# Ex.14.2 Q6 Statistics Solution - NCERT Maths Class 10

## Question

A student noted the number of cars passing through a spot on a road for 100 periods each of 3 minutes and summarised it in the table given below. Find mode of the data.

Number of cars |
\(0 -10\) | \(10 - 20\) | \(20 - 30\) | \(30 - 40\) | \(40 - 50\) | \(50 - 60\) | \(60 - 70\) | \(70 - 80\) |

Frequency |
\(7\) | \(14\) | \(13\) | \(12\) | \(20\) | \(11\) | \(15\) | \(8\) |

## Text Solution

**What is known?**

The number of cars passing through a spot on a road for \(100\) periods each of \(3\) minutes

**What is unknown?**

The Mode of the data.

**Reasoning:**

Modal Class is the class with highest frequency.

Mode \( = l + \left( {\frac{{{f_1} - {f_0}}}{{2{f_1} - {f_0} - {f_2}}}} \right) \times h\)

Where,

Class size,\(h\)

Lower limit of modal class,\(l\)

Frequency of modal class,\(f_1\)

Frequency of class preceding modal class,\(f_0\)

Frequency of class succeeding the modal class,\(f_2\)

**Steps:**

From the table, it can be observed that the maximum class frequency is \(20,\) belonging to class interval \(40 − 50\)

Therefore, Modal class\(=40 − 50\)

Class size,\(h=10\)

Lower limit of modal class,\(l=40\)

Frequency of modal class,\(f_1=20\)

Frequency of class preceding modal class,\(f_0=12\)

Frequency of class succeeding the modal class,\(f_2=11\)

Mode \( = l + \left( {\frac{{{f_1} - {f_0}}}{{2{f_1} - {f_0} - {f_2}}}} \right) \times h\)

\[\begin{array}{l}

= 40 + \left( {\frac{{20 - 12}}{{2 \times 20 - 12 - 11}}} \right) \times 10\\

= 40 + \left( {\frac{8}{{40 - 23}}} \right) \times 10\\

= 40 + \frac{8}{{17}} \times 10\\

= 40 + 4.705\\

= 40.705\\

= 40.7

\end{array}\]

Hence the mode is \(40.7\)