Hey I’m trying to get through an assignment for stats. I’m kind of stuck on a question. I know how to calculate mean and standard deviation, my problem is I don’t know how to work backwards with these numbers. Which formula would I use to solve the following?
In a national study of educational attainment, you draw a sample of 540 persons and find that the distribution of years of schooling is normally distributed with a mean of 13.5 and a standard deviation of 2.2. With the information given above, please answer the questions as follows:
(a) How many persons in your sample have dropped out of high school, (i.e., have completed less than 12 years of schooling?
I’m really stuck on this so a point in the right direction would help alot.
I would suggest trying to use the information given to find a z-score. Depending on the level of the course you are taking and how sadistic your instructor is, you may find a normal distribution chart helpful in finding a percentage. Otherwise, you will have to either use some calculus and evaluate an integral or a computational method such as the trapezoid rule to get an approximation.
[quote]Jlabs wrote:
Hey I’m trying to get through an assignment for stats. I’m kind of stuck on a question. I know how to calculate mean and standard deviation, my problem is I don’t know how to work backwards with these numbers. Which formula would I use to solve the following?
In a national study of educational attainment, you draw a sample of 540 persons and find that the distribution of years of schooling is normally distributed with a mean of 13.5 and a standard deviation of 2.2. With the information given above, please answer the questions as follows:
(a) How many persons in your sample have dropped out of high school, (i.e., have completed less than 12 years of schooling?
I’m really stuck on this so a point in the right direction would help alot.
[/quote]
I would say 24.77%
[quote]Dr.Matt581 wrote:
you may find a normal distribution chart helpful in finding a percentage[/quote]
Do you have a chart in your book or that your teacher gave you? Cuz she will probably be looking for whatever the chart says.
To get the z-score you do x minus mu over sigma, or in this case technically x minus x bar over s.
So you take the data point (12) minus the average (13.5) and get -1.5 and then you divide that by the standard deviation (2.2). So you get a z-score of .68181818181818181818181818 for a while lol.
On most of the charts for a normal distribution that I’m seeing .68 z score will give you 24.83%
Thanks brothas, yeah i started looking at the distribution and zscores, I’ve calculated them before I find them a bit tricky but thanks alot, I knew it was obvious but didn’t know where to start. I’m going to try computing it in spss as well I just though I’d see if there was an easy equation to plug number into.
Csulli I’m taking your word for that answer, you mess up I’m spamming your log some more ahha.
[quote]Jlabs wrote:
Thanks brothas, yeah i started looking at the distribution and zscores, I’ve calculated them before I find them a bit tricky but thanks alot, I knew it was obvious but didn’t know where to start. I’m going to try computing it in spss as well I just though I’d see if there was an easy equation to plug number into.
Csulli I’m taking your word for that answer, you mess up I’m spamming your log some more ahha.[/quote]
Well it said how many people so make sure you say 134 people and not the percent lol. In fact the accuracy of the answer hardly matters, because any way you slice it you’re going to end up rounding to 134 persons regardless of the third and fourth decimal place.
[quote]Dr.Matt581 wrote:
you may find a normal distribution chart helpful in finding a percentage[/quote]
Do you have a chart in your book or that your teacher gave you? Cuz she will probably be looking for whatever the chart says.
To get the z-score you do x minus mu over sigma, or in this case technically x minus x bar over s.
So you take the data point (12) minus the average (13.5) and get -1.5 and then you divide that by the standard deviation (2.2). So you get a z-score of .68181818181818181818181818 for a while lol.
On most of the charts for a normal distribution that I’m seeing .68 z score will give you 24.83%
So actually you should probably say that.[/quote]
Thanks a million can you just plainly type the equation again so I know for future use, I’ll look it up in the notes as well.
[quote]Jlabs wrote:
Thanks a million can you just plainly type the equation again so I know for future use, I’ll look it up in the notes as well.[/quote]
[quote]Jlabs wrote:
Thanks brothas, yeah i started looking at the distribution and zscores, I’ve calculated them before I find them a bit tricky but thanks alot, I knew it was obvious but didn’t know where to start. I’m going to try computing it in spss as well I just though I’d see if there was an easy equation to plug number into.
Csulli I’m taking your word for that answer, you mess up I’m spamming your log some more ahha.[/quote]
SPSS is not the direction you want to go. Create a deviation score in the numerator (your score-Mean of the population) and then divide by the error (in this case the population standard deviation).
