Ex.14.1 Q9 Statistics Solution - NCERT Maths Class 10

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Question

The following table gives the literacy rate (in percentage) of \(35\) cities. Find the mean literacy rate.

Literacy rate (in %) \(45 – 55\) \(55 – 65\) \(65 – 75\) \(75 – 85\) \(85 – 95\)
Number of cities \(3\) \(10\) \(11\) \(8\) \(3\)

Text Solution

 

What is known?

The Literacy rate (in percentage) of \(35\) cities

What is unknown?

The Mean literacy rate.

Reasoning:

We solve this question by assumed mean method. 

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

Steps:

We know that,

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

Taking assumed mean\(, a=70\)

Literacy rate No of cities\((f_i)\) \( X_i\) \( d_i = x_i -a \) \( f_id_i \)
\(45 – 55\) \(3\) \(50\) \(-20\) \(-60\)
\(55 – 65\) \(10\) \(60\) \(-10\) \(-100\)
\(65 – 75\) \(11\) \(70(a)\) \(0\) \(0\)
\(75 – 85\) \(8\) \(80\) \(10\) \(80\)
\(85 – 95\) \(3\) \(90\) \(20\) \(60\)
  \(\Sigma f_i=35 \)     \(\Sigma f_id_i=-20\)

From the table,we obtain

\[\begin{array}{l}
\sum {{f_i} = 35} \\
\sum {{f_i}{d_i}}  =  - 20
\end{array}\]

\[\begin{align} \text { Mean }(\overline{{x}}) &={a}+\left(\frac{\Sigma f_{i} d_{i}}{\Sigma {f}_{{i}}}\right) \\ \overline{{x}} &=70+\left(\frac{-20}{35}\right) \\ \overline{{x}} &=70-\frac{4}{7} \\ \overline{{x}} &=70-0.57 \\ \overline{{x}} &=69.43 \end{align}\]

Thus, the mean literacy rate is \(69.43\)%

  
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