Important Questions Class 10 Maths Chapter 14

Important Questions Class 10 Mathematics Chapter 14 – Statistics 

With higher grades Mathematics becomes a bit tougher and it’s important for students to have clear concepts about various topics studied in primary classes. Hence, students must learn the art of tackling it and understanding it effectively right from an early age to cope with Mathematical examinations. Therefore, students are advised to take expert advice and have access to good study materials while preparing for Mathematics.

The chapter ‘Statistics’ helps students learn how to collect and study a large amount of data and interpret it on the basis of various statistical tools, so they can be efficientin handling data. The vital topics included in the Chapter 14 Class 10 Mathematics important questions are:

• Mean of Grouped Data
• Mode of Grouped Data
• Median of Grouped Data
• Graphical Representation of Cumulative Frequency Distribution

 

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Important Questions Class 10 Mathematics Chapter 14 – With Solutions

The following important questions and their solutions are included in the Mathematics Class 10 Chapter 14 important questions:

 

Question 1. In the given formula

x̄ = a + (fidi)/fi,

For finding out the required mean of the grouped data, di’s are deviations from an of

(A) Lower limits of the given classes

(B) Upper limits of the given classes

(C) Midpoints of the given classes

(D) Frequencies of the given class marks

Answer 1. (C) Midpoints of the given classes

Explanation:We know,

di = xi – a

here,

xi is the data, and ‘a’ is the required assumed mean

So, di is the given deviations from the midpoints of the classes.

Hence, option (C) is correct

 

Question 2. While computing the required mean for grouped data, we assume that the given frequencies are

(A) Evenly distributed over all the given classes

(B) Centred on the class marks of the given classes

(C) Centred on the upper limits of the given classes

(D) Centred on the lower limits of the given classes

Answer 2. (B) Centred on the class marks of the given classes

 

Explanation: In computing the mean for grouped data, the required frequencies are centred at the class marks of the given classes.

Therefore, option (B) is correct.

 

Question 3. If xi‘s are the given midpoints of the class intervals of the grouped data, fi’s are the given corresponding frequencies, and x is the mean, (fixi – ) is equal to

(A)0 (B) –1 (C) 1 (D) 2

Answer 3. (A) 0

Explanation:Mean (x) = Sum of all the given observations/ Number of the given observation

x = (f1x1 + f2x2 + …..+ fnxn) / f1 + f2 +……+ fn

x = Σfixi / Σfi, Σfi = n

x = Σfixi / n

n x = Σfixi …………… (1)

Σ (fixi – x) = (f1x1 – x) + (f2x2 – x)+ …..+ (fnxn – x)

Σ (fixi – x) = (f1x1 + f2x2 + …..+ fnxn) – (x +x +….n times)

Σ (fixi – x) = Σfixi –nx

Σ(fixi – x) = nx – nx ( From eq1)

Σ(fixi – x) = 0

Hence, option (A) is correct

 

Question 4. In the formula x = a + h(fiui/fi), for finding out the required mean of grouped frequency distribution, ui =

(A) (xi+a)/h

(B) h (xi – a)

(C) (xi –a)/h

(D) (a – xi)/h

Answer 4. (C) (xi –a)/h

Explanation:According to the given question,

x = a + h(fiui/fi),

The above formula is the step deviation formula.

In the above-given formula,

xi is data values,

a is assumed mean,

h is class size,

When class size is same, we simplify the calculations of the given mean by computing the coded mean of u1,u2,u3…..,

here ui = (xi – a)/h

Hence, option (C) is correct

 

Question 5. The abscissa of the given point for intersection of the less than type and of the more than type, the cumulative frequency curves for the given grouped data give its____.

(A) mean 

(B) median

(C) mode 

(D) all three above

Answer 5. (B) median

 

Question 6. For the following given distribution:

Class 0-05 5-10 10-15 15-20 20-25

Frequency 10 15 12 20 9

the sum of the required lower limits of the median class as well as the modal class is_____.

(A)35 

(B) 25 

(C) 30 

(D) 15

Answer 6. (B) 25

Explanation:

Class	Frequency	Cumulative Frequency
0-5	10	10
5-10	15	25
10-15	12	37
15-20	20	57
20-25	9	66

From the table given, N/2 = 66/2 = 33, which lies in the interval 10 – 15.

Therefore, the lower limit for the median class is 10.

here,  the highest given frequency is 20, which lies between the intervals of 15 – 20.

hence, the lower limit of the modal class is 15.

So, required sum is 10 + 15 = 25.

Hence, option (B) is correct.

 

Question 7. Consider the following given frequency distribution:

Class 0-05 6-11 12-17 18-23 24-29

Frequency 13 10 15 8 11

The upper limit for the given median class is_____.

(A)18 

(B) 17.5 

(C) 17 

(D) 18.5

Answer 7. (B) 17.5

Explanation: As per the given question,

Classes are not continuous; hence, we make the data continuous by subtracting 0.5 from the lower limit and adding 0.5 to the upper limit of each class.

