GSEB Class 10 Maths Solutions Chapter 14 Statistics Exercise 14.3

 

Question 1. The following frequency distribution gives the monthly consumption of electricity of 68 consumers of a locality. Find the median, mean and mode of the data and compare them.

Answer:
**Median:**
First, we make a cumulative frequency table for the given data.
Here, the total number of consumers is \( n = 68 \).
So, \( \frac { n }{ 2 } = \frac { 68 }{ 2 } = 34 \).
The cumulative frequency just greater than 34 is 42, and the class interval that goes with it is 125-145.
Therefore, the median class is 125-145.
We know the median formula is:
Median \( = l + \left( \frac { \frac { n }{ 2 } - c.f. }{ f } \right) \times h \)
Here,
Lower limit of median class, \( l = 125 \)
Frequency of median class, \( f = 20 \)
Cumulative frequency of class before median class, \( c.f. = 22 \)
Class size, \( h = 20 \)
Substituting these values, we get:
Median \( = 125 + \left( \frac { 34 - 22 }{ 20 } \right) \times 20 \)
\( = 125 + \frac { 12 }{ 20 } \times 20 \)
\( = 125 + 12 \)
\( = 137 \) units

**Mean:**
We will use the Step-Deviation method to determine the mean.
Here, assumed mean \( a = 135 \) and class size \( h = 20 \).
The formula for mean is:
Mean \( \bar { x } = a + \left( \frac { \sum f_i u_i }{ \sum f_i } \right) \times h \)
\( = 135 + \left( \frac { 7 }{ 68 } \right) \times 20 \)
\( = 135 + \frac { 140 }{ 68 } \)
\( = 135 + 2.0588 \)
\( = 137.06 \) units (approximately)

**Mode:**
The class with the highest frequency is the modal class. From the table, the maximum frequency is 20, which matches the class 125-145.
So, the modal class is 125-145.
The formula for mode is:
Mode \( = l + \left( \frac { f_1 - f_0 }{ 2 f_1 - f_0 - f_2 } \right) \times h \)
Here,
Lower limit of modal class, \( l = 125 \)
Frequency of modal class, \( f_1 = 20 \)
Frequency of class before modal class, \( f_0 = 13 \)
Frequency of class after modal class, \( f_2 = 14 \)
Class size, \( h = 20 \)
Substituting these values, we get:
Mode \( = 125 + \left( \frac { 20 - 13 }{ 2 \times 20 - 13 - 14 } \right) \times 20 \)
\( = 125 + \left( \frac { 7 }{ 40 - 27 } \right) \times 20 \)
\( = 125 + \frac { 7 }{ 13 } \times 20 \)
\( = 125 + \frac { 140 }{ 13 } \)
\( = 125 + 10.769 \)
\( = 135.77 \) units (approximately)

**Comparison:**
Median \( = 137 \) units
Mean \( = 137.06 \) units
Mode \( = 135.77 \) units
All three measures of central tendency are quite close to each other.
In simple words: We first calculated the median, mean, and mode for the electricity consumption data. The median was 137 units, the mean was about 137.06 units, and the mode was about 135.77 units. These three values are all very similar, which shows that the data is fairly balanced.

Exam Tip: Remember to clearly show all steps for calculating median, mean, and mode, including the formulas used and the identification of the relevant class interval. Pay attention to decimal places for accurate comparison.

 

Question 2. If the median of the distribution given below is 28.5, find the values of x and y. \( \sum f_i = 60 \).

Answer:
Let's first create a cumulative frequency table from the given data.
From the table, we know that the total frequency is \( n = \sum f_i = 60 \).
Also, the last cumulative frequency should match the total frequency, so:
\( 45 + x + y = 60 \)
\( x + y = 60 - 45 \)

\( \implies x + y = 15 \) ... (Equation 1)

