CBSE Class 10 Maths HOTs Statistics Set 06

Very Short Answer Type Questions 

Question. Find the mode of the given data:

Answer: 52

Question. Write the frequency distribution table for the following data :

Answer:

Question. Find the mode of the following data.

Answer: 34.7

Question. The mean of the following data is 7.5. Find the value of \( p \).

Answer: \( p = 3 \)

Question. A survey conducted on 20 households in a locality by a group of students resulted in the following frequency table for the number of family members in a household.

Find the mode for the data above :
Answer: \( \text{Mode} = \frac{23}{7} \)

Question. Find the mean of the following frequency distribution :

Answer: \( \text{Mean} = 25.2 \)

Question. Convert the following data to a less than type distribution.

Answer:

Question. Find mean.

Answer: \( \text{Mean} = 4.2 \)

Question. Write the following distribution as more than type cumulative frequency distribution:

Answer:

Question. Find the mode of the following data :

Answer: \( \text{Mode} = 26.25 \)

Question. Write a frequency distribution table for the following data :

Answer:

Question. Write the following distribution as less than type cumulative frequency distribution:

Answer:

Question. Find the median of the following data :

Answer: \( \text{Median} = 46.67 \)

Question. Find the mode of the following distribution of marks obtained by 60 students.

Answer: \( \text{Mode} = 35.88 \)

Short Answer Type Questions 

Question. During the medical check up of 35 students of a class, their weights were recorded as follows. Draw a less than type ogive for the given data. Hence obtain Median weight from the graph.

Answer: \( \text{Median} = 46.5 \)

Question. Find unknown entries \( a,b,c,d,e,f \) in the following distribution of heights of students in a class and the total number of students in the class in 50.

Answer: \( a = 12, b = 13, c = 35, d = 8, e = 5, f = 50 \)

Question. Find the mode of the following frequency distribution :

Answer: \( \text{Mode} = 45 \)

Question. The mean of the following frequency distribution is 54. Find the value of \( p \) :

Answer: \( p = 11 \)

Question. Find the median for the following frequency distribution :

Answer: \( \text{Median} = 40.6 \)

Long Answer Type Question

Question. Following distribution shows the marks obtained by the class of 100 students.

Draw less than ogive for the above data. Find median graphically and verify the result by actual method.
Answer: 38.33

SECTION - E : VALUE BASED QUESTIONS

Question. Atul donate Rs. 1000, per month to a cow shelter, Rs. 2000 per month to blind school, Rs. 3000 per month to a charitable Hospital and Rs. 4000 per month to a welfare society and remaining for his own purpose. Find average of his donations. Values observed.
Answer: \( \text{Average} = \frac{1000 + 2000 + 3000 + 4000}{4} = \text{Rs. } 2500. \)

Values

• Help society for better tomorrow.
• Keep helping regularly.
• Understand the needs of needy ones.
• For growth of nation.

Question. To help Delhi Police in a road safety week, 50 boys from Govt. school, 60 girls from public school, 20 professors from universities and 30 doctors from Private Hospitals were selected. Find average participation from one institution. Write value this problem indicates.
Answer: \( \text{Average} = \frac{50 + 60 + 20 + 30}{4} = 40. \)

Values

• It increases relations and participations of public with police.
• People from different branches will understand the problem of working of police department.
• It will also show that how people from different departments helps the police.
• With this type of activities people will be trained and crime can be reduced.

Question. A family need 10 eggs cost Rs. 5 each, 2 litres of milk at the rate Rs. 40 per litre, one pack of bread cost Rs. 20 for their breakfast. Another family has same number of family members in first one need 5 burgers cost Rs. 20 each two plates of noodles Rs. 60 each and 2 packs of preserved juice each Rs. 40. Find total expenses of two family separately. Find values.
Answer: Expenses of first family = 50 + 80 + 20 = Rs. 150
Expenses of second family = 100 + 120 + 80 = Rs. 300

Values

• We should eat healthy food for keeping health good.
• We should avoid cooked food from out side as per basis of hygiene.
• We should be in habit of saving money for better tomorrow.
• Junk food is not so good for our health.

Question. A students answered 20 questions correctly out of 30 questions given in question paper with honesty and scored 20 marks in exam and scored 15 marks out of 20 in interview. Another student B answered 10 questions of his own and 15 question by cheating and scored 5 marks in interview. Why student A is selected? What is use of values?
Answer: Student A is selected because of his ability and fare selection of interview committee.

Values

• Using unfair means can never lead to success.
• Honesty, sincerity and hard work always pay in life.
• God help those who works with honesty.
• By cheating we can not lead to a long race.

Question. Name the central tendency which is obtained by point of intersection of more than ogive and less than ogive.
Answer: Median

Question. Write empirical formula.
Answer: \( \text{Mode} = 3 \text{ Median} - 2 \text{ Mean} \)

Question. Find arithmetic mean of first \( n \) natural numbers.
Answer: Sum of first \( n \) natural numbers \( = \frac{n(n+1)}{2} \)
Mean \( = \frac{\text{Sum}}{\text{Number of observations}} \)

\( \implies \) Mean \( = \frac{n(n+1)}{2n} \)
\( \implies \) Mean \( = \frac{n+1}{2} \)

Question. Explain how you can find class size ?
Answer: Class size is calculated by finding the difference between the upper limit and the lower limit of a class interval.
Class size \( = \text{Upper limit} - \text{Lower limit} \)

Question. If each of the observation of some distribution is multiplied by 5 then what will be the change in its mean?
Answer: If each observation is multiplied by 5, the mean of the distribution will also be multiplied by 5.

