Ex.14.1 Q7 Statistics Solution - NCERT Maths Class 10

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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 SO2 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 \(0.099\).

  
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