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>
Answer:
A. 3(t+2)
Step-by-step explanation:
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The result is
A. 3(t+2)
Answer:
1/26
Step-by-step explanation:
There is only one 6 of hearts and only one ace of diamonds.
The probability of selecting the 6 of hearts is thus 1/52, and that of selecting the ace of diamonds is also 1/52.
The probability of selecting the 6 of hearts or the ace of diamonds is the SUM of these two results: 1/52 + 1/52 = 2/52 = 1/26.
Quotients can sometimes be irrational so it is necessary to estimate them, products on the other hand are only rational, provided they aren't multiplied by an irrational, so it is not necessary to estimate
