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STATISTICS - Q&A

EXERCISE 13.1

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 number of plants, we first find the class mark (xi) for each interval.
Class mark (xi) = (Upper limit + Lower limit) / 2

We can use the direct method: Mean = Σfixi / Σfi

Table:
Class Interval | Frequency (fi) | Class Mark (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
Total | Σfi = 20 | | Σfixi = 162

Mean = 162 / 20 = 8.1
Answer: The mean number of plants per house is 8.1.
(Direct method was used because the numerical values of fi and xi are small).


2. Consider the following distribution of daily wages of 50 workers of a factory. Find the mean daily wages of the workers of the factory by using an appropriate method.

Daily wages (in ₹): 500-520, 520-540, 540-560, 560-580, 580-600
Number of workers: 12, 14, 8, 6, 10

Solution:
Let us use the Step-Deviation Method to simplify calculations.
Assumed Mean (a) = 550 (middle xi value)
Class height (h) = 20

Table:
Class Interval | fi | xi | ui = (xi - a)/h | fiui
---|---|---|---|---
500 - 520 | 12 | 510 | -2 | -24
520 - 540 | 14 | 530 | -1 | -14
540 - 560 | 8 | 550 | 0 | 0
560 - 580 | 6 | 570 | 1 | 6
580 - 600 | 10 | 590 | 2 | 20
Total | 50 | | | -12

Mean = a + (Σfiui / Σfi) × h
Mean = 550 + (-12 / 50) × 20
Mean = 550 + (-12 / 5) × 2
Mean = 550 - 4.8 = 545.2
Answer: The mean daily wages of the workers is ₹ 545.20.


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 ₹): 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:
Given Mean = 18.
Using Direct Method:

Table:
Class Interval | fi | xi | fixi
---|---|---|---
11 - 13 | 7 | 12 | 84
13 - 15 | 6 | 14 | 84
15 - 17 | 9 | 16 | 144
17 - 19 | 13 | 18 | 234
19 - 21 | f | 20 | 20f
21 - 23 | 5 | 22 | 110
23 - 25 | 4 | 24 | 96
Total | 44 + f | | 752 + 20f

Mean = Σfixi / Σfi
18 = (752 + 20f) / (44 + f)
18(44 + f) = 752 + 20f
792 + 18f = 752 + 20f
20f - 18f = 792 - 752
2f = 40
f = 20
Answer: The missing frequency f is 20.


4. Thirty women were examined in a hospital by a doctor and the number of heartbeats per minute were recorded and summarised as follows. Find 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

Solution:
Using Assumed Mean Method.
Assumed Mean (a) = 75.5
Class height (h) = 3

Table:
Class Interval | fi | xi | di = xi - a | fidi
---|---|---|---|---
65 - 68 | 2 | 66.5 | -9 | -18
68 - 71 | 4 | 69.5 | -6 | -24
71 - 74 | 3 | 72.5 | -3 | -9
74 - 77 | 8 | 75.5 | 0 | 0
77 - 80 | 7 | 78.5 | 3 | 21
80 - 83 | 4 | 81.5 | 6 | 24
83 - 86 | 2 | 84.5 | 9 | 18
Total | 30 | | | 12

Mean = a + (Σfidi / Σfi)
Mean = 75.5 + (12 / 30)
Mean = 75.5 + 0.4 = 75.9
Answer: The mean heartbeats per minute is 75.9.


5. In a retail market, fruit vendors were selling mangoes kept in packing boxes. These boxes contained varying number of mangoes. The following was the distribution of mangoes according to the number of boxes. Find the mean number of mangoes kept in a packing box. Which method of finding the mean did you choose?

