ABCDEFGHIJKLMNOPQRSTUVWXYZ
familiar with box and whisker plots right ?
easy - start by sorting the given data in increasing order
Excellent ! below is the format for five number summary : `minimum -- IQR1 -- median -- IQR3 -- maximum`
finding minimum and maximum is trivial, just take the first value for minimum and the last value for maximum
Right, what about the other three measures ? know how to find them ?
Inter Quartile Range 1 I think its also called as 25th percentile or Q1
Q1, median and Q2 are just numbers may be watch this short video from khan : https://www.youtube.com/watch?v=BXq5TFLvsVw It explains clearly how to calculate the 3 numbers
Okay, I'll try... lets find the `median` first
median = middle value of the data set
15, 27, `28, 34`, 42, 52
take average of those two middle numbers to get the median
median = \(\large \dfrac{28+34}{2} = ?\)
Yes !
Q1 = middle value of first half
`15, 27, 28`, 34, 42, 52
whats the middle value of first half ?
Yep !! Q3 = middle value of second half 15, 27, 28, 34, 42, 52 Q3 = ?
Correct ! its called Q3
Looks like we got all the 5 numbers ?
arrange them in below format : `minimum -- Q1-- median -- Q3-- maximum`
Looks good ^
we're done with part a
mean = (sum of all values)/(number of values)
Add up all the values, then divide by the number of values
how ?
check ur addition again 28 +34 +27+ 42+ 52+ 15 = 198 NOT 128
Oh I see, I should have told you what to add lol
Exactly !
Yes, mean = 33
right !
yes please
Notice that there are 3 columns in the given table
Fill the first column with the given data values
x -------------- 28 34 27 42 52 15
order doesn't matter... just take the given data set as it is..
yea, lets figure out what to put in second column
\(\overline{X} = \text{mean} = 33\)
yes !
put it in the table next to x column
x x-33 -------------- 28 -5 34 1 27 -6 42 9 52 19 15 -18
square each value in second column
right
Looks good
add them and you're done !
25 1 36 81 361 324 -----------
we use the result from part C for calculating standard dviation
standard deviation = \(\large \sqrt{\dfrac{(x-\overline{x})^2}{n}}\)
\(n\) = number of values in the data set
how ?
standard deviation = \(\large \sqrt{\dfrac{(x-\overline{x})^2}{n}} = \sqrt{\dfrac{828}{6}} = \sqrt{138} \approx 11.75\)
standard deviation = \(\large \sqrt{\dfrac{\color{red}{\sum} (x-\overline{x})^2}{n}} = \sqrt{\dfrac{828}{6}} = \sqrt{138} \approx 11.75 \)
np :)
I'll try, ask...
Join our real-time social learning platform and learn together with your friends!