We are given that the median of the distribution is 28.5.
Since 28.5 falls within the class interval 20-30, the median class is 20-30.
Now, we can use the median formula:
Median \( = l + \left( \frac { \frac { n }{ 2 } - c.f. }{ f } \right) \times h \)
Here,
Lower limit of median class, \( l = 20 \)
Total frequency, \( n = 60 \)
So, \( \frac { n }{ 2 } = \frac { 60 }{ 2 } = 30 \)
Cumulative frequency of class before median class, \( c.f. = 5 + x \)
Frequency of median class, \( f = 20 \)
Class size, \( h = 10 \)
Substitute these values into the median formula:
\( 28.5 = 20 + \left( \frac { 30 - (5 + x) }{ 20 } \right) \times 10 \)
\( 28.5 - 20 = \left( \frac { 30 - 5 - x }{ 20 } \right) \times 10 \)
\( 8.5 = \left( \frac { 25 - x }{ 20 } \right) \times 10 \)
\( 8.5 = \frac { 25 - x }{ 2 } \)
\( 8.5 \times 2 = 25 - x \)
\( 17 = 25 - x \)
\( x = 25 - 17 \)

\( \implies x = 8 \)

Now, substitute the value of \( x = 8 \) into Equation 1:
\( 8 + y = 15 \)
\( y = 15 - 8 \)

\( \implies y = 7 \)

Therefore, the values are \( x = 8 \) and \( y = 7 \).
In simple words: We were given a data table with some missing frequencies (x and y) and the median value. First, we wrote down the total frequency using the last cumulative frequency value. Then, by using the median formula and the given median, we found the value of x. After that, we put x back into the equation for the total frequency to get the value of y. We found that x is 8 and y is 7.

Exam Tip: When solving for missing frequencies (like x and y), always form two equations: one from the total frequency and another from the median formula. This approach helps in solving such problems systematically.

 

Question 3. A life insurance agent found the following data for the distribution of ages of 100 policyholders. Calculate the median age, if polices are only given to persons having age 18 years onwards but less than 60 years.

Answer:
First, we need to convert the given "less than" cumulative frequency distribution into a normal frequency distribution with class intervals.
Since policies are issued to individuals aged 18 years onwards, the first class interval will be 15-20, not 0-20 (assuming intervals of 5 years as seen in the data 20, 25, 30...).
Here, the total number of policyholders is \( n = 100 \).
So, \( \frac { n }{ 2 } = \frac { 100 }{ 2 } = 50 \).
The cumulative frequency just greater than 50 is 78, and the class interval associated with it is 35-40.
Therefore, the median class is 35-40.
Now we use the median formula:
Median \( = l + \left( \frac { \frac { n }{ 2 } - c.f. }{ f } \right) \times h \)
Here,
Lower limit of median class, \( l = 35 \)
Frequency of median class, \( f = 33 \)
Cumulative frequency of class before median class, \( c.f. = 45 \)
Class size, \( h = 5 \)
Substitute these values into the median formula:
Median \( = 35 + \left( \frac { 50 - 45 }{ 33 } \right) \times 5 \)
\( = 35 + \left( \frac { 5 }{ 33 } \right) \times 5 \)
\( = 35 + \frac { 25 }{ 33 } \)
\( = 35 + 0.7575 \)
\( = 35.76 \) years (approximately)
In simple words: First, we changed the 'below age' data into clear age groups and then found how many people were in each group. We used this to make a table with cumulative frequencies. Next, we found the middle value (n/2) to identify the median age group, which was 35-40 years. Finally, using the median formula with the values from this group, we determined the median age to be about 35.76 years for all the policyholders.

Exam Tip: When dealing with "less than" or "more than" type distributions, always convert them into continuous class intervals before applying median, mean, or mode formulas. Be careful with calculating the frequencies for each interval.

 

Question 4. The lengths of 40 leaves of a plant are measured correct to the nearest millimeter, and the data obtained is represented in the following table:

