# Ex.14.1 Q7 Statistics Solution - NCERT Maths Class 10

## Question

To find out the concentration of \(SO_2\) in the air (in parts per million, i.e., ppm), the data was collected for \(30\) localities in a certain city and is presented below:

Concentration of \(SO_2\) (in ppm) |
Frequency |

\(0.00 - 0.04\) | \(4\) |

\(0.04 - 0.08\) | \(9\) |

\(0.08 - 0.12\) | \(9\) |

\(0.12 - 0.16\) | \(2\) |

\(0.16 - 0.20\) | \(4\) |

\(0.20 - 0.24\) | \(2\) |

Find the mean concentration of \(SO_2\) in the air.

## Text Solution

**What is known?**

The Concentration of \(SO_2\) in the air (in parts per million, i.e., ppm) for \(30\) localities in a certain city:

**What is unknown?**

The mean concentration of \(SO_2 \)in the air

**Reasoning:**

We solve this question by step deviation method.

Mean, \(\overline x = a + \left( {\frac{{\sum {{f_i}{u_i}} }}{{\sum {{f_i}} }}} \right) \times h\)

**Steps:**

We know that,

Class mark,\({x_i} = \frac{{{\text{Upper class limit }} + {\text{ Lower class limit}}}}{2}\)

Class size\(, h = 0.04\)

Taking assumed mean\(,a = 0.14\)

Content of SO_{2} |
Frequency \((f_i)\) |
\( x_i\) | \( d_i = x_i -a \) | \(\begin{align}{u_i}=\frac{ d_i}{h}\end{align}\) | \( f_iu_i \) |

\(0.00-0.04\) | \(4\) | \(0.02\) | \(-0.12\) | \(-3\) | \(-12\) |

\(0.04-0.08\) | \(9\) | \(0.06\) | \(-0.08\) | \(-2\) | \(-18\) |

\(0.08-0.12\) | \(9\) | \(0.10\) | \(-0.04\) | \(-1\) | \(-9\) |

\(0.12-0.16\) | \(2\) | \(0.14(a)\) | \(0\) | \(0\) | \(0\) |

\(0.16-0.20\) | \(4\) | \(0.18\) | \(0.04\) | \(1\) | \(4\) |

\(0.20-0.24\) | \(2\) | \(0.22\) | \(0.08\) | \(2\) | \(4\) |

\(\Sigma f_i=30 \) | \(\Sigma f_iu_i=-31\) |

From the table, we obtain

\[\begin{array}{l}\sum {{f_i} = 30} \\\sum {{f_i}{u_i}} = - 31\end{array}\]

\[\begin{align} \text { Mean }(\overline{{x}}) &={a}+\left(\frac{\sum {f}_{{i}} {u}_{{i}}}{{f}_{{i}}}\right) {h} \\ \overline{{x}} &=0.14+\left(\frac{-31}{30}\right) 0.04 \\ \overline{{x}} &=0.14-| 0.04133 \\ \overline{{x}} &=0.099 \end{align}\]

The mean concentration of \(SO_2\) in the air is \(0.099\).