10th Maths Statistics Exercise 14.4 Solutions

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Exercise 14.4 Solutions – Class X Mathematics

Exercise 14.4 Solutions

From Class X Mathematics textbook by the State Council of Educational Research and Training, Telangana, Hyderabad

Question 1

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

Daily income (in Rupees)	250-300	300-350	350-400	400-450	450-500
Number of workers	12	14	8	6	10

Convert the distribution above to a less than type cumulative frequency distribution, and draw its ogive.

Solution:

Step 1: Convert to less than type cumulative frequency distribution:

Daily income less than (in Rupees)	300	350	400	450	500
Cumulative frequency	12	12+14=26	26+8=34	34+6=40	40+10=50

Question 2

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

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

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

Solution:

Step 1: The data is already in less than type cumulative frequency form.

Step 2: Drawing the ogive:

Ogive (Less than type) for weights of students

Step 3: Finding median from the graph:

Total number of students (n) = 35

Median position = n/2 = 17.5

From the graph, the x-coordinate corresponding to y=17.5 is approximately 46.5 kg.

Step 4: Verifying using the formula:

The median class is 46-48 (since 17.5 falls in the cumulative frequency of 28)

Using the formula:

\[ \text{Median} = L + \left(\frac{\frac{n}{2} – cf}{f}\right) \times h \]

Where:
L = 46 (lower limit of median class)
cf = 14 (cumulative frequency before median class)
f = 14 (frequency of median class)
h = 2 (class width)

\[ \text{Median} = 46 + \left(\frac{17.5 – 14}{14}\right) \times 2 = 46 + \left(\frac{3.5}{14}\right) \times 2 = 46 + 0.5 = 46.5 \text{ kg} \]

This matches our graphical estimate.

Question 3

The following table gives production yield per hectare of wheat of 100 farms of a village.

Production yield (Quintal/Hectare)	50-55	55-60	60-65	65-70	70-75	75-80
Number of farmers	2	8	12	24	38	16

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

Solution:

Step 1: Convert to more than type cumulative frequency distribution:

Production yield more than (Quintal/Hectare)	50	55	60	65	70	75
Cumulative frequency	100	100-2=98	98-8=90	90-12=78	78-24=54	54-38=16

Step 2: Drawing the ogive (more than type):

Ogive (More than type) for production yield

10th Maths Statistics Exercise 14.3 Solutions

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Exercise 14.3 Solutions – Class X Mathematics

Exercise 14.3 Solutions

Class X Mathematics – State Council of Educational Research and Training, Telangana, Hyderabad

1. Electricity Consumption Problem

Monthly consumption	65-85	85-105	105-125	125-145	145-165	165-185	185-205
Number of consumers	4	5	13	20	14	8	4

Solution:

Median:

First, we calculate cumulative frequencies:

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	8	64
185-205	4	68

Median class is where cumulative frequency ≥ n/2 = 34 → 125-145

Using median formula:

Median = L + [(n/2 – CF)/f] × h

Where L = 125, n = 68, CF = 22, f = 20, h = 20

Median = 125 + [(34 – 22)/20] × 20 = 125 + 12 = 137

Mean:

Using assumed mean method with A = 135:

Class	Midpoint (x)	Frequency (f)	d = (x – A)/h	f × d
65-85	75	4	-3	-12
85-105	95	5	-2	-10
105-125	115	13	-1	-13
125-145	135	20	0	0
145-165	155	14	1	14
165-185	175	8	2	16
185-205	195	4	3	12
Total				7

Mean = A + (Σfd/Σf) × h = 135 + (7/68) × 20 ≈ 137.06

Mode:

Modal class is 125-145 (highest frequency = 20)

Using mode formula:

Mode = L + [(f₁ – f₀)/(2f₁ – f₀ – f₂)] × h

Where L = 125, f₁ = 20, f₀ = 13, f₂ = 14, h = 20

Mode = 125 + [(20 – 13)/(40 – 13 – 14)] × 20 ≈ 125 + (7/13) × 20 ≈ 135.77

Comparison: Mean (137.06) > Median (137) > Mode (135.77)

2. Median Problem with Missing Frequencies

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

Given median = 28.5, n = 60

Solution:

Total observations: 5 + x + 20 + 15 + y + 5 = 60 ⇒ x + y = 15

Median class is where cumulative frequency ≥ n/2 = 30 → 20-30

Using median formula:

28.5 = 20 + [(30 – (5 + x))/20] × 10

8.5 = (25 – x)/2 ⇒ 17 = 25 – x ⇒ x = 8

Since x + y = 15 ⇒ y = 7

Solution: x = 8, y = 7

3. Insurance Policy Holders Age Distribution

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

Solution:

First convert to frequency distribution:

Class	Frequency	Cumulative Frequency
18-20	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

Median class is where cumulative frequency ≥ n/2 = 50 → 35-40

Using median formula:

Median = 35 + [(50 – 45)/33] × 5 ≈ 35 + 0.76 ≈ 35.76 years

4. Leaves Length Measurement

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

Solution:

Convert to continuous classes as suggested:

Class	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

Median class is where cumulative frequency ≥ n/2 = 20 → 144.5-153.5

Using median formula:

Median = 144.5 + [(20 – 17)/12] × 9 = 144.5 + 2.25 = 146.75 mm

5. Neon Lamps Life Time

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

Solution:

Calculate cumulative frequencies:

Class	Frequency	Cumulative Frequency
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

Median class is where cumulative frequency ≥ n/2 = 200 → 3000-3500

Using median formula:

Median = 3000 + [(200 – 130)/86] × 500 ≈ 3000 + 406.98 ≈ 3406.98 hours

6. Surnames Letters Distribution

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

Solution:

Median:

First make classes continuous and calculate cumulative frequencies:

Class	Frequency	Cumulative Frequency
0.5-4.5	6	6
4.5-7.5	30	36
7.5-10.5	40	76
10.5-13.5	16	92
13.5-16.5	4	96
16.5-19.5	4	100

Median class is where cumulative frequency ≥ n/2 = 50 → 7.5-10.5

Median = 7.5 + [(50 – 36)/40] × 3 = 7.5 + 1.05 = 8.55

Mean:

Using midpoint method:

Class	Midpoint (x)	Frequency (f)	f × x
1-4	2.5	6	15
4-7	5.5	30	165
7-10	8.5	40	340
10-13	11.5	16	184
13-16	14.5	4	58
16-19	17.5	4	70
Total		100	832

Mean = Σfx/Σf = 832/100 = 8.32

Mode:

Modal class is 7-10 (highest frequency = 40)

Mode = L + [(f₁ – f₀)/(2f₁ – f₀ – f₂)] × h

Where L = 7, f₁ = 40, f₀ = 30, f₂ = 16, h = 3

Mode = 7 + [(40 – 30)/(80 – 30 – 16)] × 3 ≈ 7 + (10/34) × 3 ≈ 7.88

7. Students Weight Distribution

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

Solution:

Calculate cumulative frequencies:

Class	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

Median class is where cumulative frequency ≥ n/2 = 15 → 55-60

Using median formula:

Median = 55 + [(15 – 13)/6] × 5 ≈ 55 + 1.67 ≈ 56.67 kg

10th Maths Statistics Exercise 14.2 Solutions

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Exercise 14.2 Solutions – Class X Mathematics

Exercise 14.2 Solutions

Class X Mathematics
State Council of Educational Research and Training, Telangana, Hyderabad

Problem 1

The following table shows the ages of the patients admitted in a hospital on a particular day:

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

Find the mode and the mean of the data given above. Compare and interpret the two measures of central tendency.

Solution:

Mode Calculation:

The modal class is the class with the highest frequency. Here, the highest frequency is 23, which corresponds to the class 35-45.

Using the mode formula:

Mode = L + [(f₁ – f₀)/(2f₁ – f₀ – f₂)] × h

Where:
L = lower limit of modal class = 35
f₁ = frequency of modal class = 23
f₀ = frequency of class preceding modal class = 21
f₂ = frequency of class succeeding modal class = 14
h = class width = 10

Mode = 35 + [(23 – 21)/(2×23 – 21 – 14)] × 10
= 35 + [2/(46 – 35)] × 10
= 35 + (2/11) × 10
= 35 + 1.818 ≈ 36.82

Mean Calculation:

Class Interval	Midpoint (x_i)	Frequency (f_i)	f_ix_i
5-15	10	6	60
15-25	20	11	220
25-35	30	21	630
35-45	40	23	920
45-55	50	14	700
55-65	60	5	300
Total		80	2830

Mean = \(\frac{\sum f_ix_i}{\sum f_i} = \frac{2830}{80} = 35.375\)

Comparison and Interpretation:

Mode = 36.82 years, Mean = 35.38 years

The mode (36.82) is slightly higher than the mean (35.38), indicating that the most common age group of patients is slightly older than the average age of all patients. Both values suggest that most patients admitted are in their mid-30s to mid-40s.

Problem 2

The following data gives the information on the observed life times (in hours) of 225 electrical components:

Lifetimes (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.