Class	Frequency	Cumulative Frequency
0.5-5.5	13	13
6.5-11.5	10	23
11.5-17.5	15	38
17.5-23.5	8	46
23.5-29.5	11	57

According to the given question,

N/2 = 57/2 = 28.5

28.5 lies in between the intervals 11.5 – 17.5.

Thus, the upper limit is 17.5.

So, option (B) is correct.

 

Question 8. For the following given distribution:

Marks Number of the students

Below 10 3

Below 20 12

Below 30 27

Below 40 57

Below 50 75

Below 60 80

The modal class is

(A)50-60 

(B) 20-30 

(C) 30-40 

(D) 10-20

Answer 8. (C) 30-40

Explanation:

Marks	Number of the students	Cumulative Frequency
Below 10	3=3	3
10-20	(12 – 3) = 9	12
20-30	(27 – 12) = 15	27
30-40	(57 – 27) = 30	57
40-50	(75 – 57) = 18	75
50-60	(80 – 75) = 5	80

where we see that the highest frequency is 30, which lies in the interval 30 – 40.

Thus, option (C) is correct.

 

Question 9. Consider the data:

Class	65-85	85-105	105-125	125-145	145-165	165-185	185-205
Frequency	4	5	13	20	14	7	4

The difference between the given upper limit of the median class and the given lower limit of the modal class is______.

(A)38 

(B) 19 

(C) 20 

(D) 0

Answer 9. (C) 20

Explanation:

Class	Frequency	Cumulative Frequency
65-85	4	4
85-105	5	9
105-125	13	22
125-145	20	42
145-165	14	56
165-185	7	63
185-205	4	67

 

here, N/2 = 67/2 = 33.5 which lies in the interval 125 – 145

So, the upper limit of the median class is 145.

Here, we see that the given highest frequency is 20, which lies in 125 – 145.

Hence, the lower limit of the modal class is 125.

Required difference = Upper limit of the given median class – The lower limit of modal class

                                  = 145 – 125 = 20

Hence, option (C) is correct

 

Question 10. The times, in seconds, taken by the group of 150 athletes to run a 110 m hurdle race are tabulated below

Class	13.8-14	14-14.2	14.2-14.4	14.4-14.6	14.6-14.8	14.8-15
Frequency	2	4	5	71	48	20

The number of the given athletes who completed the race in less than 14.6 seconds is :

(A)130 

(B) 71 

(C) 82 

(D) 11

Answer 10. (C) 82

Explanation:The number of the given athletes who completed the race in less than 14.6 second= 2 + 4 + 5 + 71 = 82

So, option (C) is correct

 

Question 11. Consider the following given distribution :

Marks obtained by the number of students

More than or equal to 0 63

More than or equal to 10 58

More than or equal to 20 55

More than or equal to 30 51

More than or equal to 40 48

More than or equal to 50 42

The frequency of classes 30-40 is_.

(A) 3 

(B) 4 

(C) 48 

(D) 51

Answer 11. (A) 3

Explanation:

Marks Obtained	Number of students	Cumulative Frequency
0-10	(63 – 58) = 5	5
10-20	(58 – 55) = 3	3
20-30	(55 – 51) = 4	4
30-40	(51 – 48) = 3	3
40-50	(48 – 42) = 6	6
50<	42 = 42	42

Thus, the frequency in the class interval 30 – 40 is 3.

So, option (A) is correct.

 

Question 12. A group of students surveyed as a part of their environment awareness program. They collected the following data regarding number for plants in 20 houses in a locality. Find out the mean number of the plants per house.

Number of the Plants	0-2	2-4	4-6	6-8	8-10	10-12	12-14
Number of the Houses	1	2	1	5	6	2	3

Which method would you use for finding the mean, and why?

Answer 12. To find out the required mean value, we will use the direct method as the numerical value for fi and xi are small.

Find out the midpoint of the given interval using the following formula.

Midpoint (xi) = (upper limit + lower limit)/2

No. of the plants (Class interval)	No. of the houses Frequency (fi)	Mid-point (xi)	fixi
0-2	1	1	1
2-4	2	3	6
4-6	1	5	5
6-8	5	7	35
8-10	6	9	54
10-12	2	11	22
12-14	3	13	39
	Sum fi = 20		Sum fixi = 162

The formula for finding out the mean is:

Mean = x̄ = ∑fi xi /∑fi 

= 162/20

= 8.1

So, the mean number of the given plants per house is 8.1

 

Question 13. Consider the following distribution of the daily wages of 50 workers of a factory.

Daily wages (in Rs.)	100-120	120-140	140-160	160-180	180-200
Number of the workers	12	14	8	6	10

Find the mean daily wages for the workers for the factory by using an appropriate method.

Answer 13. Find out the midpoint for the given interval using the following formula.

