CBSE Class 10 Mathematics Statistics Assignment Set 17

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Short Answer Type Questions

Question 1. Compute the mode for the following frequency distribution:
Size of items (in cm)0 - 44 - 88 - 1212 - 1616 - 2020 - 2424 - 28
Frequency5791712106

Answer:
Maximum frequency = 17, so the modal class is 12 - 16.
\( l = 12,\ h = 4,\ f_1 = 17,\ f_0 = 9,\ f_2 = 12 \).
Mode \( = l + \frac{f_1 - f_0}{2f_1 - f_0 - f_2} \times h = 12 + \frac{17 - 9}{2(17) - 9 - 12} \times 4 = 14.46 \).
The mode is about 14.46 cm.
In simple words: Highest frequency 17 gives the modal class 12 - 16.

Teacher's Note:
1. Locate the modal class first: it is the class with the highest frequency.
2. Take care to use f0 for the class before and f2 for the class after the modal class.

 

Question 2. Find the mode of the following frequency distribution:
Class15 - 2020 - 2525 - 3030 - 3535 - 4040 - 45
Frequency3891032

Answer:
Maximum frequency = 10, so the modal class is 30 - 35.
\( l = 30,\ h = 5,\ f_1 = 10,\ f_0 = 9,\ f_2 = 3 \).
Mode \( = l + \frac{f_1 - f_0}{2f_1 - f_0 - f_2} \times h = 30 + \frac{10 - 9}{2(10) - 9 - 3} \times 5 = 30.625 \).
In simple words: Highest frequency 10 gives the modal class 30 - 35.

Teacher's Note:
1. Locate the modal class first: it is the class with the highest frequency.
2. Take care to use f0 for the class before and f2 for the class after the modal class.

 

Question 3. The mode of the following data is 67. Find the missing frequency x.
Class40-5050-6060-7070-8080-90
Frequency5x15127

Answer:
The mode 67 lies in the class 60 - 70, the modal class.
\( l = 60,\ h = 10,\ f_1 = 15,\ f_0 = x,\ f_2 = 12 \).
\( 67 = 60 + \frac{15 - x}{30 - x - 12} \times 10 \Rightarrow 0.7(18 - x) = 15 - x \Rightarrow 12.6 - 0.7x = 15 - x \Rightarrow 0.3x = 2.4 \).
Hence x = 8.
In simple words: Use the mode formula with f0 = x.

Teacher's Note:
1. Locate the modal class first: it is the class with the highest frequency.
2. Take care to use f0 for the class before and f2 for the class after the modal class.

 

Question 4. The monthly incomes of 100 families are given below.
Income (in Rs.)0 - 50005000 - 1000010000 - 1500015000 - 2000020000 - 2500025000 - 3000030000 - 3500035000 - 40000
Number of families82641163321
Calculate the modal income.
Answer:
Maximum frequency = 41, so the modal class is 10000 - 15000.
\( l = 10000,\ h = 5000,\ f_1 = 41,\ f_0 = 26,\ f_2 = 16 \).
Mode \( = l + \frac{f_1 - f_0}{2f_1 - f_0 - f_2} \times h = 10000 + \frac{41 - 26}{2(41) - 26 - 16} \times 5000 = 11875 \).
The modal income is Rs. 11875.
In simple words: The class 10000 - 15000 has the highest frequency 41.

Teacher's Note:
1. Locate the modal class first: it is the class with the highest frequency.
2. Take care to use f0 for the class before and f2 for the class after the modal class.

 

Question 5. Calculate the missing frequency a from the following distribution, it being given that the median of the distribution is 24.
No. of balls0-1010-2020-3030-4040-50
No. of children525a187

Answer:
Cumulative frequencies: 5, 30, \( 30 + a \), \( 48 + a \), \( 55 + a \). So \( N = 55 + a \).
The median 24 lies in the class 20 - 30, so \( l = 20,\ c.f. = 30,\ f = a,\ h = 10 \).
\( 24 = 20 + \frac{\frac{55 + a}{2} - 30}{a} \times 10 \Rightarrow 4a = 5(55 + a) - 300 = 5a - 25 \Rightarrow a = 25 \).
Check: \( N = 80 \), \( \frac{N}{2} = 40 \), which lies in the class 20 - 30. Hence a = 25.
In simple words: Use the median class 20 - 30 and the median formula to find a.