Solution:

The modal class is the class with the highest frequency. Here, the highest frequency is 61, which corresponds to the class 60-80.

Using the mode formula:

Mode = L + [(f₁ – f₀)/(2f₁ – f₀ – f₂)] × h

Where:
L = lower limit of modal class = 60
f₁ = frequency of modal class = 61
f₀ = frequency of class preceding modal class = 52
f₂ = frequency of class succeeding modal class = 38
h = class width = 20

Mode = 60 + [(61 – 52)/(2×61 – 52 – 38)] × 20
= 60 + [9/(122 – 90)] × 20
= 60 + (9/32) × 20
= 60 + 5.625 = 65.625

Modal lifetime of components = 65.63 hours

Problem 3

The following data gives the distribution of total monthly household expenditure of 200 families of Gummadidala village. Find the modal monthly expenditure of the families. Also, find the mean monthly expenditure:

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

Solution:

Mode Calculation:

The modal class is the class with the highest frequency. Here, the highest frequency is 40, which corresponds to the class 1500-2000.

Using the mode formula:

Mode = L + [(f₁ – f₀)/(2f₁ – f₀ – f₂)] × h

Where:
L = lower limit of modal class = 1500
f₁ = frequency of modal class = 40
f₀ = frequency of class preceding modal class = 24
f₂ = frequency of class succeeding modal class = 33
h = class width = 500

Mode = 1500 + [(40 – 24)/(2×40 – 24 – 33)] × 500
= 1500 + [16/(80 – 57)] × 500
= 1500 + (16/23) × 500
= 1500 + 347.83 ≈ 1847.83

Mean Calculation:

We’ll use the step-deviation method with assumed mean a = 2750 and class width h = 500

Class Interval	Midpoint (x_i)	Frequency (f_i)	d_i = x_i – a	u_i = d_i/h	f_iu_i
1000-1500	1250	24	-1500	-3	-72
1500-2000	1750	40	-1000	-2	-80
2000-2500	2250	33	-500	-1	-33
2500-3000	2750 (a)	28	0	0	0
3000-3500	3250	30	500	1	30
3500-4000	3750	22	1000	2	44
4000-4500	4250	16	1500	3	48
4500-5000	4750	7	2000	4	28
Total					-35

Mean = \(a + \left(\frac{\sum f_iu_i}{\sum f_i}\right) \times h = 2750 + \left(\frac{-35}{200}\right) \times 500 = 2750 – 87.5 = 2662.5\)

Modal monthly expenditure = ₹1847.83
Mean monthly expenditure = ₹2662.50

Problem 4

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

Number of students per teacher	15-20	20-25	25-30	30-35	35-40	40-45	45-50	50-55
Number of States	3	8	9	10	3	0	0	2

Solution:

Mode Calculation:

The modal class is the class with the highest frequency. Here, the highest frequency is 10, which corresponds to the class 30-35.

Using the mode formula:

Mode = L + [(f₁ – f₀)/(2f₁ – f₀ – f₂)] × h

Where:
L = lower limit of modal class = 30
f₁ = frequency of modal class = 10
f₀ = frequency of class preceding modal class = 9
f₂ = frequency of class succeeding modal class = 3
h = class width = 5

Mode = 30 + [(10 – 9)/(2×10 – 9 – 3)] × 5
= 30 + [1/(20 – 12)] × 5
= 30 + (1/8) × 5
= 30 + 0.625 = 30.625

Mean Calculation:

Class Interval	Midpoint (x_i)	Frequency (f_i)	f_ix_i
15-20	17.5	3	52.5
20-25	22.5	8	180
25-30	27.5	9	247.5
30-35	32.5	10	325
35-40	37.5	3	112.5
40-45	42.5	0	0
45-50	47.5	0	0
50-55	52.5	2	105
Total		35	1022.5

Mean = \(\frac{\sum f_ix_i}{\sum f_i} = \frac{1022.5}{35} \approx 29.21\)

Interpretation:

Mode = 30.62 students per teacher
Mean = 29.21 students per teacher

The mode (30.62) is slightly higher than the mean (29.21), indicating that the most common student-teacher ratio is slightly higher than the average ratio across all states. This suggests that while most states have about 30-31 students per teacher, some states with lower ratios bring the average down slightly.

Problem 5

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

Runs	3000-4000	4000-5000	5000-6000	6000-7000	7000-8000	8000-9000	9000-10000	10000-11000
Number of batsmen	4	18	9	7	6	3	1	1

Find the mode of the data.

Solution:

The modal class is the class with the highest frequency. Here, the highest frequency is 18, which corresponds to the class 4000-5000.

