A bakery's histogram for the weights of \(200\) loaf orders is grouped into five unequal classes, A to E. The histogram shows classes A, B and C; classes D ...
Question 1 Report
A bakery's histogram for the weights of \(200\) loaf orders is grouped into five unequal classes, A to E. The histogram shows classes A, B and C; classes D (\(300 \lt w \le 350\)) and E (\(350 \lt w \le 450\)) are not drawn.
- Work out the frequency represented by each of classes A, B and C. (3)
- Given that the frequency of class D is twice the frequency of class E, form and solve equations to find the frequency of each of classes D and E. (3)
- Work out the frequency density for class E. (1)
Answer Details
On a histogram, frequency equals frequency density multiplied by class width, so each shown class's frequency is found this way; once the total for the shown classes is known, the two remaining classes' combined frequency is found by subtracting from the grand total, and their given ratio turns this into a pair of equations.
- Class A (width \(50\)) has frequency density \(0.6\): \(0.6\times50=30\). Class B (width \(50\)) has frequency density \(1.2\): \(1.2\times50=60\). Class C (width \(100\)) has frequency density \(0.5\): \(0.5\times100=50\). Together: \[ 30+60+50=140 \] [3 marks]
- Let the frequency of class E be \(e\), so class D is \(2e\). The two remaining classes account for the rest of the \(200\) loaf orders: \[ D+E = 200-140 = 60 \] \[ 2e+e = 60 \] giving \[ e=20 \] so class E \(=20\) and class D \(=2\times20=40\) [3 marks]
- Class E spans \(350\lt w\leq450\), a width of \(100\): \[ \text{frequency density} = 20\div100 = 0.2 \] [1 mark]
Multiplying frequency density by width rather than reading a bar's height directly is essential whenever classes have unequal widths, as they do here; two bars can look similar in height on the page yet represent very different numbers of loaves if their widths differ.
Download The App On Google Playstore
Everything you need to excel in your exams
