Respuesta :

Answer: Option C. $17


Solution:

The data is:

$46, $70, $36, $43, $45, $60, $52


1) Sort the data from the least to the greatest:

$36, $43, $45, $46, $52, $60, $70


2) Find the median (middle value)

$36, $43, $45, $46, $52, $60, $70

   1        2      3      4      5       6     7

We have seven values, the middle value is the fourth value (three values to the left and three values to the right), then the median is $46.


3) Find the middle of the first half of the data (without the median): Quartile 1 (Q1)

$36, $43, $45

   1       2       3

The middle of the first half of the data is the second value: Q1=$43


3) Find the middle of the second half of the data (without the median): Quartile 3 (Q3)

$52, $60, $70

   5      6      7

The middle of the second half of the data is the sixth value: Q3=$60


4) The interquartile range (IQR) is the difference between Quartile 3 and Quartile 1:

IQR=Q3-Q1

IQR=$60-$43

IQR=$17

alinakincsem

Answer:

Option C. $17

Step-by-step explanation:

Interquartile is the measure which tells us about where the middle fifty of the data set lies.

To find the interquatile range, we must arrange the data set in an ascending order first and find the median.

[tex]36, 43, 45, 46, 52, 60, 70[/tex]

Median = [tex]\frac{n+1}{2} = \frac{7+1}{2} = \frac{8}{2} = 4th value[/tex]

Therefore, median is 46.

Now divide the remaining values in two groups with reference to the median: one set at the left of the median and one at the right of the median.

[tex](36, 43, 45)[/tex] and [tex](52, 60, 70)[/tex]

Find median from both these sets and take their difference to get the interquartile range:

[tex]60-43 = 17[/tex]

Therefore, $17 is the interquartile range for this data set.

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