CBSE Class 10 Mathematics Statistics Worksheet Set 06

Question. The median of first 10 prime numbers is
(a) 12.5
(b) 13
(c) 11
(d) 12
Answer: (d) 12
Explanation: The first 10 prime numbers are 2, 3, 5, 7, 11, 13, 17, 19, 23, 29. Here \( n = 10 \), which is even number. \( \therefore \text{Median} = \frac{1}{2} \left[ \left( \frac{n}{2} \right)^{\text{th}} \text{term} + \left( \frac{n}{2} + 1 \right)^{\text{th}} \text{term} \right] = \frac{1}{2} [5^{\text{th}} \text{term} + 6^{\text{th}} \text{term}] = \frac{1}{2} [11 + 13] = \frac{1}{2} \times 24 = 12 \)

Question. The mean of 25 observations is 36. If the mean of first 13 observations is 32 and that of the last 13 observations is 39, then the 13th observation is
(a) 13
(b) 23
(c) 32
(d) 36
Answer: (b) 23
Explanation: Let terms be \( x_1, x_2, x_3, \dots, x_{25} \). According to the question, \( \frac{x_1+x_2+x_3+\dots+x_{25}}{25} = 36 \Rightarrow x_1+x_2+x_3+\dots+x_{25} = 900 \dots(i) \). And \( \frac{x_1+x_2+x_3+\dots+x_{13}}{13} = 32 \Rightarrow x_1+x_2+x_3+\dots+x_{13} = 416 \dots(ii) \). Also, \( \frac{x_{13}+x_{14}+x_{15}+\dots+x_{25}}{13} = 39 \Rightarrow x_{13}+x_{14}+x_{15}+\dots+x_{25} = 507 \dots(iii) \). Adding eq. (ii) and (iii), we get, \( x_1+x_2+\dots+x_{13}+x_{13}+x_{14}+\dots+x_{25} = 416+507 = 923 \Rightarrow (x_1+x_2+\dots+x_{25}) + x_{13} = 923 \Rightarrow 900 + x_{13} = 923 \Rightarrow x_{13} = 23 \)

Question. To represent ‘the more than type’ graphically, we plot the ____________ on the x – axis.
(a) class marks
(b) lower limits
(c) upper limits
(d) class size
Answer: (b) lower limits
Explanation: The lower limit for every class is the smallest value in that class on the other hand the upper limit for every class is the greatest value in that class. To represent ‘the more than type’ graphically, we plot the lower limits on the x-axis and cumulative frequency on the y-axis to find the median.

Question. If \( \sum f_i x_i = 1860 \) and \( \sum f_i = 30 \), then the value of \( \bar{x} \) is
(a) 26
(b) 63
(c) 64
(d) 62
Answer: (d) 62
Explanation: \( \bar{x} = \frac{\sum f_i x_i}{\sum f_i} = \frac{1860}{30} = 62 \)

Question. The empirical relationship between the three measures of central tendencies is
(a) 3 mode = mean + 2 median
(b) Mode= 3 median - 2 mean
(c) 3 mean = median + 2 mode
(d) None of the options
Answer: (b) Mode= 3 median - 2 mean
Explanation: The empirical relationship between the three measures of central tendencies is \( 3 \text{ Median} = \text{Mode} + 2 \text{ Mean} \). The relationship is as per observation. a distribution in which the values of mean, median and mode coincide (i.e. \( \text{mean} = \text{median} = \text{mode} \)) is called symmetrical distribution. conversely when the values of mean, median, mode are not equal, the distribution is called asymmetrical or skewed. knowing any two values, the third can be computed by this formula: \( 3 \text{ median} = 2 \text{ mean} + \text{mode} \), \( 2 \text{ mean} = 3 \text{ median} - \text{mode} \), \( \text{mean} = \frac{1}{2} \{3 \text{ median} - \text{mode}\} \).

Question. From the following frequency distribution, find the median class :

Answer:

\( \frac{\sum f}{2} = \frac{52}{2} = 26 \). Cumulative frequency just greater than 26 is 44, which belongs to class interval 1700 - 1850. \( \Rightarrow \text{Median class} = 1700 - 1850 \)

Question. Find the mode of the following data:
25, 16, 19, 48, 19, 20, 34, 15, 19, 20, 21, 24, 19, 16, 22, 16, 18, 20, 16, 19
Answer:

We observe that the value 19 has the maximum frequency i.e. 5. Therefore, mode of the given data is 19.

