stu132057
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The following are the ages of 13 history teachers in a school district. 24, 27, 29, 29, 35, 39, 43, 45, 46, 49, 51, 51, 56 Notice that the ages are ordered from least to greatest. Give the five-number summary and the interquartile range for the data set. Five-number summary

Minimum:
Lower quartile:
Median:
Upper quartile:
Maximum:
Interquartile range:

Sagot :

The minimum value of data is 24,lower quartile is 29,median is 41, upper quartile is 50 and maximum value is 56 and the interquartile range is 21.

Given a data about ages of 13 history teacher as under:

24, 27, 29, 29, 35, 39, 43, 45, 46, 49, 51, 51, 56.

We are required to find the minimum value, lower quartile,median,upper quartile,maximum value, interquartile range.

The minmum value is 24.

Lower quartile=(n+1)/4 th term

=(13+1)/4

=7/2

=3.5

Lower quartile=(29+29)/2

=29

Median=(n/2)th term

=13/2 th term

=6.5 th term

Median=(39+43)/2

=82/2

=41

Upper quartile=3(n+1)/4   th term

=3(13+1)/4

=3*14/4

=10.5 th term

Upper quartile=(49+51)/2=100/2=50

Inter quartile range=Upper quartile- lower quartile

=50-29

=21

Hence the minimum value of data is 24,lower quartile is 29,median is 41, upper quartile is 50 and maximum value is 56 and the interquartile range is 21.

Learn more about quartiles at https://brainly.com/question/10005803

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