What is the upper quartile, Q3, of the following data set? 54, 53, 46, 60, 62, 70, 43, 67, 48, 65, 55, 38, 52, 56, 41
scZoUnD [109]
The original data set is
{<span>54, 53, 46, 60, 62, 70, 43, 67, 48, 65, 55, 38, 52, 56, 41}
Sort the data values from smallest to largest to get
</span><span>{38, 41, 43, 46, 48, 52, 53, 54, 55, 56, 60, 62, 65, 67, 70}
</span>
Now find the middle most value. This is the value in the 8th slot. The first 7 values are below the median. The 8th value is the median itself. The next 7 values are above the median.
The value in the 8th slot is 54, so this is the median
Divide the sorted data set into two lists. I'll call them L and U
L = {<span>38, 41, 43, 46, 48, 52, 53}
U = {</span><span>55, 56, 60, 62, 65, 67, 70}
they each have 7 items. The list L is the lower half of the sorted data and U is the upper half. The split happens at the original median (54).
Q3 will be equal to the median of the list U
The median of U = </span>{<span>55, 56, 60, 62, 65, 67, 70} is 62 since it's the middle most value.
Therefore, Q3 = 62
Answer: 62</span>
The rotation is were you keep one point at the same spot and move the others
Let that certain number be 'n'
50% (n) = 39
50% ⇒ 0.5
0.5 (n) = 39
Divide both sides by 0.5
0.5n = 39
/0.5 /0.5
n = 78.
So that certain number is actually <em><u>78! :)</u></em>
CROSS CHECK:
50% of 78
0.5 × 78
= 39
Hope this helps! :)
42=6x7 then 60=6 x 10 so the least number cover the square should be 6x7x10, that is 420
