CBSE Class 10 Mathematics Statistics Worksheet Set 04

Key Notes

Statistics

• Collection of Data
• Presentation of Data
• Graphical Representation of Data
• Measures of Central Tendency
• Statistics is the area of study that deals with the collection presentation, analysis and interpretation of data.
• Data: Facts or figures, collected with a definite purpose, are called data.
• There are two types of data (i) Primary (ii) Secondary
• We can represent the data by (i) ungrouped and grouped frequency distribution.
• Data can also represent by (i) bar graph (ii) Histogram (iii) Frequency polygons
• Class mark of grouped data is \( \frac{\text{lower limit + upper limit}}{2} \)
• Measure of central tendencies by mean, median, mode.
• Mean: \( (\bar{x}) = \frac{\text{sum of all observations}}{\text{Total no. of observations}} \)
• If observations denoted by \( x_i \) and their occurrence i.e. frequency is denoted by \( f_i \) then mean is \( (\bar{x}) = \frac{\sum f_i x_i}{\sum f_i} \)
• Median: Arrange the observations in ascending or descending order then if numbers of observations (n) are odd then then median is \( \frac{n + 1}{2} \) term.
  If no. of observations (n) are even, then median is average of \( \frac{n}{2} \)th and \( \frac{n}{2} + 1 \)th terms.
• Mode: The observation whose frequency is greatest.
• Mode = 3 median - 2 mean.

Question. To find out the concentration of \( \text{SO}_2 \) in air, (in parts per million i.e., ppm), data was collected from 30 localities in a certain city. Calculate the mean concentration of \( \text{SO}_2 \) in the air.

Answer: 0.099 ppm.

Question. A survey was conducted by a group of students as a part of their environmental awareness programme, in which they collected the following data regarding the number of plants in 20 houses of a locality. Find the mean number of plants per house. 

Answer: 8 plants.

Question. Thirty women were examined in a hospital by a doctor and the number of heart-beats per minute were recorded and summarised as follows. Find the mean heart-beats per minute.

Answer: 75.9 heart-beats per minute.

Question. The following distribution shows the daily pocket allowance given to the children of a multi-storey building. The average pocket allowance is Rs. 18. Find out the missing frequency

Answer: 20.

Question. In the retail market, fruit vendors were selling mangoes kept in packing boxes. These boxes contained varying number of mangoes. The following was their distribution according to the number of boxes. Find the mean number of mangoes kept in a packing box. 

Answer: 57.

Question. The table below shows the daily expenditure on food of 25 households in a locality. Find the mean expenditure.

Answer: Rs. 211.

Question. A class teacher has the following absentee record of 40 students for the whole term. Find the mean number of days a student remained absent. 

Answer: 12.475 days.

Question. The following table gives the literacy rate (in percentage) of 35 cities. Find the mean literacy rate. 

Answer: 69.43%.

Question. The following is the distribution of the marks obtained by 74 students in a class test. Find the mean of the distribution.

Answer: 19.77 \( \approx \) 20.

Question. Find the mean of the following distribution :

Answer: 52.

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

Answer: \( x = 8, y = 7 \).

Question. A survey was conducted related to the heights (in cm) of 51 girls of class X and the following data was obtained : Determine the median height.

Answer: 149.03 cm.

Question. A life insurance agent found the following data for distribution of ages of 100 policy holders. Calculate the median age, if policies are only given to persons in the age group of 18 to 60 years.

Answer: 35.76 years.

Question. The lengths of 40 leaves of a plant are measured correct to the nearest millimetre and the data obtained is represented in the following table. Find the median length of the leaves.

Answer: 146.75 mm.

Question. The following table gives the distribution of the life time (in hours) of 400 neon lamps. Find the median life time of a lamp 

Answer: 3406.98 hours.

Question. The following distribution gives the weight (in kg) of 30 students in a class. Find the median weight of the students. 

Answer: 56.67 kg.

Question. The following data gives the distribution of the total household expenditure (in rupees) of manual workers in a city : Find the mode of the average expenditure of the families. 

Answer: Rs. 1847.83.

Question. The following table shows the ages of the patients admitted to a hospital during a year. Find the mean and mode of the data given above. 

Answer: Mean = 35.375, Mode = 36.8.

Question. The following distribution provides information about the observed lifetime (in hours) of 225 electrical components. Determine the modal lifetime of the components. 

Answer: 65.625 hours.

Question. The following distribution gives the state-wise teacher-student ratio in higher-secondary schools of India. Find the mean and mode from the data. 

Answer: Mean = 29.2, Mode = 30.6.

Question. The following data shows the number of runs scored in ODI cricket by some of the top batsmen of the world : Find the mode of the data.

Answer: 4608.7 runs.

Question. A student noted the number of cars passing through a spot on a road for 100 periods, each of 3 minutes and summarized it in the table given below. Find the mode.

