman et al. Sections 6.4 and 6.5 of Kvanli et al.	Self-directed
In-class discussion	The Standard Normal Curve (below in this page) General Approach to Solving Problems (also below)	Instructor-directed
In-class worksheet	Worksheet - Probabilities and the Standard Normal Curve
Practice problems	Practice Problems (below in this page)	Self-directed
Personal activities	Reflection and review of the learning objectives (below)	Self-directed

Learning Objectives

(Levels of understanding to be gained)

Learning objectives for this topic

Level of Understanding	Objective(s)
Very best	Can you solve problems like Practice Problem 3 and Practice Problem 4?
Highly satisfactory	Are you able to find the area under the standard normal curve for any situation you might be given, i.e. using the principles employed in Example 5, Example 6, Example 7, Example 8, Example 9, Example 10 and Example 11? Can you solve problems like Practice Problem 1 and Practice Problem 2?
Satisfactory	Can you use the standard normal table (Table A.4 in Kvanli et al.) to find the area under the curve between 0 and z, as in Example 3 and Example 4?
Maybe just enough to pass	Do you understand the basic concept of the standard normal curve? e.g. Can you repeat Example 1 and Example 2 on your own?

Lecture Notes: The Standard Normal Curve

These notes are intended to facilitate an introductory discussion to the concept of the standard normal curve.

The Standard Normal Curve

Let's start right where we left off in the topic Continuous and Normal Random Variables - looking at the meaning of z-score.

Consider the following plot of a normal curve against the random variable X:

Example 1

Can you draw the corresponding standard normal curve below this one? This is the same curve plotted against Z instead of X.

Solution
Notice how the y-axis of this plot goes through the mean of the earlier plot? Also notice that the values 1 and 2 correspond to values of  and 2 of the earlier plot

This is known as the standard normal curve. It takes the original normal curve, that has mean  and standard deviation , and standardizes it so it has a mean of 0 and standard deviation of 1. As you know, from the topic Measures of Position and Shape, any value of z for the standard curve is obtained from the value of x in the original curve using the following equation:

Example 2

a) If you have a normal curve described by  = 3 and  = 2, then what is the z-score corresponding to x = 4?

b) Can you draw the original normal curve and the standard normal curve, showing x and z?

Solution to a)
Using the above equation: z = (4-3)/2 = 0.50

Solution to b)

Note

One can also talk about the standard normal curve concepts in terms of either the population itself, or a sample of the population (as we have here). If talking about a sample, the above equation would look as follows:

where  is the sample mean (analogous but not equal to the population mean, ) and s is the population standard deviation (analogous to the population standard deviation, ). We'll look at this in more detail later. In the meantime, jump ahead to here if you want explore this in more detail. Go back to here if you need a reminder of the difference between populations and samples. 

Probabilities from the Standard Normal Curve

You can get a standard normal curve from any normal curve you are given. In other words, given  and s for a problem involving a normal random variable X, you can get the corresponding standard normal curve corresponding to a random variable Z.

Once you have the standard normal curve, it's fairly straightforward to answer probability questions.

In fact, there is a fairly simple table for it which gives the following area under the standard normal curve:

In Kvanli et al., this is Table A.4.In Bowerman et al. it is Table A.3.

The shaded area corresponds to: P( 0 < Z < z ). In other words, the shaded area corresponds to the probability that the standardized random variable lies between 0 and z.

From our earlier discussion, you should see that this is the same as the probability P( <  X < (x-)/ ).

This is often made more clear using examples.

Example 3

a) Use the standard normal table to find the area under the curve for 0 < z < 1.41

b) Sketch the situation

Solution to a)
From the table, P( 0 < z < 1.41 ) = 0.4207

Solution to b)

Example 4

What is the area under the curve for -1.41 < z < 0?

Solution
Because the standard normal curve is symetrical, you should see that this area is the same as the area as in the last example P( -1.41 < z < 0 ) = P( 0 < z < 1.41 ) = 0.4207 Or, in a sketch:

Finding any Area Under the Curve

To find any area under the curve, it helps to recognize the probabilities corresponding to the following areas. What are they?

