Threads merged


Hello, my name is Evelyn and I am taking a Statistics class during the summer in preparation for a full time Ph.D program beginning in Fall. I have been out of school for over a decade and I have questions and answers that I need clarification for in order to build a solid foundation. I truly would like to understand so that I build upon my learning I will not be so lost. Any help would be greatly appreciated.

Question: If scores on the WAIS III are normally distributed, what is the probability that someone will score in what test developers consider the average range or Full Scale IQ scores ranging from 90 to 110?

Answer: .4972. To solve, we need to define the area under the curve that falls between the z values corresponding to IQ = 90 and 110. Thus,
z = (x – mu)/sigma (symbols do not copy and paste)
z = (110 – 100)/15 = .67
Because the normal distribution is symmetrical, the z score associated with a score of 90 is - .67. There are a number of ways to solve this from the table, but we can take the mean to z score x 2 or .2486 x 2 = .4972. It makes intuitive sense that nearly half the distribution of IQ scores should fall in a range we term “average.”

I understand the formula and how to arrive at the z=.67; however, I do not understand where the .2486 number came from. I have reviewed the materials and bought study guides but I have seem to get stuck with understanding where that number came from. I would so appreciate your help. Evelyn

The 0.2486 number comes from the cumulative distribution function, which is the function of the area under the standard distribution from z = 0 (i.e the center of the bell shaped curve) to whatever z-score you want. I bet there is a table of CDFs in your text book, looking something like the attached figure. In this problem you're looking for the area between z = -0.67 and z = +0.67. If you look up the CDF value for z=0.67 you get 0.2486. Since the bell curve is symmetric about z=0, the area under the curve from -0.67 to +0.67 is twice this value.

http://images.tutorvista.com/cms/ima...core-table.PNG

Hello, I am working on a problem that requires to use the z scale. When I look at the scale provided to me by my professor I am not understanding how to use it correctly. I have the problem and answer to the question. I only need clarification to the following:

Question: What Verbal Score on the SAT corresponds to the 83rd Percentile Rank if SAT scores are normally distributed?

Answer: 595. To solve we need to find the z score below which .83 of the area under the curve falls and then to use that to solve for the unknown Verbal score on the SAT. From the table, the area under the curve is closest to .82 when z = .95 (z = .96 is also fine to solve this problem).
z = (x – )/
.95 = (x – 500)/100
95 = (x – 500)
x = 595
I am not sure how .95 was found on the Table of the Standard Normal Cumulative Distribution Function Scale. I would imagine there are different for different scales for different measures. How do I read it correctly and how to I know the correct scale to use? Thank you for your help. Evelyn

Evelyn: did you see my answer to your other question:
https://www.askmehelpdesk.com/math-s...em-755714.html ?

Look at the table, and see that the entry with 0.33 as its entry (actually it's 0.3289, which is the closest to 0.33) is z=0.95. The way to read it is look down the first column for the first digit of the z-score (0.9), then across the top row for the second digit 0.05, and the intersecting box is the CDF for z = 0.9 + 0.05 = 0.95. Hope this helps.

The problem that was answered by ebaines was very helpful. Could you please clarify the following:
Question: What Verbal Score on the SAT corresponds to the 83rd Percentile Rank if SAT scores are normally distributed?

Answer: 595. To solve we need to find the z score below which .83 of the area under the curve falls and then to use that to solve for the unknown Verbal score on the SAT. From the table, the area under the curve is closest to .82 when z = .95 (z = .96 is also fine to solve this problem).
z = (x – )/
.95 = (x – 500)/100
95 = (x – 500)
x = 595
My Question: Can you please clarify in your answer to how to arrive at .95, when you stated look at the table and see the entry with 0.33 as its entry (actually it's 0.3289, which is closest to 0.33) is z=95. I looked at the table but I am not understanding how you arrived at 0.33 and the number 0.3289. I am trying to learn this so that I can build a strong foundation in stats. Thanks for your help. Evelyn

I am working on a problem and I have the answer. I need to understand the why?

Problem: Imagine that the individual in problem EC2 received a score of 17 on the Vocabulary subtest of the WAIS III. Using the standard discussed in the Power Point lectures, is this an unlikely or unusual event? Why?

Answer: Yes, because if p < .05, the event is generally regarded as unusual.
To solve, we need to find the area under the curve that falls above the z score associated with a Wechsler subscale score = 17. If the area is < .05 (i.e. the probability of that score occurring is < .05), we will regard the event as unusual. Although the concept of unusual has a number of meanings in psychology (e.g. pathological, statistically significant), most of our statistical tests will be run with  = .05 and any finding with p < .05 will be considered significant or unusual or unlikely to have occurred by chance alone. Thus,
z = (x – )/
z = (17 – 10)/3 = 2.33
From the table, the area under the curve that falls above z = 2.33 is .0099. Thus, p < .05, the event is unusual or unlikely.

I am trying to follow the logic; however if you can simplify even further or break it down into parts as to why this is unusual so I can digest easier, I would greatly appreciate it. Thank you, Evelyn

I am asking that your 4 questions be combined so that ebaines doesn't have to search them out.