Skip to content
doubtly

The following frequency distribution table shows the classification of the number of vehicles and the volume of petrol filled in them. To find the mode of the volume of petrol filled, complete the following activity :Class (Petrol filled in Liters)Fr…

Q.3 (A)(i) The following frequency distribution table shows the classification of the number of vehicles and the volume of petrol filled in them. To find the mode of the volume of petrol filled, complete the following activity :

Class (Petrol filled in Liters)Frequency (Number of Vehicles)
0.5–3.533
3.5–6.540
6.5–9.527
9.5–12.518
12.5–15.512

Activity :

From the given table,

Modal class = ____

\(\therefore\) Mode = ____ \(+ \left[ \dfrac{f_1 - f_0}{2f_1 - f_0 -}\) ____ \(\right] \times h\)

\(\therefore\) Mode = 3.5 \(+ \left[ \dfrac{40 - 33}{2(40) - 33 - 27} \right] \times\) ____

\(\therefore\) Mode = 3.5 \(+ \left[ \dfrac{7}{80 - 60} \right] \times 3\)

\(\therefore\) Mode = ____

\(\therefore\) The mode of the volume of petrol filled is ____.

1 answer

Best answer

Answer: Modal class 3.5–6.5; L = 3.5, f2 = 27, h = 3; Mode = 4.55, so the mode of the volume of petrol filled is 4.55 litres.

The greatest frequency is 40, so the modal class is 3.5–6.5.

\(L = 3.5\), \(f_1 = 40\), \(f_0 = 33\), \(f_2 = 27\), \(h = 3\).

Mode \(= L + \left[ \dfrac{f_1 - f_0}{2f_1 - f_0 - f_2} \right] \times h = 3.5 + \left[ \dfrac{7}{80 - 60} \right] \times 3\)

\(= 3.5 + \dfrac{7}{20} \times 3 = 3.5 + 1.05 = 4.55\).

Blanks: modal class \(= 3.5 - 6.5\); \(L = 3.5\); the third term in the denominator is \(f_2 = 27\); \(h = 3\); Mode \(= 4.55\).

Doubtly AI · 25 Oct 2026 AI-verified · Oct 2026 Report
Was this answer helpful?

Your answer

Log in or create a free account to answer this question.