Question. Find the median for the frequency distribution:

Answer: \( n = 120 \). \( \Rightarrow \frac{n}{2} = 60 \). Median is average of 60th and 61st observation. Both 60th and 61st observations lie in the cumulative frequency 65, corresponding to \( x = 5 \). \( \text{Median} = \frac{5+5}{2} = 5 \). So, Median is 5.

Question. Find the mode of the data, using a empirical formula, when it is given that median = 41.25 and mean = 33.75
Answer: We know, \( \text{Mode} = 3(\text{Median}) - 2(\text{Mean}) \). Given, \( \text{Median} = 41.25 \) and \( \text{Mean} = 33.75 \). Therefore, \( \text{Mode} = 3 \times 41.25 - 2 \times 33.75 = 123.75 - 67.50 = 56.25 \). So, mode is 56.25.

Question. Find the mode of the following data:
120, 110, 130, 110, 120, 140, 130, 120, 140, 120
Answer: Let us first form the frequency table for the given data:

We observe that the value 120 has the maximum frequency. Hence, the mode or modal value is 120.

Question. Thirty women were examined in a hospital by a doctor and the number of heart beats per minute was recorded and summarized as follows. Find the mean heartbeats per minute for these women, choosing a suitable method.

Answer: Take \( a = 75.5, h = 3 \)

Using the step-deviation method, \( \bar{x} = a + \frac{\sum f_i u_i}{\sum f_i} \times h = 75.5 + \left[ \frac{4}{30} \right] \times 3 = 75.5 + 0.4 = 75.9 \). Hence, the mean heart beats per minute are 75.9.

Question. Find the unknown entries a, b, c, d in the following distribution of heights of students in a class :

Answer: From the table, \( 12 + a = 25 \Rightarrow a = 25 - 12 = 13 \). \( 25 + 10 = b \Rightarrow b = 35 \). \( b + c = 43 \Rightarrow c = 43 - b = 43 - 35 = 8 \). And \( 48 + 2 = d \Rightarrow d = 50 \).

Question. The data regarding marks obtained by 48 students of a class in a class test is given below. Calculate the modal marks of students.

Answer: Modal class is 30 - 35 because it has maximum frequency 25. Here \( l = 30, f_1 = 25, f_0 = 10, f_2 = 7, h = 5 \).
\( \text{Mode} = l + \left( \frac{f_1 - f_0}{2f_1 - f_0 - f_2} \right) \times h \)
\( \text{or, Mode} = 30 + \left( \frac{25 - 10}{2(25) - 10 - 7} \right) \times 5 = 30 + \frac{15}{50 - 10 - 7} \times 5 = 30 + \frac{15}{33} \times 5 = 30 + 0.454 \times 5 \)
\( = 30 + 2.27 = 32.27 \text{ approx.} \)

Question. The following distribution shows the daily pocket allowance given to the children of a multistorey building. The average pocket allowance is Rs.18.00. Find out the missing frequency.

Answer: Let the missing frequency be 'x'.

\( \text{Mean} = \frac{\sum f_i x_i}{\sum f_i} \Rightarrow 18 = \frac{756 + 20x}{44 + x} \Rightarrow 792 + 18x = 756 + 20x \Rightarrow 2x = 36 \Rightarrow x = 18 \).
(Note: The solution table in the screenshot has minor arithmetic differences but follows this logic. Re-calculating correctly: sum = 84+88+144+234+110+96 = 756. Correcting from OCR: sum=752 in OCR is likely 756).
\( 18(44+x) = 756 + 20x \Rightarrow 792 + 18x = 756 + 20x \Rightarrow 2x = 36 \Rightarrow x = 18 \).

Question. The mean of the following frequency distribution is 62.8 and the sum of all the frequencies is 50. Compute the missing frequency \( f_1 \) and \( f_2 \).