Question. \( Mode + \frac{2}{3}(Mean - Mode) = \)
(a) Mode
(b) Median
(c) Mean
(d) None of the options
Answer: (b) Median
Explanation: Since, \( 3 \text{ Median} = \text{ Mode} + 2 \text{ Mean} \)
\( \Rightarrow \text{ Median} = \frac{\text{ Mode} + 2 \text{ Mean}}{3} \)
\( \Rightarrow \text{ Median} = \frac{\text{ Mode}}{3} + \frac{2}{3} \text{ Mean} \)
\( \Rightarrow \text{ Median} = \text{ Mode} - \frac{2}{3} \text{ Mode} + \frac{2}{3} \text{ Mean} \)
\( \Rightarrow \text{ Median} = \text{ Mode} + \frac{2}{3} (\text{ Mean} - \text{ Mode}) \)

Question. Construction of cumulative frequency table is useful to determine
(a) mean
(b) All of the options
(c) median
(d) mode
Answer: (c) median
Explanation: A cumulative frequency distribution is the sum of the class and all classes below it in a frequency distribution. Construction of cumulative frequency table is useful to determine Median.

Question. For the following distribution

the modal class is
(a) 50 - 60
(b) 40 - 50
(c) 20 - 30
(d) 30 - 40
Answer: (d) 30 - 40
Explanation: According to the question, converting cumulative frequency to actual frequency:

Here Maximum frequency is 30. Therefore, the modal class is \( 30 - 40 \).

Question. The mean of the first 10 natural numbers is
(a) 4.5
(b) 5
(c) 6
(d) 5.5
Answer: (d) 5.5
Explanation: The first 10 natural numbers are 1, 2, 3, ............, 10
\( \therefore \text{ Mean} = \frac{\text{ Sum of first 10 natural numbers}}{10} \)
\( = \frac{1+2+3+.......+10}{10} \)
\( = \frac{55}{10} = 5.5 \)

Question. The marks obtained by 9 students in Mathematics are 59, 46, 30, 23, 27, 44, 52, 40 and 29. The median of the data is
(a) 35
(b) 29
(c) 30
(d) 40
Answer: (d) 40
Explanation: Arranging the given data in ascending order: 23, 27, 29, 30, 40, 44, 46, 52, 59
Here \( n = 9 \), which is odd.
\( \therefore \text{ Median} = \left( \frac{n+1}{2} \right)^{\text{th}} \text{ term} \)
\( = \left( \frac{9+1}{2} \right)^{\text{th}} \text{ term} = 5^{\text{th}} \text{ term} = 40 \)

Question. Find the mode of the given data 3, 3, 7, 4, 5, 3, 5, 6, 8, 9, 5, 3, 5, 3, 6, 9, 7, 4.
Answer:

We observe that the value 3 has the maximum frequency i.e 5. The mode of data is 3.

Question. If the median of a series exceeds the mean by 3, find by what number the mode exceeds its mean?
Answer: Given, \( \text{ Median} = \text{ Mean} + 3 \)
Since, \( \text{ Mode} = 3 \text{ Median} - 2 \text{ Mean} \)
\( = 3(\text{ Mean} + 3) - 2 \text{ Mean} \)
\( = 3 \text{ Mean} + 9 - 2 \text{ Mean} \)
\( \Rightarrow \text{ Mode} = \text{ Mean} + 9 \)
Hence Mode exceeds Mean by 9.

Question. If the values of mean and median are 26.4 and 27.2, what will be the value of mode?
Answer: We know that \( \text{ Mode} = 3 \text{ median} - 2 \text{ mean} \)
\( = 3(27.2) - 2(26.4) \)
\( = 81.6 - 52.8 = 28.8 \)
\( \text{ Mode} = 28.8 \)

Question. In the following frequency distribution, find the median class.

Answer:

\( N = 100 \Rightarrow \frac{N}{2} \text{ th term} = \frac{100}{2} = 50\text{th term} \).
Hence, Median class is 155 - 160.

Question. Find median of the data, using an empirical relation when it is given that Mode = 12.4 and Mean = 10.5.
Answer: \( \text{ Mode} = 3 \text{ median} - 2 \text{ mean} \)
\( \text{ Mode} = 12.4 \) and \( \text{ mean} = 10.5 \)
\( \text{ Median} = \frac{1}{3} \text{ Mode} + \frac{2}{3} \text{ Mean} \)
\( = \frac{1}{3}(12.4) + \frac{2}{3}(10.5) \)
\( = \frac{12.4}{3} + \frac{21}{3} = \frac{12.4 + 21}{3} = \frac{33.4}{3} = 11.13 \)
So, median is 11.13.

Question. Find the mode of the following distribution.

