Stem and leaf display statistics

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Homework Help Overview

The discussion revolves around a statistics assignment involving the analysis of weights of a sample of horses. Participants are tasked with constructing a stem-and-leaf display, calculating the median, mean, variance, and standard deviation, as well as creating a frequency distribution based on the provided data.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the construction of a stem-and-leaf display and the calculation of the median weight. Questions arise regarding the definitions of variance and standard deviation, and under what conditions one can be larger than the other. There is also an exploration of how to create a frequency distribution based on the data.

Discussion Status

The discussion is ongoing, with some participants providing guidance on how to approach the frequency distribution and clarifying definitions related to variance and standard deviation. There is acknowledgment of progress made, but participants express uncertainty about the calculations and seek further assistance.

Contextual Notes

Participants note their struggles with statistics as a non-primary field of study, which may impact their confidence in handling the assignment. There is a mention of the requirement to use specific class intervals for the frequency distribution.

auzie
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I am working on statistics assignment and since statistics is not my primary field of studies I have difficulties with it. can someone please help me out?? I did stem and leaf display which I think looks OK but am not sure.

Homework Statement



The following data represent the weights (in kilograms) of a sample of 25 horses:

164 148 137 157 173 156 177 172 169 165
145 168 163 162 174 152 156 168 154 151
174 146 134 140 171

Use these data to answer the following questions:
a. Construct a stem-and-leaf display for the weights.
b. Find the median weight.
c. Compute the sample mean weight.
d. Compute the sample variance and sample standard deviation.
e. Construct a frequency distribution for the data, using five class intervals and the value 130 as the lower limit of the first class.



Homework Equations



it possible for the standard deviation of a data set to be larger than its variance? Explain.


The Attempt at a Solution

 
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What have you done with these? Don't you have any work to show?

As for your final question, what are the definitions of "variance" and "standard deviation"? Under what conditions can one number be larger than the other?
 


I first sorted numbers from smallest to the largest. Based on that I believe median weight is 162kg because is the 13th in the row. My stem and leaf looks something like this;

STEM LEAF
13 4,7
14 0,5,6,8
15 1,2,4,6,6,7
16 2,3,4,5,8,8,9
17 1,2,3,4,4,7

Mean weight I calculated by adding all numbers together and then deviding by 25 which give me a sum of 159.04kg.
This is where I got stuck. Any suggestions??
 


auzie said:
I am working on statistics assignment and since statistics is not my primary field of studies I have difficulties with it. can someone please help me out?? I did stem and leaf display which I think looks OK but am not sure.

Homework Statement



The following data represent the weights (in kilograms) of a sample of 25 horses:

164 148 137 157 173 156 177 172 169 165
145 168 163 162 174 152 156 168 154 151
174 146 134 140 171

Use these data to answer the following questions:
a. Construct a stem-and-leaf display for the weights.
b. Find the median weight.
c. Compute the sample mean weight.
d. Compute the sample variance and sample standard deviation.
e. Construct a frequency distribution for the data, using five class intervals and the value 130 as the lower limit of the first class.



Homework Equations



it possible for the standard deviation of a data set to be larger than its variance? Explain.


The Attempt at a Solution


auzie said:
I first sorted numbers from smallest to the largest. Based on that I believe median weight is 162kg because is the 13th in the row. My stem and leaf looks something like this;

STEM LEAF
13 4,7
14 0,5,6,8
15 1,2,4,6,6,7
16 2,3,4,5,8,8,9
17 1,2,3,4,4,7

Mean weight I calculated by adding all numbers together and then deviding by 25 which give me a sum of 159.04kg.
This is where I got stuck. Any suggestions??

Okay, good so far! For the last one, "Construct a frequency distribution for the data, using five class intervals and the value 130 as the lower limit of the first class." I would note that all the data lies between 130 and 180 and that 180- 130= 50 so that "five class intervals" would divide that into 10 pound intervals. Basically, you can just look at your stem and leaf chart to see how many data points lie in each interval.

The only things left then are the variance and standard distribution. Again, what are their definitions?
 


HallsofIvy,

Thanks for your help so far. At least I got something right. My understanding is that standard deviation is a measure of the set values. I think I should deduct mean average, which in this case is 159, from each horse to get deviation of each number from the mean.
Each deviation should then be squared to get any negative values positive.

Next step (I think) is finding mean of those squared deviations by adding all deviations and deviding by 25. Square root of that would be standard deviation. Does this makes any sense??
 


auzie said:
HallsofIvy,

Thanks for your help so far. At least I got something right. My understanding is that standard deviation is a measure of the set values.
This is meaningless- every statistic is "a measure of the set values"!

I think I should deduct mean average, which in this case is 159, from each horse to get deviation of each number from the mean.

Next step (I think) is finding mean of those squared deviations by adding all deviations and deviding by 25. Square root of that would be standard deviation. Does this makes any sense??
The variation is defined as the sum of the squares of differences from the mean divided, in this case, by 25. The standard deviation is the square root of the variation.
 
Last edited by a moderator:


Hi HallsofIvy,

Thanks again for your valuable input. This is just the beginning of my strugle and there is worst to come. Luckilly I have only one unit of statistics to study which is bad enough.

Thanks again and good luck.
 

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