Midpoint (xi) = (upper limit + lower limit)/2

In the given case, the mid-point (xi) value is very large. Thus let us assume the given mean value, A = 150 and the class interval is h = 20.

thus, ui = (xi – A)/h = UI  = (xi – 150)/20

Substitute the values and find out the values as follows:

Daily wages (Class interval)	Number of workers frequency (fi)	Mid-point (xi)	ui = (xi – 150)/20	fiui
100-120	12	110	-2	-24
120-140	14	130	-1	-14
140-160	8	150	0	0
160-180	6	170	1	6
180-200	10	190	2	20
Total	Sum fi = 50			Sum fiui = -12

Thus, the formula for finding out the mean will:

Mean = x̄ = A + h∑fiui /∑fi =150 + (20 × -12/50) = 150 – 4.8 = 145.20

Therefore, the mean daily wage of the worker = Rs. 145.20

 

Question 14. The following distribution shows daily pocket allowance of children of a locality. The mean pocket allowance is Rs 18. Find the missing frequency f.

Daily Pocket Allowance(in c)	11-13	13-15	15-17	17-19	19-21	21-23	23-35
Number of children	7	6	9	13	f	5	4

Answer 14. To finding out the missing frequency, use the mean formula.

Where the value of mid-point (xi)  mean x̄ = 18

Class interval	Number of children (fi)	Mid-point (xi)	fixi
11-13	7	12	84
13-15	6	14	84
15-17	9	16	144
17-19	13	18 = A	234
19-21	f	20	20f
21-23	5	22	110
23-25	4	24	96
Total	fi = 44+f		Sum fixi = 752+20f

The mean formula is given as;

Mean = x̄ = ∑fixi /∑fi = (752+20f)/(44+f)

then substitute the given values and equate them to find out the missing frequency (f)

⇒ 18 = (752+20f)/(44+f)

⇒ 18(44+f) = (752+20f)

⇒ 792+18f = 752+20f

⇒ 792+18f = 752+20f

⇒ 792 – 752 = 20f – 18f

⇒ 40 = 2f

⇒ f = 20

Thus, the missing frequency, f = 20.

 

Question 15. A doctor examined thirty women in a hospital, and the number of heartbeats per minute was recorded and summarised as follows. Find out the mean heartbeats per minute for these women, choosing a suitable method.

Number of heartbeats per minute	65-68	68-71	71-74	74-77	77-80	80-83	83-86
Number of women	2	4	3	8	7	4	2

Answer 15. From the given data above, let us assume the mean as A = 75.5

xi = (Upper limit + Lower limit)/2

Class size (h) = 3

Now, find out the UI and fiui as follows:

Class Interval	Number of women (fi)	Mid-point (xi)	ui = (xi – 75.5)/h	fiui
65-68	2	66.5	-3	-6
68-71	4	69.5	-2	-8
71-74	3	72.5	-1	-3
74-77	8	75.5	0	0
77-80	7	78.5	1	7
80-83	4	81.5	3	8
83-86	2	84.5	3	6
	Sum fi= 30			Sum fiui = 4

Mean = x̄ = A + h∑fiui /∑fi 

= 75.5 + 3×(4/30)

= 75.5 + 4/10

= 75.5 + 0.4

= 75.9

Thus, the mean heart beats per minute for these women is 75.9

 

Question 16. In a retail market, fruit vendors sell mangoes kept in packing boxes. These boxes contained varying numbers of mangoes. The following was the distribution of the mangoes according to the number of the boxes.

Number of mangoes	50-52	53-55	56-58	59-61	62-64
Number of the boxes	15	110	135	115	25

Finding out the mean number of mangoes kept in a packing box. Which method of finding out the mean did you choose?

Answer 16. As the given data is not continuous, thus we add 0.5 to the upper limit and subtract 0.45 from lower limit because the gap between the two intervals is 1

Where the assumed mean (A) = 57

Class size (h) = 3

Where the step deviation is used as the frequency values are big.

Class Interval	Number of boxes (fi)	Mid-point (xi)	di = xi – A	fidi
49.5-52.5	15	51	-6	90
52.5-55.5	110	54	-3	-330
55.5-58.5	135	57 = A	0	0
58.5-61.5	115	60	3	345
61.5-64.5	25	63	6	150
	Sum fi = 400			Sum fidi = 75

The formula to find out the required mean is:

Mean = x̄ = A +h ∑fidi /∑fi 

= 57 + 3(75/400)

= 57 + 0.1875

= 57.19

Thus, the mean number of mangoes kept in the packing box is 57.19

 

Question 17. The table below shows daily expenditure on the food of 25 households in a locality. Find out the mean daily expenditure on food by a suitable method.

Daily expenditure(in c)	100-150	150-200	200-250	250-300	300-350
The number of households	4	5	12	2	2

Answer 17. Find out the midpoint of the given interval using the formula.

