andreapgomezr

13.08.2020 •

Mathematics

The most appropriate measure of center for this data is the
Weight of Male Dogs (Ib.)
DONE
8|1 means 81
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Ответ:

mckadams02

01.08.2020

Mathematics

The median of Group A is greater than the median of Group B.

Step-by-step explanation:

Given

$\begin{array}{cccccccccc}{Group\ A} & 4.5 & 4.8 & 4.6 & 5.0 & 4.8 & 4.4 & 4.7 & 5.2 & 3.9 \ \\ {Group\ B} & 5.5 & 4.9 & 4.0 & 4.2 & 4.8 & 4.1 & 3.5 & 4.6 & 4.3 \ \end{array}$

Required

Which of the options is true

(a) Group A complete in less time

To do this, we calculate the average of both using:

$\bar x = \frac{\sum x}{n}$

Where

n = 9

So, we have:

$\bar x_A = \frac{4.5 + 4.8 + 4.6 + 5.0 + 4.8 + 4.4 + 4.7 + 5.2 + 3.9}{9}$

$\bar x_A = \frac{41.9}{9}$

$\bar x_A = 4.66$

$\bar x_B =\frac{5.5 + 4.9 + 4.0 + 4.2 + 4.8 + 4.1 + 3.5 + 4.6 + 4.3}{9}$

$\bar x_B =\frac{39.9}{9}$

$\bar x_B =4.43$

The average time of Group A is higher than that of B, this means that Group A spend more time, on average.

(b) Group A has a greater median

First, we sort the given data in ascending order

$\begin{array}{cccccccccc}{Group\ A} & 3.9 & 4.4 & 4.5 & 4.6 & 4.7 & 4.8 & 4.8 & 5.0 & 5.2 \ \\ {Group\ B} & 3.5 & 4.0 & 4.1 & 4.2 & 4.3 & 4.6 & 4.8 & 4.9 & 5.5 \ \end{array}$

The median is then calculated using:

$Median = \frac{n + 1}{2}th$

This gives:

$Median = \frac{9 + 1}{2}th$

$Median = \frac{10}{2}th$

Median = 5th

The median is the fifth item for both groups.

So, we have:

A =4.7

B =4.3

4.7 is greater than 4.3

Hence, (b) is true

Since (b) is true and only one option is correct, then (c) and (d) are incorrect

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The most appropriate measure of center for this data is the Weight of Male Dogs (Ib.) DONE 8|1 means 81 Intro