Teacher's Note:
1. Build the cumulative frequency column first and find N/2 before choosing the median class.
2. The value of c.f. in the formula belongs to the class before the median class.

 

Question 6. The median of the following data is 16. Find the missing frequencies a and b, if the total of the frequencies is 70.
Class0-55-1010-1515-2020-2525-3030-3535-40
Frequency12a1215b664

Answer:
Total frequency: \( 12 + a + 12 + 15 + b + 6 + 6 + 4 = 70 \Rightarrow a + b = 15 \) ...(i).
\( \frac{N}{2} = 35 \). The cumulative frequency up to 10 - 15 is \( 24 + a \); the median 16 lies in the class 15 - 20, so \( l = 15,\ c.f. = 24 + a,\ f = 15,\ h = 5 \).
\( 16 = 15 + \frac{35 - 24 - a}{15} \times 5 \Rightarrow 1 = \frac{11 - a}{3} \Rightarrow a = 8 \).
From (i), \( b = 15 - 8 = 7 \). Hence a = 8 and b = 7.
In simple words: Two conditions (total frequency and median) give two equations in a and b.

Teacher's Note:
1. Build the cumulative frequency column first and find N/2 before choosing the median class.
2. The value of c.f. in the formula belongs to the class before the median class.

 

Question 7. Find the median of the following data:
Height (in cm)Less than 120Less than 140Less than 160Less than 180Less than 200
Number of students1226344050

Answer:
Taking the lowest class as 100 - 120, the frequencies are 12, 14, 8, 6 and 10 (differences of consecutive cumulative frequencies).
ClassFrequencyCumulative frequency
100 - 1201212
120 - 1401426
140 - 160834
160 - 180640
180 - 2001050

\( N = 50 \), so \( \frac{N}{2} = 25 \). The cumulative frequency just greater than this is 26, so the median class is 120 - 140.
\( l = 120,\ c.f. = 12,\ f = 14,\ h = 20 \).
Median \( = l + \frac{\frac{N}{2} - c.f.}{f} \times h = 120 + \frac{25 - 12}{14} \times 20 = 138.57 \).
The median height is about 138.57 cm.
In simple words: Convert 'less than' data to class frequencies, then use the median formula.

Teacher's Note:
1. Build the cumulative frequency column first and find N/2 before choosing the median class.
2. The value of c.f. in the formula belongs to the class before the median class.

 

Long Answer Type Questions

Question 8. If the mean of the following frequency distribution is 50, then find the values of the missing frequencies \( f_1 \) and \( f_2 \).
Class0-2020-4040-6060-8080-100Total
Frequency17\( f_1 \)32\( f_2 \)19120

Answer:
Total: \( 17 + f_1 + 32 + f_2 + 19 = 120 \Rightarrow f_1 + f_2 = 52 \) ...(i).
\( \sum f_ix_i = 17(10) + 30f_1 + 32(50) + 70f_2 + 19(90) = 3480 + 30f_1 + 70f_2 \).
Mean \( = \frac{3480 + 30f_1 + 70f_2}{120} = 50 \Rightarrow 30f_1 + 70f_2 = 2520 \Rightarrow 3f_1 + 7f_2 = 252 \) ...(ii).
From (i), \( f_1 = 52 - f_2 \). Then \( 156 - 3f_2 + 7f_2 = 252 \Rightarrow 4f_2 = 96 \Rightarrow f_2 = 24 \) and \( f_1 = 28 \).
Hence \( f_1 = 28 \) and \( f_2 = 24 \).
In simple words: Two equations (total and mean) give the two unknown frequencies.

Teacher's Note:
1. Choose the method by the size of the numbers: direct method for small values, assumed mean or step-deviation when the class marks are large.
2. Always check that the total frequency and the class marks are right before dividing.