Answer:

Given, sum of frequency = 50 \( \Rightarrow 5 + f_1 + 10 + f_2 + 7 + 8 = 50 \Rightarrow f_1 + f_2 = 20 \Rightarrow 3f_1 + 3f_2 = 60 \dots (1) \).
And mean = 62.8 \( \Rightarrow \frac{30f_1 + 70f_2 + 2060}{50} = 62.8 \Rightarrow 30f_1 + 70f_2 = 3140 - 2060 \Rightarrow 30f_1 + 70f_2 = 1080 \Rightarrow 3f_1 + 7f_2 = 108 \dots (2) \).
Subtract eq. (1) from (2): \( 4f_2 = 48 \Rightarrow f_2 = 12 \). Put \( f_2 \) in \( f_1 + f_2 = 20 \Rightarrow f_1 = 8 \). Hence \( f_1 = 8, f_2 = 12 \).

Question. The arithmetic mean of the following data is 14. Find the value of k.

Answer:

Given Mean = 14 \( \Rightarrow \frac{\sum f_i x_i}{\sum f_i} = 14 \Rightarrow \frac{10k + 360}{k + 24} = 14 \Rightarrow 10k + 360 = 14k + 336 \Rightarrow 4k = 24 \Rightarrow k = 6 \).

Question. The following table gives the number of children of 150 families in a village

Find the average number of children per family.
Answer: Let the assumed mean (A) = 2.

Average number of children per family = \( \frac{352}{150} = 2.35 \) (approx)

Question. Calculate the median from the following data:

Answer:

\( N = 250 \Rightarrow \frac{N}{2} = 125 \). The cumulative frequency just greater than 125 is 127. hence, median class is 50 - 60. \( \text{median} = l + \frac{\frac{N}{2} - F}{f} \times h \). Here, \( l = 50, f = 31, F = 96, h = 10 \). \( \text{median} = 50 + \frac{125 - 96}{31} \times 10 = 50 + \frac{29 \times 10}{31} = 50 + 9.35 = 59.35 \)

Question. Calculate the average daily income (in Rs) of the following data about men working in a company:

Answer:

Mean = \( \frac{\sum x_i f_i}{\sum f_i} = \frac{11000}{50} = 220 \). \( \therefore \) Average daily income = Rs 220

Question. On annual day of a school, 400 students participated in the function. Frequency distribution showing their ages is as shown in the following table :

Find mean and median of the above data.
Answer:

\( a = \text{assumed mean} = 12 \). we know that, \( \text{Mean} = \bar{x} = a + \frac{\sum f_i u_i}{\sum f_i} \times h = 12 + \frac{-366}{400} \times 2 = 12 - \frac{183}{100} = 12 - 1.83 = 10.17 \).
\( \text{Median Class} = \frac{N}{2} \text{th term} = \frac{400}{2} = 200\text{th term} \). i.e. 09 – 11.
\( \text{Median} = l + \left( \frac{\frac{N}{2} - c.f}{f} \right) \times h \Rightarrow \text{Median} = 9 + \frac{200 - 190}{32} \times 2 = 9 + 0.625 = 9.625 \)

Question. The wickets taken by a bowler in 10 cricket matches are 2, 6, 4, 5, 0, 2, 1, 3, 2, 3. The median of the data is
(a) 2.5
(b) 1
(c) 2
(d) 3
Answer: (a) 2.5
Explanation: Arranging the given data in ascending order, we get 0, 1, 2, 2, 2, 3, 3, 4, 5, 6
Here, \( n = 10 \), which is even.
\( \therefore \text{Median} = \frac{1}{2} \left[ \left( \frac{n}{2} \right)^{\text{th}} \text{term} + \left( \frac{n}{2} + 1 \right)^{\text{th}} \text{term} \right] \)
\( = \frac{1}{2} [\text{5th term} + \text{6th term}] \)
\( \Rightarrow \text{Median} = \frac{1}{2} [2 + 3] = \frac{5}{2} = 2.5 \)

Question. The measure of central tendency that can be obtained graphically is
(a) None of the options
(b) median
(c) mode
(d) mean
Answer: (b) median
Explanation: The measure of central tendency that can be obtained graphically is Median. The median is less affected by outliers and skewed data distribution i.e. when the distribution is not symmetrical.

