Math  /  Data & Statistics

QuestionHere are the scores of 13 students on an algebra test. 59,63,68,68,77,79,81,82,83,86,88,90,9259,63,68,68,77,79,81,82,83,86,88,90,92
Notice that the scores are ordered from least to greatest. Give the five-number summary and the interquartile range for the data set. \begin{tabular}{|l|} \hline \multicolumn{1}{|l|}{ Five-number summary } \\ Minimum: \\ Lower quartile: \\ Median: \\ Upper quartile: \\ Maximum: \\ \hline Interquartile range: \\ \hline \end{tabular}

Studdy Solution

STEP 1

What is this asking? We need to find the five-number summary (minimum, lower quartile, median, upper quartile, and maximum) and the interquartile range of a given set of algebra test scores. Watch out! Remember that the lower quartile isn't the lowest score, and the upper quartile isn't the highest score!
Also, don't mix up the interquartile range with the overall range.

STEP 2

1. Find the Minimum and Maximum
2. Find the Median
3. Find the Lower Quartile
4. Find the Upper Quartile
5. Calculate the Interquartile Range

STEP 3

The **minimum** is the smallest value in the dataset.
In our ordered list, that's 59\bf{59}.
The **maximum** is the largest value, which is 92\bf{92}.
Easy peasy!

STEP 4

The **median** is the middle value when the data is ordered.
Since we have 13\bf{13} scores (an odd number), the median is the 7th\bf{7^{th}} value.
Counting up from 5959, we find the median is 81\bf{81}.

STEP 5

The **lower quartile** is the median of the lower half of the data.
Since our median is 8181, we look at the values *before* 8181, which are 59,63,68,68,77,7959, 63, 68, 68, 77, 79.
There are 6\bf{6} values here (an even number).

STEP 6

To find the median of an even number of values, we take the average of the two middle values.
In this case, the two middle values are 6868 and 6868.
So, the lower quartile is 68+682=1362=68\frac{68 + 68}{2} = \frac{136}{2} = \bf{68}.

STEP 7

The **upper quartile** is the median of the upper half of the data.
We look at the values *after* the median (8181), which are 82,83,86,88,90,9282, 83, 86, 88, 90, 92.
Again, there are 6\bf{6} values.

STEP 8

The two middle values are 8686 and 8888.
So, the upper quartile is 86+882=1742=87\frac{86 + 88}{2} = \frac{174}{2} = \bf{87}.

STEP 9

The **interquartile range (IQR)** is the difference between the upper quartile and the lower quartile.
It tells us how spread out the middle 50%50\% of the data is.
In our case, the IQR is 8768=1987 - 68 = \bf{19}.

STEP 10

Five-number summary: Minimum: 5959 Lower quartile: 6868 Median: 8181 Upper quartile: 8787 Maximum: 9292
Interquartile range: 1919

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