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Mathematics

Marks obtained by 40 students in a short assessment is given below, where a and b are two missing data.

MarksNumber of students
56
6a
716
813
9b

If the mean of the distribution is 7.2, find a and b.

Measures of Central Tendency

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Answer

Marks (x)Number of students (f)fx
5630
6a6a
716112
813104
9b9b
TotalΣf = a + b + 35Σfx = 246 + 6a + 9b

Total 40 students.

∴ a + b + 35 = 40

⇒ a + b = 40 - 35

⇒ a + b = 5

⇒ a = 5 - b ……..(1)

Given, mean = 7.2

∴7.2=ΣfxΣf⇒7.2=246+6a+9b40\therefore 7.2 = \dfrac{Σfx}{Σf} \\[1em] \Rightarrow 7.2 = \dfrac{246 + 6a + 9b}{40}

Substituting value of a from (1) in above equation :

⇒7.2=246+6(5−b)+9b40⇒7.2×40=246+30−6b+9b⇒288=276+3b⇒3b=288−276⇒3b=12⇒b=123⇒b=4.\Rightarrow 7.2 = \dfrac{246 + 6(5 - b) + 9b}{40} \\[1em] \Rightarrow 7.2 \times 40 = 246 + 30 - 6b + 9b \\[1em] \Rightarrow 288 = 276 + 3b \\[1em] \Rightarrow 3b = 288 - 276 \\[1em] \Rightarrow 3b = 12 \\[1em] \Rightarrow b = \dfrac{12}{3} \\[1em] \Rightarrow b = 4.

a = 5 - b = 5 - 4 = 1.

Hence, a = 1 and b = 4.

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