o    Hi Class.  Please see the attached cartoon (mathstore).  This is the sort of smart-aleck answer I would give to the carpenter's question, "How high do you want the door?"  Please read my answer in the cartoon before reading on in this post.

     Now how would I get the data to answer the question?  Do you really think I would stop 100 males at random, ask them their heights (and perhaps get faulty answers) or measure them, rank the heights from low to high, and choose the 95th largest number?

     No, I would use a theoretical distribution, namely a bell-shaped ("Normal") curve.  And if I knew from government data that for example the mean for adult males is 70 inches and the standard deviation is 3 inches, then I could use that information to calculate the 95th percentile height.  If I were located in a city where the residents were significantly taller or shorter than average, then I might  judgmentally adjust the mean before I did the calculation.

     And would I be exactly right?  Probably not, although I don't think I would be off that much.  And have I recognized shifting racial and ethnic trends in my city, changing income and nutrition levels, and lots of other biological and societal factors?  No, except to the extent they have already been included in the data, or in the judgmental adjustment I made to the mean.  However, do you think I would have better off sampling 100 men at random?  And how serious is the consequence if I am off by an inch or two?

     As an actuary, I like "real" data as much as anyone.  But sometimes I think it is OK to use theoretical distributions.  And starting with section 6.1, we will be using theoretical distributions for much of the remainder of the course.

     Any comments?

 

 

 

 

 

 

 

 

 

 

 

 

o    You have probably seen the Normal, or so-called bell distribution.  The Normal probability distribution dates back to at least 1733, possibly earlier. 

     Many variables in nature and business seem to have a distribution shape approximately similar to this distribution.  See page 243, especially the first graph, which is a sample of just 50 values.

     I am not saying we should force data into bell-shaped curves.  I am not going to give 5% of the class A's and 5% F's because a bell-shaped curve says to.  I am simply saying that many variables just naturally appear bell-shaped.  Heights of third graders are probably approximately bell-shaped, for example.

     This cartoon is supposed to be a distribution of the heights of a 16 student class.  For this sample, the mean is 60, and the standard deviation is 1.  If I am willing to believe the distribution is approximately a Normal distribution, then I can use some of the properties of the normal distribution to make inferences and forecasts.  For example, suppose it is August, this is last year's class, and I need to buy costume materials for my upcoming class which has not yet begun and for students whose heights I don't know.  If I bought materials that are usable for students between 58 and 62 inches tall, then 95% of the students should be able to use them.  <insert art4>

     Can you think of a real-world example in your major or occupation that probably fits a Normal distribution reasonably well?  Its distribution would need to be roughly bell-shaped, symmetrical around the mean, with the most of the frequency clustered in the middle.  How about a real-world example that probably does not?

     In my work, an example that will not fit a Normal distribution is dollar amount of insurance claims.  Think about automobile collision claims.  There are many more small claims (fender benders) than there are large claims (total losses), so a distribution of claim amounts is not symmetrical, and rather is positively skewed - Figure 2-5 c on page 56.

 

 

 

 

 

 

 

 

 

 

 

o    The table of normal probabilities shows z values going down the left column to one decimal place, the second decimal place of those z values going across the top row, and then cumulative probabilities in the body of the table.  Cumulative means these are probabilities from negative infinity to the particular z value.

     A common error students make is confusing a z value and its probability. 

     How would you explain to another student what a z value means, and how to distinguish it from a probability?

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

o    Here is another video with two variations on the basic Normal probability calculation.  In this video I discuss finding a Normal probability (or area) to the right such as P(Z > 1.27); and also finding a Normal probability in a range such as between -1.23 and + 1.27, denoted P(-1.23 < Z < 1.27).  (Please disregard the last minute of the video which talks about Statcrunch, a technology we no longer use in this course.)  Let me know if you have any questions.

 

https://www.youtube.com/watch?v=WB5Dz3iJR-A

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

o    An important concept in chapter 6 is the correspondence between areas under continuous probability distribution curves, and probabilities: an area equals a probability.  The total area under any continuous probability distribution equals one, and probabilities range between zero and one.

     Suppose a curve represents the distribution of birth weights with a mean of 3420 g.  Suppose using that curve, the probability (area) of a birth weight being less than 4390 g is .975, and the probability (area) of a birth weight being less than 2450 g is .025.  Draw a quick sketch of this.  What is the probability (area) when you subtract the second from the first, and which interval of birth weights does this represent?

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

o    There are two kinds of normal distribution problems, and they are the converses of each other:

     Problem 1 -   Finding the normal probability when you know the z score:

     Please read pages 247-252 and try to follow the steps.  You should be able to do calculations like these:

1. P(z < 1.27) = .8980 from Table A-2.

 2. P(z > -1.00) = 1 - P(z < -1.00) = 1 - .1587 = .8413, where the .1587 is from the table .Here, if P(z < - 1.00) = .1587, then P(z > - 1.00) must be 1 - .1587; either you are less than - 1 or you are greater than - 1, and the sum of those probabilities or areas has to equal 1.

3. P(-2.50 < z < -1.00) = P(z < -1.00) - P(z < -2.50) = .1587 - .0062 = .1525, where the .1587 and the .0062 come from the table.  (Draw a quick sketch of the graph and convince yourself the subtraction is exactly the area you want.)

     In all problems like these, you should draw a little picture of the graph, and shade in the area you are trying to find. 
     Here is the second of the two kinds of normal distribution problems:

     Problem 2 -  Finding the z score when you know the probability:

     Please read pages 250-255 and try to follow the steps.  You should be able to do calculations like these:

1. The z score for the 95th percentile. The body of the table does not contain .9500, but .9500 is midway between .9495 (whose z score is 1.64) and  .9505 (whose z score is 1.65).  So we will use 1.645 as the z score for .95.

2. The z scores at the bottom 2.5% and the top 2.5%.  For the bottom z score, search the body of the table for a probability of .025; its z score is -1.96.  For the top z score since we want the top 2.5%, but the table gives cumulative areas from the left, we search the body of the table for 1 - .025 = .975; its z score is 1.96.

3. We will use z _α  to denote the z score with area α to the right, so z.₀₂₅  equals 1.96.

     Here is a video on some of this using the tables, Excel, and Statdisk:

https://www.youtube.com/watch?v=7VQhqncKGZY

     You need to spend some time on these normal probability calculations.  It may take more than just one reading.  In this week's work you need to understand these calculations - in next week's work we will assume you understand them when we use them to solve problems.
     If you have questions, please ask me.

     Try to reproduce these answers using the tables, using Excel, and using Statdisk.  Which do you prefer and why?

 

 

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