Stats Help Using Standard deviation and a population

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Homework Help Overview

The discussion revolves around a statistics problem involving standard deviation and normal distribution, specifically related to race finishing times. The original poster is trying to determine how many women finished faster than Joan, who completed the race 1.75 standard deviations faster than the average for her age group.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the interpretation of standard deviations and how to apply them to the population of runners. There is confusion about the distribution of scores within standard deviations and how to calculate the number of faster runners based on z-scores.

Discussion Status

Participants are actively engaging with the problem, questioning assumptions about the distribution of scores and the calculations involved. Some guidance has been offered regarding the use of standard normal distribution tables and the interpretation of z-scores, but there is no explicit consensus on the final calculation.

Contextual Notes

There is mention of a brief lecture on standard deviation, which may have left some participants unclear about the concepts. The discussion includes attempts to clarify the relationship between z-scores and the total population of runners, as well as the steps needed to arrive at a solution.

Kingyou123
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Homework Statement


Joan’s finishing time for the Bolder Boulder 10K race was 1.75 standard deviations faster than the women’s average for her age group. There were 405 women who ran in her age group. Assuming a normal distribution, how many women ran faster than Joan.

Homework Equations



The Attempt at a Solution


I know Joan ran in between the 1st and 2nd standard deviation I'm just confused on how to split up the population correctly. Would the first deviation be 68% of the population?
 
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Kingyou123 said:

Homework Statement


Joan’s finishing time for the Bolder Boulder 10K race was 1.75 standard deviations faster than the women’s average for her age group. There were 405 women who ran in her age group. Assuming a normal distribution, how many women ran faster than Joan.

Homework Equations



The Attempt at a Solution


I know Joan ran in between the 1st and 2nd standard deviation I'm just confused on how to split up the population correctly. Would the first deviation be 68% of the population?
No. About 34% of the population will have scores between z = 0 and z = 1. Between z = -1 and z = 1 (i.e., within 1 st. deviation of the mean), there is about 68% of the population. You need to figure out how many are included between z = 0 and z = 1.75 -- a standard normal distribution table will be helpful. Don't forget to add the half of the race participants whose z scores are less than or equal to 0.
 
Mark44 said:
No. About 34% of the population will have scores between z = 0 and z = 1. Between z = -1 and z = 1 (i.e., within 1 st. deviation of the mean), there is about 68% of the population. You need to figure out how many are included between z = 0 and z = 1.75 -- a standard normal distribution table will be helpful. Don't forget to add the half of the race participants whose z scores are less than or equal to 0.
So "z" would equal .4599, I'm assuming you would multiply it by 405 to get the number of runners who are faster than her. So the answer would be 186 runners?
 
Kingyou123 said:
So "z" would equal .4599, I'm assuming you would multiply it by 405 to get the number of runners who are faster than her. So the answer would be 186 runners?
No z = 1.75. I think what you found was P(0 < z < 1.75) = .4599.

Do you understand what z is? If not it's the variable for a standard normal distribution; one with mean 0 and standard deviation 1. The total area under the graph of the distribution is 1. This area (or probabillity) includes ##P(-\infty < z < 0)##, which is .5 and P(0 < z < 1.75). What proportion of the runners are faster than she is? I don't think your answer of 186 is correct.
 
Mark44 said:
No z = 1.75. I think what you found was P(0 < z < 1.75) = .4599.

Do you understand what z is? If not it's the variable for a standard normal distribution; one with mean 0 and standard deviation 1. The total area under the graph of the distribution is 1. This area (or probabillity) includes ##P(-\infty < z < 0)##, which is .5 and P(0 < z < 1.75). What proportion of the runners are faster than she is? I don't think your answer of 186 is correct.
My professor talked about standard deviation for a short time, maybe 8 minutes. So I understand that P(-infy<z<) and P(0 < z < 1.75) gives me the proportion of runners that are slower than her, however I'm confused on what the next step would be from there. Would it be as simple as multiplying the sum of both proportions by 405 to get a number than subtracting that number from 405 to get the number of runners that are faster than she is?
Edit:
.9599*405=388.75

405-388.75= 16.25 or 16 runners are faster than Joan
 
Kingyou123 said:
My professor talked about standard deviation for a short time, maybe 8 minutes. So I understand that P(-infy<z<) and P(0 < z < 1.75) gives me the proportion of runners that are slower than her, however I'm confused on what the next step would be from there. Would it be as simple as multiplying the sum of both proportions by 405 to get a number than subtracting that number from 405 to get the number of runners that are faster than she is?
Edit:
.9599*405=388.75

405-388.75= 16.25 or 16 runners are faster than Joan
Looks good
 

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