Answer: Here the maximum frequency is 28 then the corresponding class 40 - 50 is the modal class.
\( l = 40, h = 50 - 40 = 10, f = 28, f_1 = 12, f_2 = 20 \)
\( \text{ Mode} = l + \left( \frac{f - f_1}{2f - f_1 - f_2} \right) \times h \)
\( = 40 + \left( \frac{28 - 12}{2 \times 28 - 12 - 20} \right) \times 10 \)
\( = 40 + \frac{160}{24} = 40 + 6.67 = 46.67 \)

Question. Convert the following data into 'more than type' distribution:

Answer:

Question. Calculate the mean of the following data, using direct method:

Answer:

\( \text{ mean} = \frac{\sum f_i x_i}{\sum f_i} = \frac{1980}{40} = 49.5 \)

Question. If the median of the following frequency distribution is 46, find the missing frequencies.

Answer: Let the frequency of the class 30 - 40 be \( f_1 \) and that of the class 50 - 60 be \( f_2 \). The total frequency is 229.
\( \Rightarrow 12 + 30 + f_1 + 65 + f_2 + 25 + 18 = 229 \)
\( \Rightarrow f_1 + f_2 = 79 \)
It is given that the median is 46. Clearly, 46 lies in the class 40 - 50. So, 40 - 50 is the median class.
\( l = 40, h = 10, f = 65 \) and \( F = 12 + 30 + f_1 = 42 + f_1 \), \( N = 229 \).
\( \text{ Median} = l + \frac{\frac{N}{2} - F}{f} \times h \)
\( 46 = 40 + \frac{\frac{229}{2} - (42 + f_1)}{65} \times 10 \)
\( 6 = \frac{145 - 2f_1}{13} \Rightarrow 2f_1 = 67 \Rightarrow f_1 = 33.5 \text{ or } 34 \text{ (say)} \)
Since \( f_1 + f_2 = 79 \), \( f_2 = 79 - 34 = 45 \).
Hence, \( f_1 = 34 \) and \( f_2 = 45 \).

Question. Find median for the following data:

Answer:

\( n = 150 \Rightarrow \frac{n}{2} = 75 \). Median Class = 110 - 120.
\( l = 110, f = 45, c.f. = 45, h = 10 \).
\( \text{ Median} = l + \frac{\frac{n}{2} - cf}{f} \times h = 110 + \frac{75 - 45}{45} \times 10 = 116.67 \).

Question. Draw a pie-chart for the following data of expenditure on various items in a family.

Answer:
Total Expenditure = Rs. 14400.
Central angle for Education = \( \frac{1600}{14400} \times 360 = 40^\circ \)
Central angle for Food = \( \frac{3200}{14400} \times 360 = 80^\circ \)
Central angle for Rent = \( \frac{4000}{14400} \times 360 = 100^\circ \)
Central angle for Clothing = \( \frac{2400}{14400} \times 360 = 60^\circ \)
Central angle for Others = \( \frac{3200}{14400} \times 360 = 80^\circ \)

Question. Find the mean and mode of the following frequency distribution:

Answer:
\( \sum f_i x_i = 1640 \), \( \sum f_i = 50 \).
\( \text{ Mean} = \frac{\sum f_i x_i}{\sum f_i} = \frac{1640}{50} = 32.8 \).
For Mode, Modal class = 30 - 40.
\( l = 30, f_1 = 15, f_2 = 7, f_0 = 10, h = 10 \).
\( \text{ Mode} = l + \frac{f_1 - f_0}{2f_1 - f_0 - f_2} \times h = 30 + \frac{15 - 10}{30 - 10 - 7} \times 10 = 30 + \frac{50}{13} = 33.85 \).

Question. From the following frequency distribution, prepare the 'more than' ogive. Also, find the median.

Answer: 'More than' series points: (400, 230), (450, 210), (500, 175), (550, 135), (600, 103), (650, 79), (700, 52), (750, 34).
\( N = 230 \Rightarrow \frac{N}{2} = 115 \). The corresponding x-value for y = 115 on the ogive gives the median. From graph, Median = 590.

Question. Find the mean marks of students from the following cumulative frequency distribution:

Answer: First, convert to ordinary frequency distribution:
0-10 (3), 10-20 (5), 20-30 (7), 30-40 (10), 40-50 (12), 50-60 (15), 60-70 (12), 70-80 (6), 80-90 (2), 90-100 (8).
Assumed mean (a) = 55, h = 10. \( \sum f_i = 80 \), \( \sum f_i u_i = -26 \).
\( \bar{X} = a + h \frac{\sum f_i u_i}{N} = 55 + 10 \times \frac{-26}{80} = 55 - 3.25 = 51.75 \).
The mean number of marks is 51.75.

Question. The marks obtained by 100 students of a class in an examination are given below:

Draw cumulative frequency curves by using (i) 'less than' series and (ii) 'more than' series. Hence, find the median.
Answer:
(i) 'Less than' series: (5, 2), (10, 7), (15, 13), (20, 21), (25, 31), (30, 56), (35, 76), (40, 94), (45, 98), (50, 100).
(ii) 'More than' series: (0, 100), (5, 98), (10, 93), (15, 87), (20, 79), (25, 69), (30, 44), (35, 24), (40, 6), (45, 2).
The two curves intersect at a point where \( N/2 = 50 \). From the graph, the intersection point gives the median value of 28. Hence, median = 28.