Using the mode formula:

Mode = L + [(f₁ – f₀)/(2f₁ – f₀ – f₂)] × h

Where:
L = lower limit of modal class = 4000
f₁ = frequency of modal class = 18
f₀ = frequency of class preceding modal class = 4
f₂ = frequency of class succeeding modal class = 9
h = class width = 1000

Mode = 4000 + [(18 – 4)/(2×18 – 4 – 9)] × 1000
= 4000 + [14/(36 – 13)] × 1000
= 4000 + (14/23) × 1000
= 4000 + 608.70 ≈ 4608.70

Modal runs scored by batsmen = 4608.70 runs

Problem 6

A student noted the number of cars passing through a spot on a road for 100 periods, each of 3 minutes, and summarised this in the table given below.

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

Find the mode of the data.

Solution:

The modal class is the class with the highest frequency. Here, the highest frequency is 20, which corresponds to the class 40-50.

Using the mode formula:

Mode = L + [(f₁ – f₀)/(2f₁ – f₀ – f₂)] × h

Where:
L = lower limit of modal class = 40
f₁ = frequency of modal class = 20
f₀ = frequency of class preceding modal class = 12
f₂ = frequency of class succeeding modal class = 11
h = class width = 10

Mode = 40 + [(20 – 12)/(2×20 – 12 – 11)] × 10
= 40 + [8/(40 – 23)] × 10
= 40 + (8/17) × 10
= 40 + 4.71 ≈ 44.71

Mode number of cars passing = 44.71 cars per 3-minute period

10th Maths Statistics Exercise 14.1 Solutions

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Exercise 14.1 Solutions – Class X Mathematics

Exercise 14.1 Solutions

Class X Mathematics
State Council of Educational Research and Training, Telangana, Hyderabad

Problem 1

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

Number of plants	0 – 2	2 – 4	4 – 6	6 – 8	8 – 10	10 – 12	12 – 14
Number of houses	1	2	1	5	6	2	3

Solution:

To find the mean, we’ll use the midpoint of each class interval and multiply by frequency.

Class Interval	Midpoint (x_i)	Frequency (f_i)	f_ix_i
0-2	1	1	1
2-4	3	2	6
4-6	5	1	5
6-8	7	5	35
8-10	9	6	54
10-12	11	2	22
12-14	13	3	39
Total		20	162

Mean = \(\frac{\sum f_ix_i}{\sum f_i} = \frac{162}{20} = 8.1\)

Mean number of plants per house = 8.1

Problem 2

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

Daily wages in Rupees	200 – 250	250 – 300	300 – 350	350 – 400	400 – 450
Number of workers	12	14	8	6	10

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

Solution:

We’ll use the step-deviation method with assumed mean a = 325 and class width h = 50

Class Interval	Midpoint (x_i)	Frequency (f_i)	d_i = x_i – a	u_i = d_i/h	f_iu_i
200-250	225	12	-100	-2	-24
250-300	275	14	-50	-1	-14
300-350	325 (a)	8	0	0	0
350-400	375	6	50	1	6
400-450	425	10	100	2	20
Total					-12

Mean = \(a + \left(\frac{\sum f_iu_i}{\sum f_i}\right) \times h = 325 + \left(\frac{-12}{50}\right) \times 50 = 325 – 12 = 313\)

Mean daily wages = ₹313

Problem 3

The following distribution shows the daily pocket allowance of children of a locality. The mean pocket allowance is ₹18. Find the missing frequency \( f \).

Daily pocket allowance (in Rupees)	11 – 13	13 – 15	15 – 17	17 – 19	19 – 21	21 – 23	23 – 25
Number of children	7	6	9	13	\( f \)	5	4

Solution:

Let’s calculate the sum of frequencies and products of midpoints and frequencies.

Class Interval	Midpoint (x_i)	Frequency (f_i)	f_ix_i
11-13	12	7	84
13-15	14	6	84
15-17	16	9	144
17-19	18	13	234
19-21	20	f	20f
21-23	22	5	110
23-25	24	4	96
Total		44 + f	752 + 20f

Given mean = 18

\(\frac{752 + 20f}{44 + f} = 18\)

752 + 20f = 792 + 18f

2f = 40

f = 20

The missing frequency \( f = 20 \)

Problem 4

Thirty women were examined in a hospital by a doctor and their heart beats per minute were recorded and summarised as shown. Find the mean heart beats per minute for these women, choosing a suitable method.

Number of heart beats/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

Solution:

We’ll use the direct method to find the mean.