Question. If median of 1, 2, 3, \( x + 1 \), \( x + 2 \), 6, 7, 8 is 4.5 then find mean of data.
Answer: The data is already in ascending order: 1, 2, 3, \( x + 1 \), \( x + 2 \), 6, 7, 8. The number of observations \( n = 8 \).
Median \( = \frac{\text{Value of } (\frac{n}{2})^{th} \text{ observation} + \text{Value of } (\frac{n}{2} + 1)^{th} \text{ observation}}{2} \)
\( 4.5 = \frac{(x + 1) + (x + 2)}{2} \)

\( \implies 4.5 = \frac{2x + 3}{2} \)

\( \implies 9 = 2x + 3 \)

\( \implies 2x = 6 \)

\( \implies x = 3 \)
The observations are 1, 2, 3, 4, 5, 6, 7, 8.
Mean \( = \frac{1+2+3+4+5+6+7+8}{8} = \frac{36}{8} \)

\( \implies \) Mean \( = 4.5 \)

Question. Find mode of 3, 7, 7, 5, 2, 7, 9, 9, 9.
Answer: The frequencies of the observations are:
2: 1 time
3: 1 time
5: 1 time
7: 3 times
9: 3 times
Since 7 and 9 both have the highest frequency (3), the data is bimodal.

\( \implies \) Mode \( = 7 \) and \( 9 \)

Question. Find mean of distribution as under:

Answer:

Mean \( = \frac{\sum f_i x_i}{\sum f_i} = \frac{3900}{50} = 78 \)

Question. Mean of 25 observation is 60. If mean of first 13 observations is 70 and mean of last 13 observations is 50. Find Middle observations when arranged in ascending order.
Answer: Total observations \( = 25 \)
Sum of 25 observations \( = 25 \times 60 = 1500 \)
Sum of first 13 observations \( = 13 \times 70 = 910 \)
Sum of last 13 observations \( = 13 \times 50 = 650 \)
Middle observation (13th observation) \( = (\text{Sum of first 13}) + (\text{Sum of last 13}) - (\text{Sum of 25}) \)

\( \implies \) Middle observation \( = 910 + 650 - 1500 \)

\( \implies \) Middle observation \( = 1560 - 1500 = 60 \)

Question. Find mode for given distribution :

Answer: The class with the maximum frequency (61) is 60-80. This is the modal class.
\( l = 60, f_1 = 61, f_0 = 52, f_2 = 38, h = 20 \)
Mode \( = l + \left( \frac{f_1 - f_0}{2f_1 - f_0 - f_2} \right) \times h \)

\( \implies \) Mode \( = 60 + \left( \frac{61 - 52}{2(61) - 52 - 38} \right) \times 20 \)

\( \implies \) Mode \( = 60 + \left( \frac{9}{122 - 90} \right) \times 20 \)

\( \implies \) Mode \( = 60 + \left( \frac{9}{32} \right) \times 20 \)

\( \implies \) Mode \( = 60 + \frac{45}{8} \)

\( \implies \) Mode \( = 60 + 5.625 = 65.625 \)

Question. Draw a less than ogive for given distributions :

Answer: Construct the cumulative frequency table:

Plot the points (10, 7), (20, 16), (30, 20), (40, 23), and (50, 24) on a graph and join them with a smooth curve.

Question. Draw a more than ogive for give data :

Answer: The data is already given in "more than" cumulative frequency form.
Plot the points (0, 100), (20, 50), (40, 30), (60, 10), and (80, 2) on a graph where the lower limits are on the x-axis and the cumulative frequencies are on the y-axis. Join these points with a smooth curve.

Question. Find mean of given distribution by using step deviation method :

Answer: Let Assumed Mean \( A = 225 \) and class size \( h = 50 \).

Mean \( = A + h \times \left( \frac{\sum f_i u_i}{\sum f_i} \right) \)

\( \implies \text{Mean} = 225 + 50 \times \left( \frac{-7}{25} \right) \)

\( \implies \text{Mean} = 225 + 2 \times (-7) \)

\( \implies \text{Mean} = 225 - 14 = 211 \)

Question. Find \( f_1 \) and \( f_2 \) if mean of given distribution in 50.

Answer: Total frequency \( \sum f_i = 120 \).
\( 17 + f_1 + 32 + f_2 + 19 = 120 \)

\( \implies f_1 + f_2 = 120 - 68 \)

\( \implies f_1 + f_2 = 52 \quad \text{---(i)} \)
Mean \( = 50 \). Using direct method table:

Mean \( = \frac{\sum f_i x_i}{\sum f_i} \)
\( 50 = \frac{3480 + 30f_1 + 70f_2}{120} \)

\( \implies 6000 = 3480 + 30f_1 + 70f_2 \)

\( \implies 30f_1 + 70f_2 = 2520 \)

\( \implies 3f_1 + 7f_2 = 252 \quad \text{---(ii)} \)
Multiply (i) by 3: \( 3f_1 + 3f_2 = 156 \quad \text{---(iii)} \)
Subtract (iii) from (ii):
\( 4f_2 = 96 \)

\( \implies f_2 = 24 \)
Substitute \( f_2 = 24 \) in (i):
\( f_1 + 24 = 52 \)

\( \implies f_1 = 28 \)