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

Solution:
The class intervals are not continuous. We make them continuous by subtracting 0.5 from the lower limit and adding 0.5 to the upper limit.
New intervals: 49.5-52.5, 52.5-55.5, etc. Note that class marks (xi) remain the same.
Using Step-Deviation Method.
Assumed Mean (a) = 57
Height (h) = 3

Table:
Class Interval | fi | xi | ui = (xi - a)/h | fiui
---|---|---|---|---
49.5 - 52.5 | 15 | 51 | -2 | -30
52.5 - 55.5 | 110 | 54 | -1 | -110
55.5 - 58.5 | 135 | 57 | 0 | 0
58.5 - 61.5 | 115 | 60 | 1 | 115
61.5 - 64.5 | 25 | 63 | 2 | 50
Total | 400 | | | 25

Mean = a + (Σfiui / Σfi) × h
Mean = 57 + (25 / 400) × 3
Mean = 57 + (1 / 16) × 3
Mean = 57 + 0.1875 = 57.19 (approx)
Answer: The mean number of mangoes is 57.19. The step-deviation method was chosen to simplify calculations involving large frequencies.


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

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

Solution:
Using Step-Deviation Method.
Assumed Mean (a) = 225
Height (h) = 50

Table:
Class Interval | fi | xi | ui | fiui
---|---|---|---|---
100 - 150 | 4 | 125 | -2 | -8
150 - 200 | 5 | 175 | -1 | -5
200 - 250 | 12 | 225 | 0 | 0
250 - 300 | 2 | 275 | 1 | 2
300 - 350 | 2 | 325 | 2 | 4
Total | 25 | | | -7

Mean = 225 + (-7 / 25) × 50
Mean = 225 + (-7 × 2)
Mean = 225 - 14 = 211
Answer: The mean daily expenditure is ₹ 211.


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: Find the mean concentration of SO2 in the air.

Concentration: 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

Solution:
Using Direct Method.
Table:
Class Interval | fi | 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 | 30 | | 2.96

Mean = 2.96 / 30 = 0.0986... = 0.099 ppm (approx)
Answer: The mean concentration of SO2 is 0.099 ppm.


8. A class teacher has the following absentee record of 40 students of a class for the whole term. Find 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 students: 11, 10, 7, 4, 4, 3, 1

Solution:
Class sizes are unequal. We use the Direct Method.
Table:
Class Interval | fi | 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
Total | 40 | | 499

Mean = 499 / 40 = 12.475
Answer: The mean number of days is 12.48 days (approx).


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:
Using Step-Deviation Method.
Assumed Mean (a) = 70, h = 10.
Table:
Class Interval | fi | xi | ui | fiui
---|---|---|---|---
45 - 55 | 3 | 50 | -2 | -6
55 - 65 | 10 | 60 | -1 | -10
65 - 75 | 11 | 70 | 0 | 0
75 - 85 | 8 | 80 | 1 | 8
85 - 95 | 3 | 90 | 2 | 6
Total | 35 | | | -2

Mean = 70 + (-2 / 35) × 10
Mean = 70 - 4/7 = 70 - 0.57 = 69.43
Answer: The mean literacy rate is 69.43%.

EXERCISE 13.2

1. The following table shows the ages of the patients admitted in a hospital during a year: Find the mode and the mean of the data given above. Compare and interpret the two measures of central tendency.

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

Solution:
Mode:
Maximum frequency is 23. Modal class is 35 - 45.
l = 35, f1 = 23, f0 = 21, f2 = 14, h = 10.
Mode = l + [(f1 - f0) / (2f1 - f0 - f2)] × h
Mode = 35 + [(23 - 21) / (46 - 21 - 14)] × 10
Mode = 35 + [2 / 11] × 10 = 35 + 1.8 = 36.8 years.

Mean:
Using assumed mean a = 30, h = 10.
xi: 10, 20, 30, 40, 50, 60.
fi: 6, 11, 21, 23, 14, 5. Total = 80.
ui: -2, -1, 0, 1, 2, 3.
fiui: -12, -11, 0, 23, 28, 15. Total = 43.
Mean = 30 + (43/80) × 10 = 30 + 5.375 = 35.375 = 35.37 years.

Answer: Mode = 36.8 years, Mean = 35.37 years.
Interpretation: Maximum number of patients are 36.8 years old, while the average age of patients admitted is 35.37 years.


2. The following data gives the information on the observed lifetimes (in hours) of 225 electrical components: Determine the modal lifetimes of the components.

