Question

The weights of a sample of crates ready for shipment to Moscow, Russia are (in kilograms): 103, 97, 101, 106 and 103. What is the mean deviation?

Answer 1

Answer:

The weights of a sample of crates ready for shipment to Moscow, Russia are (in kilograms): 103, 97, 101, 106, and 103. ... The sample mean is $150 and the standard deviation is $20.

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Answer 2

Given are the weights of the sample crates, Find the mean deviation of the data.

Explanation:

• The mean deviation of a data set is the measure of the deviation of the data values about the mean of the data set.
• Mathematically the mean deviation of the data having the mean $\bar x$ and total data values $N$ is given by,      
•                                  $mean\ dev.=\frac{\sum (x_i-\bar x)}{N}$      
• Firstly, the mean of the given data is given by,
•                                  $\bar x=\frac{\sum x_i}{N}\\ ->\bar x=\frac{103+97+101+106+103}{5} \\->\bar x=102$  
• Now using the mean deviation formula mentioned above we get,
•              $->mean\ dev.=\frac{\sum (x_i-102)}{5} \\->mean \ dev.=\frac{(103-102)+(97-102)+(101-102)+(106-102)+(103-102)}{5} \\->mean\ dev.=0$    
• Mean deviation of the given weights of the sample of creates is $0$.