Question

Average Life Expectancy Animal Years Bison 15 Cow 15 Deer 8 Donkey 12 EIK 15 Goat 8 Horse 20 Moose 12 Pig 10 Sheep 12 5) Find the median 6) State the 5-number summary: Min, Q1, median, Q3, max values 7) What is the IQR?

Answer 1

Answer:

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Explanation: Provided the Data, we need to simply find the (5) (6) and (7):

(5) Median:

Median is simply a middle number or a number that is at the center of the data organized in the order of least to greatest, therefore organizing the data results in the following sequence:

$8,8,10,12,12,12,15,15,15,20$

Median is the middle number if it is odd-numbered or the average of the two middle numbers in the case of the even-numbered data sequence, the median is as follows:

$M=\frac{12+12}{2}=12$

(6) 5-number summary:

(i) Min Is the least number in the data sequence, therefore it is:

$8$

(ii) Q1 or First Quartile, is simply the median of the first half of the data, therefore it is:

$Q_1=10$

(iii) Median It has been calculated already in the (5) part of this question, therefore it is:

$12$

(iv) Q3 Third Quartile, is simply the median of the second half of the data, and likewise, it is:

$15$

(v) Max value

$20$

(7) IQR

The interquartile range has the following simple formula:

$\begin{gathered} \text{IQR}=\frac{Q_3-Q_1}{2} \\ \end{gathered}$

Therefore the interquartile range is as follows:

$\begin{gathered} \text{IQR}=\frac{Q_3-Q_1}{2} \\ \therefore\Rightarrow \\ \text{IQR}=\frac{15-10}{2}=\frac{5}{2} \\ \text{IQR}=\frac{5}{2} \end{gathered}$