GSEB Class 10 Maths Solutions Chapter 14 Statistics Exercise 14.2

 

Question 1. The following table shows the ages of the patients admitted in a hospital for a year. Find the mode and the mean of the data given above. Compare and interpret the two measures of central tendency.

Answer:
Mode: Here, the highest frequency is 23, and the class corresponding to this frequency is 35-45. So, the modal class is 35-45.
Now, size \( (h) = 10 \), lower limit \( (l) \) of modal class \( = 35 \), frequency \( (f_1) \) of the modal class \( = 23 \), frequency \( (f_0) \) of class preceding the modal class \( = 21 \), frequency \( (f_2) \) of class succeeding the modal class \( = 14 \).
\( \therefore \text{Mode} = l + \frac{f_1-f_0}{2f_1-f_0-f_2} \times h \)
\( = 35 + \frac{23-21}{2 \times 23-21-14} \times 10 \)
\( = 35 + \frac{2}{46-35} \times 10 \)
\( = 35 + \frac{2}{11} \times 10 \)
\( = 35 + \frac{20}{11} \)
\( = 35 + 1.8 \) (approx.)
\( = 36.8 \) (approx.)
Using the step-deviation method,
\( \bar{x} = a + \left(\frac{\Sigma f_i u_i}{\Sigma f_i}\right) \times h \)
\( = 40 + \left(\frac{-37}{80}\right) \times 10 \)
\( = 40 - \frac{37}{8} \)
\( = 40 - 4.63 \)
\( = 35.37 \) years
Interpretation: The highest number of patients admitted to the hospital are approximately 36.8 years old, while on average, the age of a patient admitted to the hospital is about 35.37 years.
In simple words: The mode, which is about 36.8 years, tells us the age group that has the most patients. The mean, about 35.37 years, shows the average age of all patients. Both measures are quite close, suggesting the data is fairly spread around this central age.

Exam Tip: Remember to clearly state the modal class before calculating the mode and explain the meaning of both mode and mean in context for full marks.

 

Question 2. The following data gives information on the observed lifetimes (in hours) of 225 electrical components. Determine the modal lifetimes of the components.

Answer:
Here, the highest class frequency is 61, and the class corresponding to this frequency is 60-80. So, the modal class is 60-80.
Therefore, \( h = 20, l = 60, f_1 = 61, f_0 = 52, f_2 = 38 \).
\( \text{Mode} = l + \frac{f_1-f_0}{2f_1-f_0-f_2} \times h \)
\( = 60 + \frac{61-52}{2 \times 61-52-38} \times 20 \)
\( = 60 + \frac{9}{122-90} \times 20 \)
\( = 60 + \frac{9}{32} \times 20 \)
\( = 60 + \frac{180}{32} \)
\( = 60 + \frac{45}{8} \)
\( = 60 + 5.625 \)
\( = 65.625 \) hours
Therefore, the modal lifetimes of the component are 65.625 hours.
In simple words: The mode tells us the most common lifetime for these electrical parts. In this case, 65.625 hours is the lifetime that occurs most often among the components.

Exam Tip: Always identify the modal class correctly before calculating the mode. This involves finding the class interval with the highest frequency.

 

Question 3. The following date gives the distribution of total monthly expenditure of 200 families of a village. Find the modal monthly expenditure of the families. Also, find the mean monthly expenditure.

Answer:
Mode: Since the maximum number of families have their total monthly expenditure (in Rs.) in the interval 1500-2000, the modal class is 1500-2000.
Therefore, \( l = 1500, h = 500, f_1 = 40, f_0 = 24, f_2 = 33 \).
\( \text{Mode} = l + \frac{f_1-f_0}{2f_1-f_0-f_2} \times h \)
\( = 1500 + \frac{40-24}{2 \times 40-24-33} \times 500 \)
\( = 1500 + \frac{16}{80-57} \times 500 \)
\( = 1500 + \frac{16 \times 500}{23} \)
\( = 1500 + \frac{8000}{23} \)
\( = 1500 + 347.83 \)
\( = 1847.83 \)
Therefore, the modal monthly expenditure of the families is Rs. 1841.83.

Mean: Take \( a = 3250, h = 500 \)
Using the step deviation method,
\( \bar{x} = a + \left(\frac{\Sigma f_i u_i}{\Sigma f_i}\right) \times h \)
\( = 3250 + \left(\frac{-235}{200}\right) \times 500 \)
\( = 3250 - \frac{235 \times 5}{2} \)
\( = 3250 - \frac{1175}{2} \)
\( = 3250 - 587.50 \)
\( = 2662.50 \)
Hence, the mean monthly expenditure is Rs. 2662.50.
In simple words: The modal expenditure of Rs. 1841.83 means this is the most common monthly spending range for families. The mean expenditure of Rs. 2662.50 indicates the average monthly spending across all families in the village.

