**MEAN: **The arithmetic mean of grouped frequency distribution can be obtained using:

Class Mark Method:

X = ∑fx/∑f where x is the midpoint of the class interval.

Assumed Mean Method: It is also called working mean method. X = A + (∑ Fd/∑f)

Where, d = x – A, x = class mark and A = assumed mean.

** EXAMPLE:** The numbers of matches in 100 boxes are counted and the results are shown in the table below:

Number of matches | 25 – 28 | 29 – 32 | 33 – 36 | 37 – 40 |

Number of boxes | 18 | 34 | 37 | 11 |

Calculate the mean (i) using class mark (ii) assumed mean method given that the assumed mean is 30.5.

*Solution:*

Class interval | F | X | FX | d = x – A | Fd |

25 – 28 | 18 | 26.5 | 477 | 4 | 72 |

29 – 32 | 34 | 30.5 | 1037 | 0 | 0 |

33 – 36 | 37 | 34.5 | 1276.5 | 4 | 148 |

37 – 40 | 11 | 38.5 | 423.5 | 8 | 88 |

Total | 100 | 3214 | 164 |

- Class Mark Method: X = ∑fx/∑f = 3214/100 = 32. 14 = 32 matches per box (nearest whole no)
- Assumed Mean Method: X = A + (∑ Fd/∑f)

= 30. 5 + (164/100) =30.5 + 1.64

= 32.14 = 32 matches per box (nearest whole number)

*EVALUATION:*

Calculate the mean shoe sizes of the number of shoes represented in the table below using (i) class mark (ii) assumed mean method given that the assumed mean is 42.

Shoe sizes | 30 – 34 | 35 – 39 | 40 – 44 | 45 – 49 | 50 – 54 |

No of Men | 10 | 12 | 8 | 15 | 5 |

**MODE**

The mode of a grouped frequency distribution can be determined ** geometrically** and by

*interpolation method.*Mode from Histogram: The highest bar is the modal class and the mode can be determined by drawing a straight line from the right top corner of the bar to the right top corner of the adjacent bar on the left. Draw another line from the left top corner to the bar of the modal class to the left top corner of the adjacent bar on the right.

**EVALUATION**: Calculate the modal shoe sizes and median of the number of shoes represented in the table below using interpolation and graphical method.

Shoe sizes | 30 – 34 | 35 – 39 | 40 – 44 | 45 – 49 | 50 – 54 |

No of Men | 10 | 12 | 8 | 15 | 5 |

**GENERAL EVALUATION**:

The table below gives the distribution of masses (kg) of 40 people

Masses (kg) | 1 – 5 | 6 – 10 | 11 -15 | 16 – 20 | 21 – 25 | 26 – 30 | 31 – 35 | 36 – 40 |

Frequency | 9 | 20 | 32 | 42 | 35 | 22 | 15 | 5 |

- State the modal class of the distribution and find the mode.
- Draw a cumulative frequency curve to illustrate the distribution.
- Use the curve in ‘2’ to estimate the median.
- Calculate the mean of the distribution.

**READING ASSIGNMENT**

New General Mathematics SSS2,page 160, exercise14a.

**WEEKEND ASSIGNMENT**

The table gives the frequency distribution of a random sample of 250 steel bolts according to their head diameter, measured to the nearest 0.01mm.

Diameter (mm) | 23.06 – 23.10 | 23.11 – 23.15 | 23.16 – 23.20 | 23.21 – 23.25 | 23.26-23.30 | 23.31 – 23.35 | 23.36-23.40 | 23.41-23.45 | 23.46-23.50 |

No of bolts | 10 | 20 | 28 | 36 | 52 | 38 | 32 | 21 | 13 |

- State the median class and calculate the median using interpolation method.
- Draw the histogram and use it to estimate the mode.
- Calculate the mean value using a working mean of 23.28mm.

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