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

Solution:
Maximum frequency is 61. Modal class is 60 - 80.
l = 60, f1 = 61, f0 = 52, f2 = 38, h = 20.
Mode = l + [(f1 - f0) / (2f1 - f0 - f2)] × h
Mode = 60 + [(61 - 52) / (122 - 52 - 38)] × 20
Mode = 60 + [9 / 32] × 20
Mode = 60 + [9 × 5 / 8] = 60 + 5.625
Mode = 65.625
Answer: The modal lifetime is 65.625 hours.


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

Expenditure: 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:
Maximum freq = 40. Modal class = 1500 - 2000.
l = 1500, f1 = 40, f0 = 24, f2 = 33, h = 500.
Mode = 1500 + [(40 - 24) / (80 - 24 - 33)] × 500
Mode = 1500 + [16 / 23] × 500
Mode = 1500 + 347.83 = 1847.83

Mean:
Using Step-Deviation (a=2750, h=500).
ui: -3, -2, -1, 0, 1, 2, 3, 4.
fi: 24, 40, 33, 28, 30, 22, 16, 7.
fiui: -72, -80, -33, 0, 30, 44, 48, 28. Total = -35.
Mean = 2750 + (-35 / 200) × 500
Mean = 2750 - 87.50 = 2662.50
Answer: Modal expenditure = ₹ 1847.83, Mean expenditure = ₹ 2662.50.


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.

Students per teacher: 15-20, 20-25, 25-30, 30-35, 35-40, 40-45, 45-50, 50-55
Number of states/UT: 3, 8, 9, 10, 3, 0, 0, 2

Solution:
Mode:
Max freq = 10. Modal class = 30 - 35.
l = 30, f1 = 10, f0 = 9, f2 = 3, h = 5.
Mode = 30 + [(10 - 9) / (20 - 9 - 3)] × 5
Mode = 30 + [1 / 8] × 5 = 30 + 0.625 = 30.6 (approx).

Mean:
Using Step-Deviation (a=32.5, h=5).
ui: -3, -2, -1, 0, 1, 2, 3, 4.
fi: 3, 8, 9, 10, 3, 0, 0, 2. Total = 35.
fiui: -9, -16, -9, 0, 3, 0, 0, 8. Total = -23.
Mean = 32.5 + (-23 / 35) × 5 = 32.5 - 3.28 = 29.22 (approx).
Answer: Mode = 30.6, Mean = 29.2.
Interpretation: Most states have a teacher-student ratio of 30.6, while the average ratio is 29.2.


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

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

Solution:
Max freq = 18. Modal class = 4000 - 5000.
l = 4000, f1 = 18, f0 = 4, f2 = 9, h = 1000.
Mode = 4000 + [(18 - 4) / (36 - 4 - 9)] × 1000
Mode = 4000 + [14 / 23] × 1000
Mode = 4000 + 608.7 = 4608.7
Answer: The mode is 4608.7 runs.


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

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

Solution:
Max freq = 20. Modal class = 40 - 50.
l = 40, f1 = 20, f0 = 12, f2 = 11, h = 10.
Mode = 40 + [(20 - 12) / (40 - 12 - 11)] × 10
Mode = 40 + [8 / 17] × 10
Mode = 40 + 4.7 = 44.7
Answer: The mode is 44.7 cars.

EXERCISE 13.3

1. The following frequency distribution gives the monthly consumption of electricity of 68 consumers of a locality. Find the median, mean and mode of the data and compare them.

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: N = 68, N/2 = 34.
Cumulative frequencies: 4, 9, 22, 42, 56, 64, 68.
Median class (cf > 34) is 125 - 145.
l = 125, cf = 22, f = 20, h = 20.
Median = 125 + [(34 - 22) / 20] × 20 = 125 + 12 = 137 units.

Mean: Assumed mean a = 135.
fiui: -12, -10, -13, 0, 14, 16, 12. Total = 7.
Mean = 135 + (7/68) × 20 = 135 + 2.05 = 137.05 units.

Mode: Max freq = 20. Modal class = 125 - 145.
Mode = 125 + [(20 - 13) / (40 - 13 - 14)] × 20
Mode = 125 + (7/13) × 20 = 125 + 10.76 = 135.76 units.

