NCERT Exemplar Solutions
Class 10 - Mathematics - CHAPTER 13: Statistics and Probability - Exercise 13.3
Question 6

Question. 6

6. An aircraft has 120 seats. Over 100 flights, the seats occupied were:

Seats	100–104	104–108	108–112	112–116	116–120
Frequency	15	20	32	18	15

Determine the mean number of seats occupied.

109.92 seats

Step 1: Identify the data.

We are given the number of seats occupied in groups (called classes) and how many times they occurred (called frequency).

Class Interval (Seats)	100–104	104–108	108–112	112–116	116–120
Frequency (f)	15	20	32	18	15

Step 2: Find the mid-point (class mark) of each interval.

The class mark is the average of the lower and upper limits:

For 100–104: \( \tfrac{100+104}{2} = 102 \)
For 104–108: \( \tfrac{104+108}{2} = 106 \)
For 108–112: \( \tfrac{108+112}{2} = 110 \)
For 112–116: \( \tfrac{112+116}{2} = 114 \)
For 116–120: \( \tfrac{116+120}{2} = 118 \)

Step 3: Multiply frequency by class mark.

We calculate \( f \times x \):

Step 4: Add the values.

Step 5: Apply the mean formula.

The mean is given by:

\[ \bar{x} = \dfrac{\sum f x}{\sum f} \]

Substituting the values:

\[ \bar{x} = \dfrac{10992}{100} = 109.92 \]

Final Answer: The mean number of seats occupied = 109.92 seats.