Midpoint (xi) = (upper limit + lower limit)/2

Let us assume the required mean (A) = 225

Class size (h) = 50

Class Interval	Number of households (fi)	Mid-point (xi)	di = xi – A	ui = di/50	fiui
100-150	4	125	-100	-2	-8
150-200	5	175	-50	-1	-5
200-250	12	225	0	0	0
250-300	2	275	50	1	2
300-350	2	325	100	2	4
	Sum fi = 25				Sum fiui = -7

Mean = x̄ = A +h∑fiui /∑fi

 = 225+50(-7/25)

= 225-14

= 211

Thus, the mean daily expenditure on the food is 211

 

Question 18. To finding out the concentration of SO2 in the air (in parts per million, i.e., ppm), data was collected for 30 localities in the certain city and is presented below:

The concentration of SO2 ( in ppm)	Frequency
0.00 – 0.04	4
0.04 – 0.08	9
0.08 – 0.12	9
0.12 – 0.16	2
0.16 – 0.20	4
0.20 – 0.24	2

Find out the mean concentration of SO2 in the air.

Answer 18. For finding out the mean, first, find out the midpoint of the given frequencies as follows:

The concentration of SO2 (in ppm)	Frequency (fi)	Mid-point (xi)	fixi
0.00-0.04	4	0.02	0.08
0.04-0.08	9	0.06	0.54
0.08-0.12	9	0.10	0.90
0.12-0.16	2	0.14	0.28
0.16-0.20	4	0.18	0.72
0.20-0.24	2	0.22	0.44
Total	Sum fi = 30		Sum (fixi) = 2.96

The formula to find out the required mean is

Mean = x̄ = ∑fixi /∑fi

= 2.96/30

= 0.099 ppm

Thus, the mean concentration of SO2 in the air is 0.099 ppm.

 

Question 19. The class teacher has the following absentee record of the 40 students of a class for the wholeterm. Find out the mean number of days a student was absent.

Number of days	0-6	6-10	10-14	14-20	20-28	28-38	38-40
Number of the students	11	10	7	4	4	3	1

Answer 19. Find out the midpoint of the given interval using the formula.

Midpoint (xi) = (upper limit + lower limit)/2

Class interval	Frequency (fi)	Mid-point (xi)	fixi
0-6	11	3	33
6-10	10	8	80
10-14	7	12	84
14-20	4	17	68
20-28	4	24	96
28-38	3	33	99
38-40	1	39	39
	Sum fi = 40		Sum fixi = 499

The mean formula is,

Mean = x̄ = ∑fixi /∑fi

= 499/40

= 12.48 days

So, the mean number of days an student was absent = 12.48.

 

Question 20. The following table gives the literacy rate (in percentage) of 35 cities. Find out the mean literacy rate.

Literacy rate (in %)	45-55	55-65	65-75	75-85	85-98
Number of cities	3	10	11	8	3

Answer 20. Find out the midpoint of the given interval using the formula.

Midpoint (xi) = (upper limit + lower limit)/2

In the given case, the mid-point (xi) value is very large. Thus let us assume the given mean value, A = 70 and the class interval is h = 10.

Thus, ui = (xi-A)/h = UI = (xi-70)/10

Substitute and find out the values as follows:

Class Interval	Frequency (fi)	(xi)	di = xi – a	ui = di/h	fiui
45-55	3	50	-20	-2	-6
55-65	10	60	-10	-1	-10
65-75	11	70	0	0	0
75-85	8	80	10	1	8
85-95	3	90	20	2	6
	Sum fi  = 35				Sum fiui  = -2

Thus, Mean = x̄ = A+(∑fiui /∑fi)×h

= 70+(-2/35)×10

= 69.42

Hence, mean literacy part = 69.42

 

Question 21. The following table shows the ages of the patients admitted in a hospital during a year:

Age (in years)	5-15	15-25	25-35	35-45	45-55	55-65
The number of patients	6	11	21	23	14	5

Find out the mode and the mean of the data given above. Compare and interpret the two

measures of central tendency.

Answer 21. For finding out the modal class, let us consider the class interval with the high frequency

Where, the greatest frequency = 23, hence the modal class = 35 – 45,

l = 35,

class width (h) = 10,

FM = 23,

f1 = 21 and f2 = 14

The formula to find out the mode is

Mode = l+ [(fm-f1)/(2fm-f1-f2)]×h

Substituting the values in the given formula, we get

Mode = 35+[(23-21)/(46-21-14)]×10

Mode = 35+(20/11) = 35+1.8

Mode = 36.8 year

Thus, the mode of the given data = 36.8 year

Calculation of the Mean:

First, find out the midpoint using the given formula, 

xi = (upper limit +lower limit)/2

Class Interval	Frequency (fi)	Mid-point (xi)	fixi
5-15	6	10	60
15-25	11	20	220
25-35	21	30	630
35-45	23	40	920
45-55	14	50	700
55-65	5	60	300
	Sum fi = 80		Sum fixi = 2830

The mean formula is

Mean = x̄ = ∑fixi /∑fi

= 2830/80

= 35.37 years

Thus, the mean of the given data = 35.37 years

 

Question 22. The following data gives information on the observed lifetimes (in hours) of 225

electrical components:

Lifetime (in hours)	0-20	20-40	40-60	60-80	80-100	100-120
Frequency	10	35	52	61	38	29

Determine the modal lifetimes of the components.