Find the median length of the leaves.
Answer:
First, the given data is in discontinuous class intervals. We need to change it into continuous classes by adjusting the limits.
The adjustment factor is \( \frac { \text{lower limit of next class} - \text{upper limit of current class} }{ 2 } \).
For example, for 118-126 and 127-135, the adjustment is \( \frac { 127 - 126 }{ 2 } = \frac { 1 }{ 2 } = 0.5 \).
Subtract 0.5 from lower limits and add 0.5 to upper limits.
Here, the total number of leaves is \( n = 40 \).
So, \( \frac { n }{ 2 } = \frac { 40 }{ 2 } = 20 \).
The cumulative frequency just greater than 20 is 29, and the class interval linked to it is 144.5-153.5.
Therefore, the median class is 144.5-153.5.
Now, we use the median formula:
Median \( = l + \left( \frac { \frac { n }{ 2 } - c.f. }{ f } \right) \times h \)
Here,
Lower limit of median class, \( l = 144.5 \)
Frequency of median class, \( f = 12 \)
Cumulative frequency of class before median class, \( c.f. = 17 \)
Class size, \( h = 9 \) (153.5 - 144.5)
Substitute these values into the median formula:
Median \( = 144.5 + \left( \frac { 20 - 17 }{ 12 } \right) \times 9 \)
\( = 144.5 + \left( \frac { 3 }{ 12 } \right) \times 9 \)
\( = 144.5 + \frac { 27 }{ 12 } \)
\( = 144.5 + 2.25 \)
\( = 146.75 \) mm
In simple words: We changed the given data into a continuous form first. Then, we made a table to add up the frequencies. We found the middle value of all the leaves (n/2) and used it to find the median class. Finally, we put all the numbers into the median formula, which gave us the median length of the leaves as 146.75 mm.

Exam Tip: Always remember to convert discontinuous class intervals into continuous ones before calculating median or mean. A common mistake is to overlook this step, leading to incorrect results.

 

Question 5. The following table gives the distribution of the lifetime of 400 neon lamps.

Find the median lifetime of a lamp.
Answer:
First, we construct a cumulative frequency table for the provided data.
Here, the total number of lamps is \( n = 400 \).
So, \( \frac { n }{ 2 } = \frac { 400 }{ 2 } = 200 \).
The cumulative frequency just greater than 200 is 216, and the class interval corresponding to it is 3000-3500.
Therefore, the median class is 3000-3500.
Now, we use the median formula:
Median \( = l + \left( \frac { \frac { n }{ 2 } - c.f. }{ f } \right) \times h \)
Here,
Lower limit of median class, \( l = 3000 \)
Frequency of median class, \( f = 86 \)
Cumulative frequency of class before median class, \( c.f. = 130 \)
Class size, \( h = 500 \)
Substitute these values into the median formula:
Median \( = 3000 + \left( \frac { 200 - 130 }{ 86 } \right) \times 500 \)
\( = 3000 + \left( \frac { 70 }{ 86 } \right) \times 500 \)
\( = 3000 + \frac { 35000 }{ 86 } \)
\( = 3000 + 406.9767 \)
\( = 3406.98 \) hours (approximately)
In simple words: We started by building a table to show the total frequencies for different lamp lifetimes. Since there are 400 lamps, we found the middle point (200) to find the median class. This class was 3000-3500 hours. Then, we used the median formula with the numbers from this class to calculate that the median lifetime of a lamp is approximately 3406.98 hours.

Exam Tip: In problems involving large datasets, ensure precise calculation of cumulative frequencies to correctly identify the median class. Errors in cumulative frequency can lead to an incorrect median class and, consequently, an incorrect median.

 

Question 6. 100 surnames were randomly picked up from a local telephone directory and the frequency distribution of the number of letters in the English alphabets in the surnames was obtained as follows:

Determine the median number of letters in the surnames. Find the mean number of letters in the surnames. Also, find the modal size of the surnames.
Answer:
**Median:**
First, we make a cumulative frequency table.
Here, the total number of surnames is \( n = 100 \).
So, \( \frac { n }{ 2 } = \frac { 100 }{ 2 } = 50 \).
The cumulative frequency just greater than 50 is 76, and the class interval corresponding to it is 7-10.
Therefore, the median class is 7-10.
Now, we use the median formula:
Median \( = l + \left( \frac { \frac { n }{ 2 } - c.f. }{ f } \right) \times h \)
Here,
Lower limit of median class, \( l = 7 \)
Frequency of median class, \( f = 40 \)
Cumulative frequency of class before median class, \( c.f. = 36 \)
Class size, \( h = 3 \)
Substitute these values into the median formula:
Median \( = 7 + \left( \frac { 50 - 36 }{ 40 } \right) \times 3 \)
\( = 7 + \left( \frac { 14 }{ 40 } \right) \times 3 \)
\( = 7 + \frac { 42 }{ 40 } \)
\( = 7 + 1.05 \)
\( = 8.05 \)

