Answered: During a quality assurance check, the…

Name: During a quality assurance check, the actual contents (in grams) of
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Description: During a quality assurance check, the actual contents (in grams) of six containers of protein powder were recorded as 1526, 1532, 1502, 1516 , 1534, and 1512. (a) Find the mean and the median of the contents. (b) The third value was incorrect

MATLAB: An Introduction with Applications

6th Edition

ISBN:9781119256830

Author:Amos Gilat

Publisher:Amos Gilat

Chapter1: Starting With Matlab

Section: Chapter Questions

Problem 1P

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During a quality assurance check, the actual contents (in grams) of six containers of protein powder were recorded as 1526, 1532, 1502, 1516 , 1534, and 1512.

(a) Find the mean and the median of the contents.

(b) The third value was incorrectly measured and is actually 1517.

Find the mean and the median of the contents again.

c) Which measure of central tendency, the mean or the median, was affected more by the data entry error?

Definition Definition Middle value of a data set. The median divides a data set into two halves, and it also called the 50th percentile. The median is much less affected by outliers and skewed data than the mean. If the number of elements in a dataset is odd, then the middlemost element of the data arranged in ascending or descending order is the median. If the number of elements in the dataset is even, the average of the two central elements of the arranged data is the median of the set. For example, if a dataset has five items—12, 13, 21, 27, 31—the median is the third item in ascending order, or 21. If a dataset has six items—12, 13, 21, 27, 31, 33—the median is the average of the third (21) and fourth (27) items. It is calculated as follows: (21 + 27) / 2 = 24.

Expert Solution

Step 1

From the given information

The recorded values (in grams) $x_{i} = 1526, 1532, 1502, 1516, 1534, 1512$ and n=6

Therefore the mean value is calculated as follows:

$\begin{array}{rcl} x & = & \frac{\sum_{i = 1}^{n}}{n} \\ = & \frac{1526 + 1532 + 1502 + 1516 + 1534 + 1512}{6} \\ = & 1520.33 \end{array}$

The formula for median is $M = {(\frac{n + 1}{2})}^{t h}$ observation.

Therefore,

$\begin{array}{rcl} M & = & (\frac{n + 1}{2}) \\ = & \frac{6 + 1}{2} \\ = & 3.5 \end{array}$

In this case, there are 6 observation so we have to find the average of the 3^rd and 4^th values when data is ascending order.

So, we get the median is

$\begin{array}{rcl} M & = & \frac{1516 + 1526}{2} \\ = & 1521 \end{array}$ .

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