Answer 22. From the given data, we get the modal class are 60–80.

l = 60,

The frequencies are:

FM = 61, f1 = 52, f2 = 38 as well as h = 20

formula to find out the mode is

Mode = l+ [(fm-f1)/(2fm-f1-f2)]×h

Substituting the values in given formula, we get

Mode =60+[(61-52)/(122-52-38)]×20

Mode = 60+((9 x 20)/32)

Mode = 60+(45/8) = 60+ 5.625

Thus, the modal lifetime of the components = 65.625 hours.

 

Question 23. The following data gives a distribution of total monthly household expenditure of 200

families of a village. Find out the modal monthly expenditure of the families. Also, find the

mean monthly expenditure:

Expenditure	Number of families
1000-1500	24
1500-2000	40
2000-2500	33
2500-3000	28
3000-3500	30
3500-4000	22
4000-4500	16
4500-5000	7

Answer 23. Given as:Modal class = 1500-2000,

l = 1500,

Frequencies:

FM = 40 f1 = 24, f2 = 33 and

h = 500

Mode formula:

Mode = l+ [(fm-f1)/(2fm-f1-f2)]×h

Substituting the values in given formula, we get

Mode =1500+[(40-24)/(80-24-33)]×500

Mode = 1500+((16×500)/23)

Mode = 1500+(8000/23) = 1500 + 347.83

So, the modal monthly expenditure of the families = Rupees 1847.83

The calculation of the required mean:

First, find out the midpoint using the formula, xi =(upper limit +lower limit)/2

Let us assume the mean as; A be 2750

Class Interval	fi	xi	di = xi – a	ui = di/h	fiui
1000-1500	24	1250	-1500	-3	-72
1500-2000	40	1750	-1000	-2	-80
2000-2500	33	2250	-500	-1	-33
2500-3000	28	2750	0	0	0
3000-3500	30	3250	500	1	30
3500-4000	22	3750	1000	2	44
4000-4500	16	4250	1500	3	48
4500-5000	7	4750	2000	4	28
	fi = 200				fiui = -35

The formula for calculating the mean,

Mean = x̄ = a +(∑fiui /∑fi)×h

Substituting the values in the given formula

= 2750+(-35/200)×500

= 2750-87.50

= 2662.50

Thus, the mean monthly expenditure of families = Rupees 2662.50

 

Question 24. The following distribution gives the state-wise teacher-student ratio in higher secondary schools of India. Find the mode and mean of this data.

No. of Students per teacher	Number of states / U.T
15-20	3
20-25	8
25-30	9
30-35	10
35-40	3
40-45	0
45-50	0
50-55	2

Answer 24. Given as:

Modal class = 30 – 35,

l = 30,

Class width (h) = 5,

FM = 10, f1 = 9 and f2 = 3

Mode Formula:

Mode = l+ [(fm-f1)/(2fm-f1-f2)]×h

Substituting the values in the given formula

Mode = 30+((10-9)/(20-9-3))×5

Mode = 30+(5/8) = 30+0.625

Mode = 30.625

Thus, the mode of the given data = 30.625

Calculation of the required mean:

Find out the midpoint using the formula, 

xi =(upper limit +lower limit)/2

Class Interval	Frequency (fi)	Mid-point (xi)	fixi
15-20	3	17.5	52.5
20-25	8	22.5	180.0
25-30	9	27.5	247.5
30-35	10	32.5	325.0
35-40	3	37.5	112.5
40-45	0	42.5	0
45-50	0	47.5	0
50-55	2	52.5	105.5
	Sum fi = 35		Sum fixi = 1022.5

Mean = x̄ = ∑fixi /∑fi

= 1022.5/35

= 29.2

Thus, the mean = 29.2

 

Question 25. The given distribution shows the number of runs scored by some top batsmen of the world in one-day international cricket matches.

Run Scored	Number of Batsman
3000-4000	4
4000-5000	18
5000-6000	9
6000-7000	7
7000-8000	6
8000-9000	3
9000-10000	1
10000-11000	1

Find the mode of the data.

Answer 25. Given as:

Modal class = 4000 – 5000,

l = 4000,

class width (h) = 1000,

FM = 18, f1 = 4 and f2 = 9

Mode Formula is given as:

Mode = l+ [(fm-f1)/(2fm-f1-f2)]×h

Substituting the values we get,

Mode = 4000+((18-4)/(36-4-9))×1000

Mode = 4000+(14000/23) = 4000+608.695

Mode = 4608.695

Mode = 4608.7 (approximately)

Therefore, the mode of given data is 4608.7 runs

 

Question 26. A student noted the number of cars passing through a spot on the road for 100 periods each of 3 minutes and summarised it in the table given below. Find out the mode of the data.