**Mean:**
We will use the Step-Deviation method to determine the mean. Let \( a = 8.5 \) and \( h = 3 \).
The formula for mean is:
Mean \( \bar { x } = a + \left( \frac { \sum f_i u_i }{ \sum f_i } \right) \times h \)
\( = 8.5 + \left( \frac { -6 }{ 100 } \right) \times 3 \)
\( = 8.5 - \frac { 18 }{ 100 } \)
\( = 8.5 - 0.18 \)
\( = 8.32 \)

**Mode:**
The class with the highest frequency is the modal class. From the given table, the maximum frequency is 40, which falls in the class interval 7-10.
So, the modal class is 7-10.
The formula for mode is:
Mode \( = l + \left( \frac { f_1 - f_0 }{ 2 f_1 - f_0 - f_2 } \right) \times h \)
Here,
Lower limit of modal class, \( l = 7 \)
Frequency of modal class, \( f_1 = 40 \)
Frequency of class before modal class, \( f_0 = 30 \)
Frequency of class after modal class, \( f_2 = 16 \)
Class size, \( h = 3 \)
Substitute these values into the mode formula:
Mode \( = 7 + \left( \frac { 40 - 30 }{ 2 \times 40 - 30 - 16 } \right) \times 3 \)
\( = 7 + \left( \frac { 10 }{ 80 - 46 } \right) \times 3 \)
\( = 7 + \left( \frac { 10 }{ 34 } \right) \times 3 \)
\( = 7 + \frac { 30 }{ 34 } \)
\( = 7 + 0.8823 \)
\( = 7.88 \) (approximately)
In simple words: We calculated the median, mean, and mode for the number of letters in surnames. The median was 8.05 letters, found by summing frequencies and using the median formula. The mean was 8.32 letters, calculated with the step-deviation method. The mode, representing the most common length, was about 7.88 letters, obtained from the class with the highest frequency.

Exam Tip: Always double-check class intervals when finding mean, median, and mode for grouped data. Overlapping intervals (like 1-4, 4-7) should be treated carefully as they are often meant to be exclusive or have a different interpretation for limits. In this case, standard methods apply as if they were 1-3, 4-6, 7-9 etc. (Midpoints are taken as 2.5, 5.5, 8.5, etc.)

 

Question 7. The distribution below gives the weights of 30 students of a class. Find the median weight of the students.

Answer:
First, we calculate the cumulative frequency for the given distribution.
Here, the total number of students is \( n = 30 \).
So, \( \frac { n }{ 2 } = \frac { 30 }{ 2 } = 15 \).
The cumulative frequency just greater than 15 is 19, and the class interval connected to it is 55-60.
Therefore, the median class is 55-60.
Now, we use the median formula:
Median \( = l + \left( \frac { \frac { n }{ 2 } - c.f. }{ f } \right) \times h \)
Here,
Lower limit of median class, \( l = 55 \)
Frequency of median class, \( f = 6 \)
Cumulative frequency of class before median class, \( c.f. = 13 \)
Class size, \( h = 5 \)
Substitute these values into the median formula:
Median \( = 55 + \left( \frac { 15 - 13 }{ 6 } \right) \times 5 \)
\( = 55 + \left( \frac { 2 }{ 6 } \right) \times 5 \)
\( = 55 + \frac { 10 }{ 6 } \)
\( = 55 + \frac { 5 }{ 3 } \)
\( = 55 + 1.666... \)
\( = 56.67 \) kg (approximately)
In simple words: We first made a table that shows the total number of students up to each weight group. Then, we found the middle student number (n/2) to find the weight group that contains the median, which was 55-60 kg. Using the median formula with numbers from this group, we determined that the median weight of the students is about 56.67 kg.

Exam Tip: For continuous data like weights, ensure the median class is correctly identified using the cumulative frequency. A small mistake in calculating or reading the cumulative frequency can lead to picking the wrong median class, which affects the entire median calculation.