 

Question 9. The mean of the following distribution is 18. Find the frequency f of the class 19 - 21.
Class11-1313-1515-1717-1919-2121-2323-25
Frequency36913f54

Answer:
Class marks are 12, 14, 16, 18, 20, 22, 24.
\( \sum f_i = 40 + f \) and \( \sum f_ix_i = 36 + 84 + 144 + 234 + 20f + 110 + 96 = 704 + 20f \).
\( \frac{704 + 20f}{40 + f} = 18 \Rightarrow 704 + 20f = 720 + 18f \Rightarrow 2f = 16 \)
Hence f = 8.
In simple words: Write the mean with f unknown and solve.

Teacher's Note:
1. Choose the method by the size of the numbers: direct method for small values, assumed mean or step-deviation when the class marks are large.
2. Always check that the total frequency and the class marks are right before dividing.

 

Question 10. Daily wages of 110 workers, obtained in a survey, are tabulated below.
Daily wages (in Rs.)100 - 120120 - 140140 - 160160 - 180180 - 200200 - 220220 - 240
Number of workers10152022181213
Compute the mean daily wages and modal daily wages of the workers.
Answer:
Mean.
Class mark \( x_i = \frac{\text{lower limit} + \text{upper limit}}{2} \).
Class\( f_i \)\( x_i \)\( f_ix_i \)
100 - 120101101100
120 - 140151301950
140 - 160201503000
160 - 180221703740
180 - 200181903420
200 - 220122102520
220 - 240132302990
Total\( \sum f_i = 110 \)\( \sum f_ix_i = 18720 \)

Mean \( \bar{x} = \frac{\sum f_ix_i}{\sum f_i} = \frac{18720}{110} = 170.18 \).
Mode.
Maximum frequency = 22, so the modal class is 160 - 180.
\( l = 160,\ h = 20,\ f_1 = 22,\ f_0 = 20,\ f_2 = 18 \).
Mode \( = l + \frac{f_1 - f_0}{2f_1 - f_0 - f_2} \times h = 160 + \frac{22 - 20}{2(22) - 20 - 18} \times 20 = 166.67 \).
The mean daily wage is about Rs. 170.18 and the modal daily wage is about Rs. 166.67.
In simple words: Direct method for the mean, modal-class formula for the mode.

Teacher's Note:
1. Choose the method by the size of the numbers: direct method for small values, assumed mean or step-deviation when the class marks are large.
2. Always check that the total frequency and the class marks are right before dividing.

 

Question 11. The following table gives the frequency distribution of married women by age at marriage.
Age (in years)15-1920-2425-2930-3435-3940-4445-4950-5455-5960 and above
Frequency5314098321295332
Obtain the median and describe what information it conveys.
Answer:
The classes are not continuous; make them continuous (subtract 0.5 from each lower limit, add 0.5 to each upper limit). The last class 60 and above becomes 59.5 - 64.5 (any width does not affect the median).
ClassFrequencyCumulative frequency
14.5 - 19.55353
19.5 - 24.5140193
24.5 - 29.598291
29.5 - 34.532323
34.5 - 39.512335
39.5 - 44.59344
44.5 - 49.55349
49.5 - 54.53352
54.5 - 59.53355
59.5 - 64.52357

\( N = 357 \), so \( \frac{N}{2} = 178.5 \). The cumulative frequency just greater than this is 193, so the median class is 19.5 - 24.5.
\( l = 19.5,\ c.f. = 53,\ f = 140,\ h = 5 \).
Median \( = l + \frac{\frac{N}{2} - c.f.}{f} \times h = 19.5 + \frac{178.5 - 53}{140} \times 5 = 23.98 \).
The median age at marriage is about 23.98 years. This means that half of the women were married before the age of about 24 years and half after it.
In simple words: Make the classes continuous, find N/2 = 178.5 and use the median formula.

Teacher's Note:
1. Build the cumulative frequency column first and find N/2 before choosing the median class.
2. The value of c.f. in the formula belongs to the class before the median class.