Example 5

Given the three cases above, what is the probability P( z < 1.41 )? Sketch the area it corresponds to, and find the numerical probability.

Solution
It's the area below: which is given by: P( z < 1.41 ) = 0.5 + P( 0 < z < 1.41 ) and we know P ( 0 < z < 1.41 ) from the table (see Example 3), so: P( z < 1.41 ) =  0.5 + 0.4207 = 0.9207

Example 6

What is the probability P( -1.41 < z )?

Solution
It's the area below: which is given by: P( -1.41 < z ) = P( -1.41 < 0 < z ) + 0.5 and we know P ( -1.41 < z < 0 ) = P( 0 < z < 1.41 ) which we know from the table (see Example 4), so: P( -1.41 < z ) =  0.4207 + 0.5 = 0.9207 This is the same answer as in Example 5, which should make sense to you just based on the symmetry of the situation.

Example 7

What about the probability P( 1.41 < z < 2.00 )?

Solution

It's the area below: which is given by: P( 1.41 < z < 2.00 ) = P( z < 2.00 ) - P( z < 1.41 ) We know P( 0 < z < 1.41 ) = 0.4207 from Example 4, and we can get the other probability from the table: P( 0 < z < 2.00 ) =  0.4772 So: P(1.41 < z < 2.00) = 0.4772 - 0.4207 = 0.0565 And, just so you are completely clear on how we got this, make sure you can see that it's the difference between the following two areas (both of which we obtained straight from the table):

A Few More ...

As you've probably noticed, there are many ways of slicing and dicing these normal curves. We've already seen about half of them. Let's have a quick look at the rest. Once you've seen these, you've seen almost anything that'll be thrown at you.

Example 8

What is the probability P( z > 1.41 )? Sketch the area it corresponds to, and find the numerical probability.

Solution
It's the following area: And it's given by the following: P( z > 1.41 ) = 0.5 - P( 0 < z < 1.41 ) = 0.5 - 0.4207 = 0.0793

Example 9

What is the probability P( -2.00 < z and z < 1.41 )? Sketch the area it corresponds to, and find the numerical probability.

Solution
This is the same as saying P( -2.00 < z < 1.41 ), which is the following area: And it's given by the following: P( -2.00 < z < 1.41 ) = P( -2.00 < z < 0 ) + P( 0 < z < 1.41 ) = P( 0 < z < 2.00 ) + P( 0 < z < 1.41 ) which we know from the tables (and earlier examples) is: P( -2.00 < z < 1.41 ) = 0.4772 + 0.4207 = 0.8979

Example 10

What is the probability P( z < 0.5 or z > 1.41 )? Sketch the area it corresponds to, and find the numerical probability.

Solution
It's the following area: The left area is given by the following: P( z < 0.5 ) = 0.5 + P( 0 < z < 0.5 ) = 0.5 + 0.1915 = 0.6915 and the right area is given by the following (recall Example 8): P( z > 1.41 ) = 0.5 - P( 0 < z < 1.41) = 0.5 - 0.4207 = 0.0793 and the total area is the sum of the areas on the right and the left: P( z < 0.5 or z > 1.41 ) = 0.6915 + 0.0793 = 0.7708

Example 11

What is the probability P( z < -1.41 or z > 1.41 )? Sketch the area it corresponds to, and find the numerical probability.

Solution
It's the following area: The area on the right is given by the following (recall Example 8): P( z > 1.41 ) = 0.5 - P( 0 < z < 1.41) = 0.5 - 0.4207 = 0.0793 and the area on the left is the same as the area on the right: P( z < -1.41 ) = 0.0793 and the total area is the sum of the areas on the right and the left: P( z < -1.41 or z > 1.41 ) = 0.0793 + 0.0793 = 0.1586

Lecture Notes: General Approach to Solving Problems

The following steps are recommended for solving probability problems of normal curves:

1. Write out the given information: x,  and  for the normal random variable that is described in the problem

2. Get the value for the standard normal curve by solving for z using the equation:

3. Sketch the situation like we did in the examples above

4. Write out the required probability in terms of the areas in your sketch, using the z-score variable

5. Use Table A.4 to look up the required areas

6. Solve for the required probability from step 4.

Now, try your hand at some problems...