Exam Tip: Always show both the formula and the step-by-step calculations for mode and mean. Clearly identify the modal class and the assumed mean for the step deviation method.

 

Question 4. The following distribution gives the state-wise teacher-student ratio in higher secondary schools of India. Find the mode and mean of this data. Interpret, the two measures

Answer:
Mode: Since the maximum number of states/UT have the number of students per teacher in the interval 30-35, the modal class is 30-35.
Therefore, \( l = 30, h = 5, f_1 = 10, f_0 = 9, f_2 = 3 \).
\( \text{Mode} = l + \left(\frac{f_1-f_0}{2 f_1-f_0-f_2}\right) \times h \)
\( = 30 + \left(\frac{10-9}{2 \times 10-9-3}\right) \times 5 \)
\( = 30 + \frac{1}{20-12} \times 5 \)
\( = 30 + \frac{1}{8} \times 5 \)
\( = 30 + \frac{5}{8} \)
\( = 30 + 0.6 \)
\( = 30.6 \)
Hence, the mode of the given data is 30.6.

Mean: Take \( a = 37.5 \) and \( h = 5 \).
Using the step deviation method,
\( \bar{x} = a + \left(\frac{\Sigma f_i u_i}{\Sigma f_i}\right) \times h \)
\( = 37.5 + \left(\frac{-58}{35}\right) \times 5 \)
\( = 37.5 - \frac{58 \times 5}{35} \)
\( = 37.5 - \frac{58}{7} \)
\( = 37.5 - 8.2857 \)
\( \approx 29.21 \)
Interpretation: Most states/UT have a student teacher ratio of 30.6, and on average, this ratio is 29.2.
In simple words: The mode (30.6) shows that the most common teacher-student ratio is about 30 students per teacher. The mean (29.2) indicates that, on average, there are roughly 29 students for every teacher. These figures are quite close, giving a good picture of the typical student-teacher balance.

Exam Tip: Pay careful attention to the class intervals and corresponding frequencies when identifying the modal class and calculating the mean. Small errors in setup can lead to incorrect results.

 

Question 5. Find the mode of the data. (CBSE 2016)

Answer:
Since the maximum number of batsmen have their runs scored in the interval 4000-5000, the modal class is 4000-5000.
Therefore, \( l = 4000, h = 1000, f_1 = 18, f_0 = 4, f_2 = 9 \).
\( \therefore \text{Mode} = l + \frac{f_1-f_0}{2f_1-f_0-f_2} \times h \)
\( = 4000 + \frac{18-4}{2 \times 18-4-9} \times 1000 \)
\( = 4000 + \frac{14}{36-13} \times 1000 \)
\( = 4000 + \frac{14}{23} \times 1000 \)
\( = 4000 + \frac{14000}{23} \)
\( = 4000 + 608.69 \)
\( = 4608.69 \)
Hence, the mode of the data is 4608.7 runs.
In simple words: The mode of 4608.7 runs tells us that the most common number of runs scored by batsmen in this dataset is approximately 4608.7. This means more batsmen scored around this value than any other.

Exam Tip: Remember to use the correct class interval values for \( l, h, f_0, f_1, \) and \( f_2 \) according to the maximum frequency, as a small mistake can alter the final mode value significantly.

 

Question 6. A student noted in 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:

Answer:
Here, the maximum frequency is 20, and the class corresponding to this frequency is 40-50. So, the modal class is 40-50.
Therefore, \( l = 40, h = 10, f_1 = 20, f_0 = 12, f_2 = 11 \).
\( \therefore \text{Mode} = l + \frac{f_1-f_0}{2f_1-f_0-f_2} \times h \)
\( = 40 + \frac{20-12}{2 \times 20-12-11} \times 10 \)
\( = 40 + \frac{8}{40-23} \times 10 \)
\( = 40 + \frac{8}{17} \times 10 \)
\( = 40 + \frac{80}{17} \)
\( = 40 + 4.7 \)
\( = 44.7 \) cars
Hence, the mode of the data is 44.7 cars.
In simple words: The mode is 44.7 cars, meaning that the most frequent number of cars passing by the spot during these periods was around 44.7. This shows the busiest car traffic range.

Exam Tip: Double-check your values for \( f_0, f_1, \) and \( f_2 \) to ensure they correspond to the frequency of the preceding, modal, and succeeding classes respectively, to avoid calculation errors.