Averages/Dispersion-Solutions

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Question 1
Find the mean and standard deviation of the following numbers.

5,9,3,3,4,5,6,7,9,8

First we need to find the sum of the numbers

Therefore, since there are 10 numbers, the mean is

To find the standard deviation, we will need to square all of the numbers

x2 = 25,81,9,9,16,25,36,59,81,64

and then find the sum

Therefore the standard deviation is

 

Substitute the numbers into the formula.

Write over a common denominator.

 

Square root to find the standard deviation.

Therefore
and

 

Question 2

A supermarket wanted to check the consistency of the weights of squashes it was selling. They therefore weighed a sample, putting the results in the table below. Find the mean weight and the standard deviation of the weights.

Mass (kg)
Number 4 10 22 12 2

To calculate the mean and standard deviation of this data, we need to draw up a table, including the midpoint, the midpoint squared and both of these multiplied by the frequency.

Mass (kg) Frequency Midpoint (m) m2 fm fm2
4 1.3 1.69 5.2 6.76
10 1.5 2.25 15 22.5
22 1.7 2.89 37.4 63.58
12 1.9 3.61 22.8 43.32
2 2.1 4.41 4.2 8.82
50 84.6 144.98

We can now calculate the mean and the standard deviation, using the formulæ.

Substitute in the sums from the table.
Therefore

and for the standard deviation

 

 

Substitute in the sums from the table.

Tidy up.

Square root for standard deviation.

Therefore

 

Question 3
Find the mean and standard deviation of the following data.
x
f 6 17 30 27 12 8

We first need to put together a table, containing the midpoints and the various sums that are required to calculate the mean and standard deviation.

x f m m2 fm fm2
6 2.5 6.25 15 37.5
17 7 49 119 833
30 11 121 330 3630
27 14 196 378 5292
12 18 324 216 3888
8 23 529 184 4232
100 1242 17912.5

We now just need to put the sums into the formulæ

First the mean

 

 

 

Therefore the mean is

 

Now the standard deviation

 

 

 

 

 

Substitute in the values from the table.

Tidy up.

Square root to find the standard deviation.

Therefore the standard deviation is

 

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