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 the 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 8 188
Step-by-Step Solution
Key Concept: Mode of grouped (continuous) data is found using the formula: \(\text{Mode}=l+\frac{f_1-f_0}{2f_1-f_0-f_2}\times h\), where \(l\) is the lower limit of the modal class, \(f_1\) is its frequency, \(f_0\) and \(f_2\) are frequencies of the preceding and succeeding classes respectively, and \(h\) is the class width.
1. Identify the modal class
The class with the highest frequency is the modal class. From the table, the frequencies are: \(7, 14, 13, 12, 20, 11, 15, 8\). The maximum frequency is \(20\) which corresponds to the class \(40-50\). Hence, the modal class is \(40-50\).\
2. Write down the required quantities
- Lower limit of modal class, \(l = 40\)
- Class width, \(h = 10\) (since each class interval is of size 10)\
- Frequency of modal class, \(f_1 = 20\)\
- Frequency of preceding class, \(f_0 = 12\) (class \(30-40\))\
- Frequency of succeeding class, \(f_2 = 11\) (class \(50-60\))\
3. Apply the mode formula
\[\text{Mode}= l + \frac{f_1-f_0}{2f_1-f_0-f_2}\times h \]\
Substitute the values:
\[\text{Mode}= 40 + \frac{20-12}{2\times20-12-11}\times 10 \]\
\[\text{Mode}= 40 + \frac{8}{40-23}\times 10 \]\
\[\text{Mode}= 40 + \frac{8}{17}\times 10 \]\
\[\text{Mode}= 40 + \frac{80}{17} \]\
\[\text{Mode}= 40 + 4.7059 \approx 44.7 \]\
4. State the answer
The mode of the data (to one decimal place) is approximately 44.7 cars.
Correct Answer: 44.7 (approximately)