 

Question 12. The following frequency distribution shows the sales of athletic shoes at an online store each day during the past 100 days. Find x and y if the median is 525.
Sales0-100100-200200-300300-400400-500500-600600-700700-800800-900900-1000
No. of days25x121720y974

Answer:
Total: \( 2 + 5 + x + 12 + 17 + 20 + y + 9 + 7 + 4 = 76 + x + y = 100 \Rightarrow x + y = 24 \) ...(i).
The median 525 lies in the class 500 - 600, so \( l = 500,\ f = 20,\ h = 100 \). The cumulative frequency of the previous class is \( c.f. = 36 + x \), and \( \frac{N}{2} = 50 \).
\( 525 = 500 + \frac{50 - 36 - x}{20} \times 100 \Rightarrow 25 = 5(14 - x) \Rightarrow 14 - x = 5 \Rightarrow x = 9 \).
From (i), \( y = 24 - 9 = 15 \). Hence x = 9 and y = 15.
In simple words: Total frequency gives x + y; the median formula gives x.

Teacher's Note:
1. Build the cumulative frequency column first and find N/2 before choosing the median class.
2. The value of c.f. in the formula belongs to the class before the median class.

 

Question 13. The following frequency distribution shows the marks obtained by 80 students in an examination. Find the mean marks obtained by the students.
MarksAbove 0Above 10Above 20Above 30Above 40Above 50Above 60Above 70Above 80Above 90Above 100
No. of students80777265554323161080

Answer:
The frequencies of the classes 0 - 10, 10 - 20, ..., 90 - 100 are found by subtracting the next 'above' number from the current one: 3, 5, 7, 10, 12, 20, 7, 6, 2, 8.
Class mark \( x_i = \frac{\text{lower limit} + \text{upper limit}}{2} \). Take \( a = 55 \), class size \( h = 10 \) and \( u_i = \frac{x_i - a}{h} \).
Class\( f_i \)\( x_i \)\( u_i = \frac{x_i - a}{h} \)\( f_iu_i \)
0 - 1035-5-15
10 - 20515-4-20
20 - 30725-3-21
30 - 401035-2-20
40 - 501245-1-12
50 - 60205500
60 - 7076517
70 - 80675212
80 - 9028536
90 - 100895432
Total\( \sum f_i = 80 \)\( \sum f_iu_i = -31 \)

Mean \( \bar{x} = a + h \times \frac{\sum f_iu_i}{\sum f_i} = 55 + 10 \times \frac{-31}{80} = 51.12 \).
In simple words: Convert the 'above' data to class frequencies, then use the step-deviation method.

Teacher's Note:
1. Choose the method by the size of the numbers: direct method for small values, assumed mean or step-deviation when the class marks are large.
2. Always check that the total frequency and the class marks are right before dividing.

 

Question 14. The following frequency distribution shows the heights of 50 students of a class. Find the unknown entries.
Class intervals (height in cm)FrequencyCumulative frequency
150-15512a
155-160b25
160-16510c
165-170d43
170-175e48
175-1802f
Totalg

Answer:
Cumulative frequency of the first class equals its frequency: a = 12.
\( b = 25 - 12 = 13 \).
\( c = 25 + 10 = 35 \).
\( d = 43 - 35 = 8 \).
\( e = 48 - 43 = 5 \).
\( f = 48 + 2 = 50 \).
Total frequency \( g = 12 + 13 + 10 + 8 + 5 + 2 = 50 \).
In simple words: Each cumulative frequency is the previous one plus the class frequency.

Teacher's Note:
1. Build the cumulative frequency column first and find N/2 before choosing the median class.
2. The value of c.f. in the formula belongs to the class before the median class.

 

Question 15. The maximum bowling speeds, in km per hour, of 33 players at a cricket coaching centre are given as follows. Calculate the median bowling speed.
Speed (km/h)85 - 100100 - 115115 - 130130 - 145
Number of players11985

Answer:
ClassFrequencyCumulative frequency
85 - 1001111
100 - 115920
115 - 130828
130 - 145533

\( N = 33 \), so \( \frac{N}{2} = 16.5 \). The cumulative frequency just greater than this is 20, so the median class is 100 - 115.
\( l = 100,\ c.f. = 11,\ f = 9,\ h = 15 \).
Median \( = l + \frac{\frac{N}{2} - c.f.}{f} \times h = 100 + \frac{16.5 - 11}{9} \times 15 = 109.17 \).
The median bowling speed is about 109.17 km/h.
In simple words: N/2 = 16.5 lies in the class 100 - 115.

Teacher's Note:
1. Build the cumulative frequency column first and find N/2 before choosing the median class.
2. The value of c.f. in the formula belongs to the class before the median class.

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