GCSE(H),

A **Histogram** shows the frequency of **Grouped Data** in **Class Intervals**.

On a bar chart, the height of the bar gives the frequency. On a histogram, the *area* of the bar represents the frequency, rather than the height.

The height of each histogram bar is calculated by dividing the frequency by the **Class Width**. This height is called the **Frequency Density.** For a histogram with equal intervals, the class width is constant.

The table below shows percentage scores for a maths test. The table shows the Frequency Density and the resultant histogram.

Hour | Frequency | Class width | Frequency Density |
---|---|---|---|

0 < h ≤ 20 | 3 | 20 | 0.15 |

20 < h ≤ 40 | 18 | 20 | 0.90 |

40 < h ≤ 60 | 85 | 20 | 4.25 |

60 < h ≤ 80 | 73 | 20 | 3.65 |

80 < h ≤ 100 | 23 | 20 | 1.15 |

1. The speed of cars passing a point on the road was recorded over a period of one hour. The data was plotted on a histogram. From the histogram, below, determine the number of cars that passed the point at more than 30mph.

Answer: 5 cars

The width of the class interval is 10 (10mph).

The frequency density for 30 < s ≤ 40 is 0.5

0.5 x 10 = 5 cars

2. The table below shows the amount that customers spent at a local shop. A histogram is to be plotted for this data. What is the value of frequency density for the class interval £0 < S ≤ £5?

Amount (P) | Frequency |
---|---|

0 < h ≤ 5 | 48 |

5 < h ≤ 10 | 32 |

10 < h ≤ 15 | 15 |

15 < h ≤ 20 | 17 |

20 < h ≤ 25 | 8 |

25 < h ≤ 30 | 4 |

Answer: 9.6

The frequency for this class is 48.

The class interval is 5.

Frequency density = `frac(text(frequency))(text(class interval))` = 48 ÷ 5 = 9.6

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