Topic:More about the Standard Normal Curve
From SharedExperienceProject
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Topic Highlights
(What you will learn)
- More advanced probability concepts for the standard normal curve
- How to solve the corresponding problems
Introduction and Motivation
(Why learn it)
In the topic Probabilities and the Standard Normal Curve, we had a first look at the types of problems you might run into when dealing with normal curves. For most of these, we were given the defining parameters of a normal distribution (mean and standard deviation), and asked to come up with some probability. In this topic, we consider problems with the opposite structure: those in which we are given the probability and asked to come up with either: 1) the corresponding value of the random variable, x, or 2) the corresponding mean, 
.
As you will see from the practice problems, there are some interesting real-world cases where this type of problem comes up.
Learning Activities
(How the levels of understanding will be gained)
| Type | Name | Direction |
| Reading |
| Self-directed |
| Lecture and discussion |
| Instructor-directed |
| Practice problems |
| Self-directed |
| Personal activities |
| Self-directed |
Learning Objectives
(Levels of understanding to be gained)
| Level of Understanding | Objective(s) |
| Very best |
|
| Highly satisfactory |
|
| Satisfactory |
|
| Maybe just enough to pass |
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Topic Notes: Solving for x
These notes are intended to facilitate a discussion of Section 6.5 of Kvanli et al.
What's the deal?
It's pretty simple actually. In the basic problems from last time:
- We were given
and
- We solved for z using the equation:
- And then we found the probability by looking up an area in the table
In the cases where you solve for x:
- We are given the probability and the normal distribution (
and
- We get z from the table
- We can then solve for x rearranging the above equation:
In other words, it's the reverse of what we saw before.
General Approach to Solving for x
There's a pretty straightforward approach to doing this:
1. Read the question and sketch the situation
2. Using the sketch, figure out the area corresponding to 0 < Z < z
3. Using the table, look up the z for that area
4. Solve for x, using the equation
Example 1
The overnight low (temperature) on days in November can be represented by a normal curve with an average temperature of 3 degrees and a standard deviation of 2 degrees. 16% of days will have at least what temperature for their overnight low?
| Solution |
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Using the steps above: 1. Sketch the situation:
2. Using the figure below, we figure out that the area corresponding to 0 < Z < z is 0.34:
3. Using the table to look up that area: z ~= 1.0 4. Compute x, using the values of
So, to answer the question: 16% of days will have at least 5 degrees for overnight low. |
Example 2
For the same example, 65% of overnight lows in November will be at most what temperature?
(You could also ask this as: 65% of overnight lows will be less than what temperature?)
| Solution |
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1. and 2. The following sketch applies, and shows us that the area P( 0 < Z < z ) = 0.15:
3. Using the table, we see that z ~= 0.385 4. So we solve for x
So, to answer the question: 65% of overnight lows will be at most 3.77 degrees or: 65% of overnight lows will be less than 3.77 degrees |
Topic Notes: Solving for the mean
These notes are intended to facilitate a discussion of Section 6.5 of Kvanli et al.
What's the deal?
It's almost the same as solving for x (as we did above). However, this time we're solving for .
A re-arranged form of the z-score equation applies again:
General Approach to Solving for the Mean
The steps are identical to those followed when solving for x, except for step 4:
1. Read the question and sketch the situation
2. Using the sketch, figure out the area corresponding to 0 < Z < z
3. Using the table, look up the z for that area
4. Solve for x, using the equation
You can try this out in Practice Problem 3 below.
Practice Problems
Practice Problem 1
A popular cupcake store wants to meet demand for its cupcakes 95% of the time. Measurements made by the owner have shown that demand follows a normal distribution with a mean of 305 cupcakes per day and a standard deviation of 20 cupcakes per day.
How many cupcakes should it make per day to meet its goal?
| Solution |
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This is a lot like Example 2: 1. and 2. The following sketch applies, and shows us that the area P( 0 < Z < z ) = 0.45:
3. Using the table, we see that z = 1.645 4. So we solve for x:
So, by rounding up we can conclude that the store should make 338 cupcakes in a day in order to meet demand 95% of the time. |
Practice Problem 2
A local casino has noticed that it earns the most profit from customers that show up at the craps tables at 9 PM (on average) on Saturday nights. Profit at those tables appears to be normally distributed around this mean time, with a standard devation of 2 hours.
If the casino wants to limit the time it has to keep those tables open, but still wants to capture 68.2% of the profit, then at what times should it open and close the craps tables?
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This is a variant on what you've seen, where you need to solve for two values of x. Using the steps outlined above: 1. and 2. The following sketch applies, and shows us that the area P( 0 < Z < z ) = 0.341:
3. Using the table, we see that: zmax ~= 1.00 and zmin ~= -1.00 4. So we solve for both required values of x: xmax = zmax• So, to achieve 68.2% of the profits, the casino should keep the tables open between 7 and 9 PM. Although not required to answer the question, the corresponding normal curve is shown below:
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Practice Problem 3
The amount of time spent by each statistics student preparing for the next class can be represented by a normal curve with a standard deviation of 2 hours.
If 15% of the students spend 4 hours or more studying, what is the average amount of time students spend studying?
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This is a case of having to solve for the mean (which is the same as average). Following the steps given above: 1. The following sketch applies:
2. The following shows us that the area P( 0 < Z < z ) = 0.35:
3. From the table we see that z ~= 1.035 4. We can then solve for the mean:
So, if 15% of the students spend 4 hours or more studying, then the average amount of time students spend studying is 1.93 hours. Although not required to answer the question, the corresponding normal curve is shown below:
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Additional Normal Curve Problems
Also try your hand at the following problems:
Full solutions can be found on Blackboard.
As you work through the problems, you can use the following to check you're using Table A.4 correctly...
For Question 1a
| Solution |
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For z = 0.75 (in green), we get an area of 0.2734 (in yellow), as shown below:
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For Question 1c
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The area of 0.4500 falls between 0.4495 and 0.4505 (both in green). These correspond to z = 1.64 and 1.65, as shown (in yellow) below. Since 0.4500 falls exactly in the middle of the values in the table, we can use z = 1.645.
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For Question 2a
| Solution |
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The area 0.3500 falls between 0.3485 and 0.3508 (both in green below). These correspond to z = 1.03 and 1.04, as shown below (in yellow). Since 0.3500 is closer to the middle than to either value, we can approximate z ~= 1.035.
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