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Question 1196832: PICTURE FOR PROBLEM https://imagizer.imageshack.com/img924/482/OjKL0M.jpg
a) Find the quartiles using technology.
Q1= Q2=
b) Find the quartiles using the Tukey method.
Q1= Q2=
c) Find the IQR using the quartiles from part b.
d) Find the standard deviation.
Answer by math_tutor2020(3817)   (Show Source):
You can put this solution on YOUR website!
a)
The answer will slightly vary depending on which type of technology you use. Are you using a graphing calculator? Spreadsheet? A programming language such as R or Python?
Refer to this article for more details
https://en.wikipedia.org/wiki/Quartile
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b)
Given Data
| 40.4 | 5.18 | 32.59 | 47.01 | | 37.12 | 34.61 | 63.44 | 51.53 | | 57.32 | 81.13 | 47.5 | 11.65 | | 21.83 | 42.21 | 38.77 | 32.05 | | 76.72 | 72.8 | 46.93 | 65.45 |
Let's list out the sorted values from smallest to largest
{5.18, 11.65, 21.83, 32.05, 32.59, 34.61, 37.12, 38.77, 40.4, 42.21, 46.93, 47.01, 47.5, 51.53, 57.32, 63.44, 65.45, 72.8, 76.72, 81.13}
There are n = 20 items in this data set
The median is at the midpoint of slot 10 and 11
since n/2 = 20/2 = 10
The items in the 10th and 11th slots are 42.21 and 46.93 respectively
The midpoint of which is (42.21+46.93)/2 = 44.57 and this is the median of the given set.
The median is sometimes referred to as Q2
Let's form two new sets L and U
L = lower set = items below the median 44.57
U = upper set = items above the median 44.57
L = {5.18, 11.65, 21.83, 32.05, 32.59, 34.61, 37.12, 38.77, 40.4, 42.21}
U = {46.93, 47.01, 47.5, 51.53, 57.32, 63.44, 65.45, 72.8, 76.72, 81.13}
Sets L and U have n/2 = 20/2 = 10 items each in them.
The median of set L is 33.60
The median of set U is 60.38
These are the values of Q1 and Q3 respectively.
Summary:
Q1 = 33.60
Q2 = 44.57 aka median
Q3 = 60.38
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c)
IQR = Q3 - Q1
IQR = 60.38 - 33.60
IQR = 26.78
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d)
You can use a spreadsheet or graphing calculator to compute the standard deviation in one single command.
So there isn't much for me to go over here if your teacher wants you to use technology.
However, I'll show a method to calculate the standard deviation by hand.
Check out this spreadsheet
https://docs.google.com/spreadsheets/d/1VsuY1rMMsdiF1GKgO_5x0Ts5vIOnn6hf5fhHlC4cIfw/edit?usp=sharing
The x column is the set of original values
xbar = 45.312 is the sample mean (found by adding every x value up and dividing by n = 20)
The column labeled (x-xbar)^2 has us do these two things in the order mentioned- Subtract xbar from each x value
- Square the result of the previous step
For example,
(x-xbar)^2 = (5.18 - 45.312)^2 = 1610.577424
and that result goes in the first entry of the (x-xbar)^2 column. It corresponds to x = 5.18
Once we calculate all the (x-xbar)^2 values, we add them all up to get
Sum[ (x-xbar)^2 ] = 7843.32072
In the spreadsheet provided above, I defined this value as the variable A.
Divide A over (n-1) to get the sample variance
Sample Variance = A/(n-1) = 7843.32072/(20-1) = 412.8063537
Then take the square root of that to get the sample standard deviation
SampleSD = sqrt(sampleVar)
SampleSD = sqrt(412.8063537)
SampleSD = 20.31763652
Or if you wanted the population variance, then,
popVar = A/n = 7843.32072/20 = 392.166036
and
popSD = sqrt(popVar) = sqrt(392.166036) = 19.803182471512
So we have:
sample standard deviation = 20.31763652
population standard deviation = 19.803182471512
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