Number of cars	Frequency
0-10	7
10-20	14
20-30	13
30-40	12
40-50	20
50-60	11
60-70	15
70-80	8

Answer 26. Given as:

Modal class = 40 – 50, l = 40,

Class width (h) = 10, FM = 20, f1 = 12 as well as f2 = 11

Mode = l+ [(fm-f1)/(2fm-f1-f2)]×h

Substituting the values we get,

Mode = 40+((20-12)/(40-12-11))×10

Mode = 40 + (80/17) = 40 + 4.7 = 44.7

Hence, the mode of given data is 44.7 cars

 

Question 27. The following frequency distribution gives monthly electricity consumption of 68 consumers in a locality. Find out the data’s median, mean, and mode and compare them.

Monthly consumption(in units)	No. of customers
65-85	4
85-105	5
105-125	13
125-145	20
145-165	14
165-185	8
185-205	4

Answer 27. 

Find out the cumulative frequency of the given data as follows:

Class Interval	Frequency	Cumulative frequency
65-85	4	4
85-105	5	9
105-125	13	22
125-145	20	42
145-165	14	56
165-185	8	64
185-205	4	68
	N=68

 

From given table, it is observed that n = 68 and so n/2=34

Therefore, the required median class is 125-145 with cumulative frequency = 42

here, l = 125, n = 68, Cf = 22, f = 20, h = 20

The median can be calculated as follows:

Median = l + [{(n/2)-Cf}/f] x h

=125+((34−22)/20) × 20

=125+12 = 137

Thus, median = 137

For calculating the mode:

Modal class = 125-145,

f1=20, f0=13, f2=14 & h = 20

Mode formula:

Mode = l+ [(f1-f0)/(2f1-f0-f2)]×h

Mode = 125 + ((20-13)/(40-13-14))×20

=125+(140/13)

=125+10.77

=135.77

Thus, mode = 135.77

Calculate the required mean:

Class Interval	fi	xi	di=xi-a	ui=di/h	fiui
65-85	4	75	-60	-3	-12
85-105	5	95	-40	-2	-10
105-125	13	115	-20	-1	-13
125-145	20	135	0	0	0
145-165	14	155	20	1	14
165-185	8	175	40	2	16
185-205	4	195	60	3	12
	Sum fi= 68				Sum fiui= 7

x̄ =a+h ∑fiui/∑fi =135+20(7/68)

Mean=137.05

In the given case, mean, median and mode are more/less equal in the distribution.

 

Question 28. If the distribution median given below is 28.5, find the value of x & y.

Class Interval	Frequency
0-10	5
10-20	x
20-30	20
30-40	15
40-50	y
50-60	5
Total	60

Answer 28. Given as n = 60

Median for the given data = 28.5

here, n/2 = 30

The median class is 20 – 30, with the cumulative frequency = 25+x

The lower limit of the given median class, l = 20,

Cf = 5+x,

f = 20 & h = 10

Median = l + [{(n/2)-Cf}/f] x h

Substituting the values we get,

28.5=20+((30−5−x)/20) × 10

8.5 = (25 – x)/2

17 = 25-x

Thus, x =8

Now, from the given cumulative frequency, we can identify the required value for x + y as follows:

As

60=5+20+15+5+x+y

Now, substituting the value of x to find out y

60 = 5+20+15+5+8+y

y = 60-53

y = 7

So, the value of x = 8 and y = 7.

 

Question 29. The life insurance agent found the following data for the distribution of ages of 100 policyholders. Calculate median age if policies are given only to the persons whose age is 18 years onwards but less than 60 years.

Age (in years)	Number of policyholders
Below 20	2
Below 25	6
Below 30	24
Below 35	45
Below 40	78
Below 45	89
Below 50	92
Below 55	98
Below 60	100

Answer 29. 

Class interval	Frequency	Cumulative frequency
15-20	2	2
20-25	4	6
25-30	18	24
30-35	21	45
35-40	33	78
40-45	11	89
45-50	3	92
50-55	6	98
55-60	2	100

Given as: n = 100 and n/2 = 50

Median class = 35-45

Now, l = 35, cf = 45, f = 33 & h = 5

Median = l + [{(n/2)-Cf}/f] x h

Median = 35+((50-45)/33) × 5

= 35 + (5/33)5

= 35.75

Thus, the median age is 35.75 years.

 

Question 30. The lengths of 40 leaves in the plant are measured correctly to the nearest millimetre, as well as the data obtained is represented in the following table:

Length (in mm)	Number of leaves
118-126	3
127-135	5
136-144	9
145-153	12
154-162	5
163-171	4
172-180	2

Find the median length of leaves.             

