#### Question

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

Marks | 5 | 6 | 7 | 8 | 9 |

Number of Students | 6 | a | 16 | 13 | b |

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

#### Solution 1

Marks (x) |
Number of students (f) |
fx |

5 | 6 | 30 |

6 | a | 6a |

7 | 16 | 112 |

8 | 13 | 104 |

9 | b | 9b |

`sumf = 34 + a + b` | `sumfx = 246 + 6a + 9b` |

It is given that the number of students is 40.

∴ 35 + a + b = 40

⇒ a + b – 5 = 0 ....(1)

`Mean = (sum fx)/(sum f)`

`=> (246 + 6a + 9b)/(35 + a + b) = 7.2`

`=> 246 + 6a + 9b = 7.2(35 + a + b)`

`=> 246 + 6a + 9b = 252 + 7.2a + 7.2b`

`=> 0 = 252 - 246 + 7.2a - 6a + 7.2b - 9b`

`=> 6 + 1.2a - 1.8b = 0`

`=> 10 + 2a - 3b = 0` .... (2)

Solving equations (1) and (2), we have

5a - 5 = 0

`=> a = 1`

From (1), we have b = 4

Hence, the values of a and b are 1 and 4 respectively.

#### Solution 2

Mean=`(∑ fx)/(∑f)`

⇒`7.2=(6xx5+axx6+16xx7+13xx8+bxx9)/(6+a+16+13+b)`

⇒`7.2=(246+6a+9b)/(35+a+b)`

⇒`1.2-1.8b=-6`..............(i)

Total number of students= `6+a+16+13+b `

⇒`40=35+a+b`

⇒`a+b=5`..............(ii)

Multiple equation (ii) by 1.8 and add it to equation (i)

1.8a + 1.8b = 9

1.2a - 1.8b = -6

3a = 3

⇒ a = 1

Substituting = 1 in equation (ii) we get,

1+b =5

⇒ b = 4