TG 10th Statistics

Statistics

10th Maths Statistics Exercise 14.4 Solutions

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

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:

ClassFrequencyCumulative Frequency
65-8544
85-10559
105-1251322
125-1452042
145-1651456
165-185864
185-205468

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:

ClassMidpoint (x)Frequency (f)d = (x – A)/hf × d
65-85754-3-12
85-105955-2-10
105-12511513-1-13
125-1451352000
145-16515514114
165-1851758216
185-2051954312
Total7

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:

ClassFrequencyCumulative Frequency
18-2022
20-2546
25-301824
30-352145
35-403378
40-451189
45-50392
50-55698
55-602100

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:

ClassFrequencyCumulative Frequency
117.5-126.533
126.5-135.558
135.5-144.5917
144.5-153.51229
153.5-162.5534
162.5-171.5438
171.5-180.5240

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:

ClassFrequencyCumulative Frequency
1500-20001414
2000-25005670
2500-300060130
3000-350086216
3500-400074290
4000-450062352
4500-500048400

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:

ClassFrequencyCumulative Frequency
0.5-4.566
4.5-7.53036
7.5-10.54076
10.5-13.51692
13.5-16.5496
16.5-19.54100

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:

ClassMidpoint (x)Frequency (f)f × x
1-42.5615
4-75.530165
7-108.540340
10-1311.516184
13-1614.5458
16-1917.5470
Total100832

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:

ClassFrequencyCumulative Frequency
40-4522
45-5035
50-55813
55-60619
60-65625
65-70328
70-75230

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

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 (xi) Frequency (fi) fixi
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 (xi) Frequency (fi) di = xi – a ui = di/h fiui
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 (xi) Frequency (fi) fixi
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

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 (xi) Frequency (fi) fixi
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 (xi) Frequency (fi) di = xi – a ui = di/h fiui
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 (xi) Frequency (fi) fixi
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 (xi) Frequency (fi) fixi
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 (xi) Frequency (fi) di = xi – a ui = di/h fiui
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 (xi) Frequency (fi) fixi
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 (xi) Frequency (fi) fixi
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 (xi) Frequency (fi) fixi
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 (xi) Frequency (fi) fixi
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%