Answer 30. As the given data are not continuous, reduce 0.5 in the lower limit and add 0.5 to the upper limit.

Class Interval	Frequency	Cumulative frequency
117.5-126.5	3	3
126.5-135.5	5	8
135.5-144.5	9	17
144.5-153.5	12	29
153.5-162.5	5	34
162.5-171.5	4	38
171.5-180.5	2	40

Thus, the given data obtained are:

n = 40 and n/2 = 20

Median class = 144.5-153.5

Now, l = 144.5,

cf = 17, f = 12 & h = 9

Median = l + [{(n/2)-Cf}/f] x h

Median = 144.5+((20-17)/12)×9

= 144.5+(9/4)

= 146.75 mm

here, the median length of the leaves is 146.75 mm.

 

Question 31. The following table gives a distribution of a lifetime of 400 neon lamps.

Lifetime (in hours)	Number of lamps
1500-2000	14
2000-2500	56
2500-3000	60
3000-3500	86
3500-4000	74
4000-4500	62
4500-5000	48

Find the median lifetime of a lamp.

Answer 31. 

Class Interval	Frequency	Cumulative
1500-2000	14	14
2000-2500	56	70
2500-3000	60	130
3000-3500	86	216
3500-4000	74	290
4000-4500	62	352
4500-5000	48	400

Given as:

n = 400 &n/2 = 200

Median class = 3000 – 3500

Thus, l = 3000, Cf = 130,

f = 86 & h = 500

 

Median = l + [{(n/2)-Cf}/f] x h

Median = 3000 + ((200-130)/86) × 500

= 3000 + (35000/86)

= 3000 + 406.97

= 3406.97

So, the median lifetime of the lamps = 3406.97 hours.

 

Question 32. 100 surnames were randomly picked up from a local telephone directory and the frequency distribution of the number for letters in English alphabets in the surnames, was obtained as follow:

Number of the letters	1-4	4-7	7-10	10-13	13-16	16-19
Number of the surnames	6	30	40	16	4	4

Determine the number of median letters in the surnames. Find out the number of mean letters in the surnames and the size of the modal in the surnames.

Answer 32. For calculating the median:

Class Interval	Frequency	Cumulative Frequency
1-4	6	6
4-7	30	36
7-10	40	76
10-13	16	92
13-16	4	96
16-19	4	100

Given as:

n = 100 & n/2 = 50

Median class = 7-10

Thus, l = 7, Cf = 36, f = 40 & h = 3

Median = l + [{(n/2)-Cf}/f] x h

Median = 7+((50-36)/40) × 3

Median = 7+42/40

Median=8.05

Calculate the required Mode:

Modal class = 7-10,

we know, l = 7, f1 = 40, f0 = 30, f2 = 16 & h = 3

Mode = l + {(f1-f0)/(2f1-f0-f2)} x h

Mode = 7+((40-30)/(2×40-30-16)) × 3

= 7+(30/34)

= 7.88

Hence, mode = 7.88

Calculate the required mean:

Class Interval	fi	xi	fixi
1-4	6	2.5	15
4-7	30	5.5	165
7-10	40	8.5	340
10-13	16	11.5	184
13-16	4	14.5	51
16-19	4	17.5	70
	Sum fi = 100		Sum fixi = 825

Mean = x̄ = ∑fi xi /∑fi 

Mean = 825/100 = 8.25

Thus, the mean = 8.25

 

Question 33. The distributions below give a weight of 30 students in a class. Find the median weight for a student.

Weight(in kg)	40-45	45-50	50-55	55-60	60-65	65-70	70-75
Number for students	2	3	8	6	6	3	2

Answer 33. 

Class Interval	Frequency	Cumulative frequency
40-45	2	2
45-50	3	5
50-55	8	13
55-60	6	19
60-65	6	25
65-70	3	28
70-75	2	30

Given as: n = 30 and n/2= 15

In the Median class = 55-60

l = 55, Cf = 13, f = 6 & h = 5

Median = l + [{(n/2)-Cf}/f] x h

Median = 55+((15-13)/6)×5

Median=55 + (10/6) = 55+1.666

Median =56.67

So, the median weight for the students = 56.67

 

Question 34. The following distribution gives the daily income for 50 workers in a factory. Convert the distribution above to a less, type cumulative frequency distribution and draw its ogive.

Daily income in Rupees	100-120	120-140	140-160	160-180	180-200
Number of workers	12	14	8	6	10

Answer 34. On converting the given distribution table to a less than type cumulative frequency distribution, we get

Daily income	Frequency	Cumulative Frequency
Less than 120	12	12
Less than 140	14	26
Less than 160	8	34
Less than 180	6	40
Less than 200	10	50

From the given table, plot the points corresponding to the ordered pairs consisting of (120, 12), (140, 26), (160, 34), (180, 40) and (200, 50) on the required graph paper and the plotted given points are joined to get the smooth curve, and an obtained curve is called less than type of given curve.