Question. In the given data if \( n = 230, l = 40, cf = 76, h = 10, f = 65 \), then its median is
(a) 48
(b) 40
(c) 47
(d) 46
Answer: (d) 46
Explanation: \( \text{Median} = l + \frac{\frac{n}{2} - cf}{f} \times h \)
\( = 40 + \frac{\frac{230}{2} - 76}{65} \times 10 \)
\( = 40 + \frac{115 - 76}{65} \times 10 \)
\( = 40 + \frac{39}{65} \times 10 \)
\( = 40 + \frac{390}{65} \)
\( = 40 + 6 = 46 \)

Question. The times, in seconds, taken by 75 athletes to run a 500m race are tabulated as below:

The number of athletes who completed the race in less than 125 seconds is
(a) 20
(b) 17
(c) 18
(d) 27
Answer: (d) 27
Explanation: According to the question, athletes who completed the race in less than 125 seconds belong to the classes 65-85, 85-105, and 105-125.

Therefore, the number of athletes who completed the race in less than 125 seconds is \( 4 + 5 + 18 = 27 \).

Question. In a data, if \( l = 40, h = 15, f_1 = 7, f_0 = 3, f_2 = 6 \), then the mode is
(a) 82
(b) 62
(c) 52
(d) 72
Answer: (c) 52
Explanation: \( \text{Mode} = l + \left( \frac{f_1 - f_0}{2f_1 - f_0 - f_2} \right) \times h \)
\( = 40 + \left( \frac{7 - 3}{2 \times 7 - 3 - 6} \right) \times 15 \)
\( = 40 + \frac{4}{5} \times 15 \)
\( = 40 + 12 = 52 \)

Question. If \( \sum f_i = 11, \sum f_i x_i = 2p + 52 \) and the mean of any distribution is 6, find the value of p.
Answer: \( \bar{x} = \frac{\sum f_i x_i}{\sum f_i} \)
\( \Rightarrow 6 = \frac{2p + 52}{11} \)
\( \Rightarrow 66 = 2p + 52 \)
\( \Rightarrow 2p = 66 - 52 \)
\( \Rightarrow 2p = 14 \)
\( \Rightarrow p = 7 \)

Question. The mean of a set of numbers is \( \bar{x} \). If each number is multiplied by k, then find the mean of the new set.
Answer: Let numbers be \( x_1, x_2, ...... x_n \).
\( \text{Mean} = \frac{x_1 + x_2 + ... + x_n}{n} = \bar{x} \)
New observations = \( kx_1, kx_2, ... kx_n \)
\( \text{New Mean} = \frac{kx_1 + kx_2 + ... + kx_n}{n} = \frac{k(x_1 + x_2 + ... + x_n)}{n} = k\bar{x} \)

Question. A set of numbers consists of four 5's, six 7's, ten 9's, eleven 12's, three 13's, two 14's. Find the mode of this set of numbers.
Answer:

Mode = 12, Since, It has highest frequency.

Question. Find the value of x, if the mode of the following data is 25: 15, 20, 25, 18, 14, 15, 25, 15, 18, 16, 20, 25, 20, x, 18
Answer: The frequency table of the given data is as given below:

It is given that the mode of the given date is 25. So, it must have the maximum frequency. That is possible only when x = 25.
Hence, x = 25.

Question. Following distribution gives cumulative frequencies of 'more than type' :

Change the above data to a continuous grouped frequency distribution.
Answer:

Question. The arithmetic mean of the following data is 25, find the value of k.

Answer:

Given mean = 25
\( \frac{\text{Sum}}{N} = 25 \)
\( 15k + 390 = 25(k + 14) \)
\( 15k + 390 = 25k + 350 \)
\( 25k - 15k = 390 - 350 \)
\( 10k = 40 \)
\( k = 4 \)

Question. The annual profits earned by 30 shops of a shopping complex in a locality are recorded in the table shown below:

If we draw the frequency distribution table for the above data, find the frequency corresponding to the class 20-25.
Answer: The frequency table is as follows:

The frequency corresponding to the class 20 - 25 is 4.

Question. Find the value of p for the following distribution whose mean is 16.6.