Practice Problems

Practice Problem 1

The amount of time spent by each statistics student preparing for the next class can be represented by a normal curve with an average time of 3 hours and a standard deviation of 2 hours. What is the probability that a student will spend more than 4 hours?

Solution

First recognize that  = 3 and  = 2, and that we are being asked for P( x > 4 ) Then compute the z-score: = (4-3)/2 = 0.50 So, we can say that P( x > 4 ) is the same as P( z > 0.50 ) This very situation was sketched in Example 2, and the area you are looking for is shown below: As was the case in Example 8, we know that: P( z > 0.50 ) = 0.5 - P( 0 < z < 0.5 ) And from Table A.4, we get: P( 0 < z < 0.50) = 0.1915 So, P( z > 0.50 ) = P( x > 4 ) = 0.5 - 0.1915 = 0.3085

Practice Problem 2 (Updated)

For the situation described above, what is the probability that a student will spend less than 1 hour preparing?

Solution

Again,  = 3 and  = 2. This time we are being asked for the probability that a student will spend less than 1 hour preparing. It's tempting to look at this as P( x < 1 ), but important to realize a student can't study for less than 0 hours! So, we're in fact looking for: P( 0 < x < 1 ) Then we compute the corresponding z-scores: = (1-3)/2 = -1.00 = (0-3)/2 = -1.50 So, we can say that P( 0 < x < 1 ) = P( -1.50 < z < -1.00 ) You should be able to sketch this for yourself to show the following (not to scale): In a way similar to what we did in Example 7, you should recognize that P( -1.50 < z < -1.00 ) is the difference between two probabilities: P( -1.50 < z < -1.00 ) = P( -1.50 > z > 0 ) - P( -1.00 > z > 0 ) Then we know that: P( -1.00 > z > 0 ) = P( 0 < z < 1.00 ) = 0.3413 And that: P( -1.50 > z > 0 ) = P( 0 < z < 1.50 ) = 0.4332 So, the probability we are looking for is: P( 0 < x < 1 ) = P( -1.50 < z < -1.00 ) = 0.4332 - 0.3414 = 0.0919

Practice Problem 3

For the same situation, what is the probability that a student will spend between 1 and 4 hours?

Solution

Again,  = 3 and  = 2. This time we are being asked for P( 1 < x < 4 ) We know the z-scores from parts a) and b) above, so we can say that: P( 1 < x < 4 ) is the same as P( -1.00 < z < 0.50 ) You should be able to sketch this for yourself. In a way similar to what we did in Example 9, and using Table A.4, we get: P( -1.00 < z < 0.50 ) = P( 0 < z < 1.00 ) + P( 0 < z < 0.50 ) = 0.3413 + 0.1915 = 0.5328 So, the requested probability is 0.5328

Practice Problem 4 (Updated)

For the same situation, what is the probability that a student will spend less than 1 or more than 4 hours?

Solution

Again,  = 3 and  = 2. This time we are being asked for P( 0 < x < 1 ) + P( x > 4 ) We know the z-scores from the problems above, so we can say that: P( 0 < x < 1 ) is the same as P( -1.50 < z < -1.00 ) which we found in Practice Problem 2 to be 0.0919 P( x > 4 ) is the same as P( z > 0.50 ) which we found in Practice Problem 1 to be 0.3085 You should be able to sketch this for yourself. And solve for the desired probability: P( 0 < x < 1 ) + P( x > 4 ) = P( -1.50 < z < -1.00 ) + P( z > 0.50 ) = 0.3085 + 0.0919 = 0.4004