# TOPIC 3: STATISTICS ~ MATHEMATICS FORM 3

**TOPIC 3: STATISTICS ~ MATHEMATICS FORM 3**

**Mean**

Calculating the Mean from a Set of Data, Frequency Distribution Tables and Histogram

Calculate the mean from a set of data, frequency distribution tables and histogram

Measures of central tendency:

The arithmetic mean

Example 1

The masses of some parcels are 5kg, 8kg, 20kg and 15kg. Find the mean mass of the parcels.

**Solution**

Total mass = (5 + 8 + 20 + 15) kg = 48kg

The number of parcels = 4

The mean mass = 48kg ÷ 4 = 12kg

The arithmetic mean used as measure of central tendency can be misleading as can be seen in the following example.

Example 2

John

and Mussa played for the local cricket team. In the last six batting

innings, they scored the following number of runs.John: 64, 0, 1, 2, 4,

1;Mussa: 15, 20, 13, 11 , 10, 3.Find the mean score of each player.

Which player would you rather have in your team? Give a reason.

and Mussa played for the local cricket team. In the last six batting

innings, they scored the following number of runs.John: 64, 0, 1, 2, 4,

1;Mussa: 15, 20, 13, 11 , 10, 3.Find the mean score of each player.

Which player would you rather have in your team? Give a reason.

**Solution**

John’s mean = (64 + 0 + 1 + 2 + 4 + 1) ÷ 6 = 12

Mussa’s mean = (15 + 20 + 13 + 11 + 10 + 3) ÷ 6 = 12

Each

player has the same mean score. However, observing the individual

scores suggests that they are different types of player. If you are

looking for a steady reliable player, you would probably choose Mussa.

player has the same mean score. However, observing the individual

scores suggests that they are different types of player. If you are

looking for a steady reliable player, you would probably choose Mussa.

Often it is possible to use the mean of one set of numbers to find the mean of another set of related numbers.

Suppose a number a is added to or subtracted from all the data. Then a is added to or subtracted from the mean.

Suppose

the n values are 𝑥!+𝑥! + 𝑥! ………+𝑥!. Multiply each by a, and

we obtain 𝑎𝑥!+𝑎𝑥! + 𝑎𝑥! ………+𝑎𝑥!. So we see that the mean

has been multiplied by a.

the n values are 𝑥!+𝑥! + 𝑥! ………+𝑥!. Multiply each by a, and

we obtain 𝑎𝑥!+𝑎𝑥! + 𝑎𝑥! ………+𝑎𝑥!. So we see that the mean

has been multiplied by a.

Interpreting the Mean Obtained from a Set Data, Frequency Distribution Tables and Histogram

Interpret the mean obtained from a set data, frequency distribution tables and histogram

Measures of central tendency from frequency tables

If the data has already been put into a frequency table, the calculation of the measures of central tendency is slightly easier.

Exercise 1

Juma rolled a six- sided die 50 times. The scores he obtained are summarized in the following table.Calculate the mean score

Score (x) | 1 | 2 | 3 | 4 | 5 | 6 |

Frequency (f) | 8 | 10 | 7 | 5 | 12 | 8 |

**Solution**

10 scores of 2 give a total 10 x 2 = 20

8 scores of 1 gives a total 8 x 1 = 8

And so on, giving a total score of

8 x 1 +10 x 2+7 x 3 + 5 x 4 + 12 x 5 + 8 x 6 = 177

The total frequency = 8 + 10 + 7 + 5 + 12 + 8 = 50

The mean score = 177 ÷ 50 = 3.34

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