Answer:

Mean = 16.6
\( \text{Mean} = \frac{\sum f_i x_i}{\sum f_i} \)
\( \frac{24p + 1228}{100} = 16.6 \)
\( 24p + 1228 = 1660 \)
\( 24p = 1660 - 1228 \)
\( 24p = 432 \)
\( p = \frac{432}{24} \)
\( p = 18 \)

Question. Draw an ogive to represent the following frequency distribution:

Answer: The given frequency distribution is not continuous, so we will first make it continuous and then prepare the cumulative frequency:

Plot the points (4.5, 2), (9.5, 8), (14.5, 18), (19.5, 23), (24.5, 26) by taking the upper class limit over the x-axis and cumulative frequency over the y-axis to draw the Less than Ogive.

Question. Five coins were simultaneously tossed 1000 times, and at each toss the number of heads was observed. The number of tosses during which 0, 1, 2, 3, 4 and 5 heads were obtained are shown in the table below. Find the mean number of heads per toss

Answer: Let the assumed mean (A) = 2

Mean number of per toss = \( A + \frac{\text{Sum}}{N} = 2 + \frac{470}{1000} = 2 + 0.47 = 2.47 \)

Question. Compare the modal ages of two groups of students appearing for an entrance test:

Answer:
For Group A:
Here the maximum frequency is 78, then the corresponding class 18 - 20 is modal class.
\( l = 18, h = 20 - 18 = 2, f_0 = 78, f_1 = 50, f_2 = 46 \)
\( \text{Mode} = l + \frac{f_0 - f_1}{2f_0 - f_1 - f_2} \times h \)
\( = 18 + \frac{78 - 50}{2 \times 78 - 50 - 46} \times 2 \)
\( = 18 + \frac{56}{60} = 18 + 0.93 = 18.93 \text{ years} \)
For group B:
Here the maximum frequency is 89, then the corresponding class 18 - 20 is the modal class.
\( l = 18, h = 20 - 18 = 2, f_0 = 89, f_1 = 54, f_2 = 40 \)
\( \text{Mode} = l + \frac{f_0 - f_1}{2f_0 - f_1 - f_2} \times h \)
\( = 18 + \frac{89 - 54}{2 \times 89 - 54 - 40} \times 2 \)
\( = 18 + \frac{70}{84} = 18 + 0.83 = 18.83 \text{ years} \)
Hence the modal age for the Group A is higher than that for Group B.

Question. Calculate the median from the following data:

Answer:

\( N = 140 \Rightarrow \frac{N}{2} = \frac{140}{2} = 70 \)
The cumulative frequency just greater than \( \frac{N}{2} \) is 98. \( \therefore \) median class is 55-65 .
\( l = 55, f = 40, F = 58, h = 65 - 55 = 10 \)
\( \therefore \text{Median} = l + \frac{\frac{N}{2} - F}{f} \times h \)
\( = 55 + \frac{70 - 58}{40} \times 10 \)
\( = 55 + 3 = 58 \)

Question. From the following information, construct less than and more than Ogive and find out median from it.

Answer:

The intersection point of the two curves (less than and more than ogives) gives the median. From the graph, the curves intersect at a point where wages correspond to approximately Rs 48.33.

Question. The following distribution gives the daily income of 50 workers of a factory :

Convert the distribution to a 'less than type' cumulative frequency distribution and draw its ogive. Hence obtain the median of daily income.
Answer:

From graph, \( \frac{N}{2} = \frac{50}{2} = 25 \). The x-value corresponding to y = 25 on the ogive is Rs 345.
Hence, Median of daily income = Rs 345

Question. To find out the concentration of \( SO_2 \) in the air (in parts per million, i.e., ppm), the data was collected for 30 localities in a certain city and is presented below:

find the mean concentration of \( SO_2 \) in the air.
Answer: We may find class marks for each interval by using the relation \( x = \frac{\text{upperlimit+lowerclasslimit}}{2} \).
Class size of this data = 0.04.

let a = 0.14
Mean \( \bar{x} = a + \left( \frac{\sum f_i u_i}{\sum f_i} \right) \times h \)
\( = 0.14 + (0.04) \left( \frac{-31}{30} \right) \)
\( = 0.099 \text{ ppm} \)