Class Interval	Midpoint (x_i)	Frequency (f_i)	f_ix_i
65-68	66.5	2	133
68-71	69.5	4	278
71-74	72.5	3	217.5
74-77	75.5	8	604
77-80	78.5	7	549.5
80-83	81.5	4	326
83-86	84.5	2	169
Total		30	2277

Mean = \(\frac{\sum f_ix_i}{\sum f_i} = \frac{2277}{30} = 75.9\)

Mean heart beats per minute = 75.9

Problem 5

In a retail market, fruit vendors were selling oranges kept in packing baskets. These baskets contained varying number of oranges. The following was the distribution of oranges.

Number of oranges	10-14	15-19	20-24	25-29	30-34
Number of baskets	15	110	135	115	25

Find the mean number of oranges kept in each basket. Which method of finding the mean did you choose?

Solution:

We’ll use the step-deviation method with assumed mean a = 22 and class width h = 5

Class Interval	Midpoint (x_i)	Frequency (f_i)	d_i = x_i – a	u_i = d_i/h	f_iu_i
10-14	12	15	-10	-2	-30
15-19	17	110	-5	-1	-110
20-24	22 (a)	135	0	0	0
25-29	27	115	5	1	115
30-34	32	25	10	2	50
Total					25

Mean = \(a + \left(\frac{\sum f_iu_i}{\sum f_i}\right) \times h = 22 + \left(\frac{25}{400}\right) \times 5 = 22 + \frac{125}{400} = 22 + 0.3125 = 22.3125\)

Mean number of oranges per basket = 22.31 (approx)

We chose the step-deviation method because the class intervals are equal and the frequencies are large.

Problem 6

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

Daily expenditure (in Rupees)	100-150	150-200	200-250	250-300	300-350
Number of households	4	5	12	2	2

Find the mean daily expenditure on food by a suitable method.

Solution:

We’ll use the direct method to find the mean.

Class Interval	Midpoint (x_i)	Frequency (f_i)	f_ix_i
100-150	125	4	500
150-200	175	5	875
200-250	225	12	2700
250-300	275	2	550
300-350	325	2	650
Total		25	5275

Mean = \(\frac{\sum f_ix_i}{\sum f_i} = \frac{5275}{25} = 211\)

Mean daily expenditure on food = ₹211

Problem 7

To find out the concentration of SO₂ 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:

Concentration of SO₂ in ppm	0.00-0.04	0.04-0.08	0.08-0.12	0.12-0.16	0.16-0.20	0.20-0.24
Frequency	4	9	9	2	4	2

Find the mean concentration of SO₂ in the air.

Solution:

We’ll use the direct method to find the mean.

Class Interval	Midpoint (x_i)	Frequency (f_i)	f_ix_i
0.00-0.04	0.02	4	0.08
0.04-0.08	0.06	9	0.54
0.08-0.12	0.10	9	0.90
0.12-0.16	0.14	2	0.28
0.16-0.20	0.18	4	0.72
0.20-0.24	0.22	2	0.44
Total		30	2.96

Mean = \(\frac{\sum f_ix_i}{\sum f_i} = \frac{2.96}{30} \approx 0.0987\)

Mean concentration of SO₂ = 0.099 ppm (approx)

Problem 8

A class teacher has the following attendance record of 40 students of a class for the whole term. Find the mean number of days a student was present out of 56 days in the term.

Number of days	35-38	38-41	41-44	44-47	47-50	50-53	53-56
Number of students	1	3	4	4	7	10	11

Solution:

We’ll use the direct method to find the mean.

Class Interval	Midpoint (x_i)	Frequency (f_i)	f_ix_i
35-38	36.5	1	36.5
38-41	39.5	3	118.5
41-44	42.5	4	170
44-47	45.5	4	182
47-50	48.5	7	339.5
50-53	51.5	10	515
53-56	54.5	11	599.5
Total		40	1961

Mean = \(\frac{\sum f_ix_i}{\sum f_i} = \frac{1961}{40} = 49.025\)

Mean number of days present = 49.03 days (approx)

Problem 9

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

Literacy rate in %	45-55	55-65	65-75	75-85	85-95
Number of cities	3	10	11	8	3

Solution:

We’ll use the direct method to find the mean.

Class Interval	Midpoint (x_i)	Frequency (f_i)	f_ix_i
45-55	50	3	150
55-65	60	10	600
65-75	70	11	770
75-85	80	8	640
85-95	90	3	270
Total		35	2430

Mean = \(\frac{\sum f_ix_i}{\sum f_i} = \frac{2430}{35} \approx 69.43\)

Mean literacy rate = 69.43%