Very hard distribution question

[Sirsh]

The school psychologist was researching the results of a third school and found them to closely fit a normal distribution with a mean completion time of 20.68 minutes and a standard deviation of 5.3 minutes. Unfortuantly, after he had finished calculating the results, he could not read of of the observed values.

Given there are 125students at the school, and a percentage error of ten percent determine the possible range of values for the observed frequencies, falling between 14.5 and the 19.5 minute interval.

Equations:

z(standardised score) = (x-mean/standard dev)
Percentage error = |o-e|/total frequency * 100.


I have no clue.. please help me!

 

[nate808]

Hi there,

I am happy to help you with this problem. Let's break it down step by step.

First, let's define some variables:
- n = total number of students (125)
- m = mean completion time (20.68 minutes)
- s = standard deviation (5.3 minutes)
- p = percentage error (10%)

We know that the range of observed frequencies should fall between the 14.5 and 19.5 minute intervals. Let's convert these values to standardised scores using the formula: z = (x-m)/s

For the lower end of the range (14.5 minutes), we have z = (14.5-20.68)/5.3 = -1.16
For the upper end of the range (19.5 minutes), we have z = (19.5-20.68)/5.3 = -0.22

Next, let's use the formula for percentage error to determine the range of possible values for the observed frequencies. Rearranging the formula, we get:
|o-e| = p/100 * n
where o is the observed frequency and e is the expected frequency.

For the lower end of the range, we have:
|o-125| = 10/100 * 125 = 12.5
This means that the observed frequency could be anywhere between 125-12.5 = 112.5 and 125+12.5 = 137.5.

For the upper end of the range, we have:
|o-125| = 10/100 * 125 = 12.5
This means that the observed frequency could be anywhere between 125-12.5 = 112.5 and 125+12.5 = 137.5.

Therefore, the possible range of values for the observed frequencies falling between the 14.5 and 19.5 minute intervals is 112.5 to 137.5.

I hope this helps you understand the problem better. Let me know if you have any other questions. Good luck with your research!