 

Question 35. During the medical check-up of 35 students in a class, their weights were recorded as follows:

Weight in kg	Number of students
Less than 38	0
Less than 40	3
Less than 42	5
Less than 44	9
Less than 46	14
Less than 48	28
Less than 50	32
Less than 52	35

Draw a less than type ogive for the given data. Hence obtain the median weight from the graph and verify the result using the formula.

Answer 35. From given data, to represent the table in the required form of the graph, choose the upper limits of the given class intervals on the x-axis and frequencies on the y-axis by choosing the given convenience scale. Now plot the given points corresponding in the ordered pairs is given by (38, 0), (40, 3), (42, 5), (44, 9),(46, 14), (48, 28), (50, 32) as well as (52, 35) on the graph paper as well as join them to get the smooth curve. The curve obtained is called less than type given curve.

Locate the given point 17.5 on the y-axis and draw the line parallel to the x-axis cutting the curve at the point. From the given point, draw the perpendicular line to the x-axis. intersection point perpendicular to the x-axis is the data’s median. Now, for finding out the mode by making the table.

Class interval	Number of students(Frequency)	Cumulative Frequency
Less than 38	0	0
Less than 40	3-0=3	3
Less than 42	5-3=2	5
Less than 44	9-5=4	9
Less than 46	14-9=5	14
Less than 48	28-14=14	28
Less than 50	32-28=4	32
Less than 52	35-22=3	35

Class 46 – 48 has maximum frequency. Thus, this is the modal class

Where, l = 46, h = 2, f1= 14, f0= 5 and f2 = 4

The mode formula is given as:-

Now, mode = l +[(f1-f0)/2f1-f0-f2] x h

= 46 + 0.95 = 46.95

Hence, the model is verified.

 

Question 36. The following tables give the production yield per hectare of wheat of 100 farms.

Production Yield	50-55	55-60	60-65	65-70	70-75	75-80
Number of the farms	2	8	12	24	38	16

Change the distribution to a more than type distribution and draw its ogive.

Answer 36. On converting the given distribution to a more than type distribution, we get

Production Yield (kg/ha)	Number of the farms
More than or equal to the 50	100
More than or equal to the 55	100-2 = 98
More than or equal to the 60	98-8= 90
More than or equal to 65	90-12=78
More than or equal to 70	78-24=54
More than or equal to the 75	54-38 =16

From the given table obtained, draw and give by plotting the given corresponding points here, the upper limits in the x-axis as well as the frequencies obtained in y-axis are (50, 100), (55, 98), (60, 90), (65, 78), (70, 54) as well as (75, 16) on that graph paper. The graph obtained are called more than type given curve.

 

Question 37. The maximum bowling speeds, in km per hour, of the 33 players at a cricket coaching centre are given as follows:

Speed (km/h) 85-100 100-115 115-130 130-145

Number of the players 11 9 8 5

Calculates the median bowling speed.

Answer 37. First, we construct the given cumulative frequency table

Speed ( in km/h)	Number of the players	Cumulative frequency
85 – 100	11	11
100 – 115	9	11 + 9 = 20
115 – 130	8	20 + 8 = 28
130 – 145	5	28 + 5 = 33

It is given that n = 33

 n/2 = 33/2 = 16.5

Hence, the median class is 100 – 115.

here, lower limit(l) = 100

Frequency (f) = 9

Cumulative frequency (cf) = 11

As well as the given class width(h) = 15

Median = l + [{(n/2)-Cf}/f] x h

= 100 + (16.5-11)/9 x 15

= 100 + 5.5 x 15 / 9

= 100 + 82.5/9

= 100 + 9.17

= 109.17

thus, the median bowling speed is 109.17 km/h.

 

Question 38. Weekly income of the 600 families is tabulated below :

Weekly income Number of the families

(in Rs)

0-1000 250

1000-2000 190

2000-3000 100

3000-4000 40

4000-5000 15

5000-6000 5

Total 600

Computes the median income.

Answer 38. 

Weekly Income	Number of the families (fi)	Cumulative frequency (cf)
0-1000	250	250
1000-2000	190	250 + 190 = 400
2000-3000	100	440 + 100 = 540
3000-4000	40	540 + 40 = 580
4000-5000	15	580 + 15 = 595
5000-6000	5	595 + 5 = 600

As per the given question,

n = 600

So, cn/2 = 600/2 = 300

Cumulative frequency 440 lies in the interval 1000 – 2000.

As the given lower median class, l = 1000

f = 190,

cf = 250,

Class width, h = 1000

And for the total observation, n = 600

Median = l + [{(n/2)-Cf}/f] x h

= 1000 + (300-250)/190 x 1000

= 1000 + 50/190 x 1000

= 1000 + 5000/19

= 1000 + 263.15 = 1263.15

therefore, the median income is Rs.1263.15.

 

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