Answer: Median = 137, Mean = 137.05, Mode = 135.76. All three measures are approximately equal.


2. If the median of the distribution given below is 28.5, find the values of x and y.

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

Solution:
Total frequency n = 60.
Sum of frequencies = 5 + x + 20 + 15 + y + 5 = 45 + x + y.
So, 45 + x + y = 60 ⇒ x + y = 15 ... (i)
Median is 28.5, which lies in class 20 - 30.
l = 20, f = 20, cf = 5 + x, h = 10, n/2 = 30.
Median = l + [(n/2 - cf) / f] × h
28.5 = 20 + [(30 - (5 + x)) / 20] × 10
8.5 = (25 - x) / 2
17 = 25 - x
x = 8.
Substitute x in (i): 8 + y = 15 ⇒ y = 7.
Answer: x = 8, y = 7.


3. A life insurance agent found the following data for distribution of ages of 100 policy holders. Calculate the median age, if policies are given only to persons having age 18 years onwards but less than 60 year.

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:
Convert 'Below' types to class intervals.
Class Intervals: 15-20, 20-25, 25-30, 30-35, 35-40, 40-45, 45-50, 50-55, 55-60.
Frequencies: 2, 4 (6-2), 18 (24-6), 21 (45-24), 33 (78-45), 11 (89-78), 3 (92-89), 6 (98-92), 2 (100-98).
N = 100, N/2 = 50.
Median class (cf > 50) is 35 - 40.
l = 35, cf = 45, f = 33, h = 5.
Median = 35 + [(50 - 45) / 33] × 5
Median = 35 + (5 / 33) × 5 = 35 + 25 / 33 = 35 + 0.76 = 35.76 years.
Answer: The median age is 35.76 years.


4. The lengths of 40 leaves of a plant are measured correct to the nearest millimetre, and the data obtained is represented in the following table. Find the median length of the leaves.

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:
Classes are discontinuous. Convert to continuous by subtracting 0.5 from lower limit and adding 0.5 to upper limit.
Continuous classes: 117.5-126.5, 126.5-135.5, ...
N = 40, N/2 = 20.
Cumulative frequencies: 3, 8, 17, 29, 34, 38, 40.
Median class is 144.5 - 153.5 (cf = 17, f = 12).
l = 144.5, h = 9, cf = 17, f = 12.
Median = 144.5 + [(20 - 17) / 12] × 9
Median = 144.5 + (3/12) × 9 = 144.5 + 2.25 = 146.75 mm.
Answer: The median length of the leaves is 146.75 mm.


5. The following table gives the distribution of the life time of 400 neon lamps. Find the median life time of a lamp.

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:
N = 400, N/2 = 200.
Cumulative frequencies: 14, 70, 130, 216, 290, 352, 400.
Median class is 3000 - 3500.
l = 3000, cf = 130, f = 86, h = 500.
Median = 3000 + [(200 - 130) / 86] × 500
Median = 3000 + (70 / 86) × 500 = 3000 + 406.98 = 3406.98 hours.
Answer: The median lifetime is 3406.98 hours.


6. 100 surnames were randomly picked up from a local telephone directory and the frequency distribution of the number of letters in the English alphabets in the surnames was obtained as follows. Determine the median number of letters in the surnames. Find the mean number of letters in the surnames? Also, find the modal size of the surnames.

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: N = 100, N/2 = 50.
CF: 6, 36, 76, 92, 96, 100.
Median Class: 7 - 10.
l = 7, cf = 36, f = 40, h = 3.
Median = 7 + [(50 - 36) / 40] × 3 = 7 + (14/40) × 3 = 7 + 1.05 = 8.05.

Mean: xi: 2.5, 5.5, 8.5, 11.5, 14.5, 17.5.
fixi: 15, 165, 340, 184, 58, 70. Total = 832.
Mean = 832 / 100 = 8.32.

Mode: Max freq = 40. Class 7 - 10.
Mode = 7 + [(40 - 30) / (80 - 30 - 16)] × 3
Mode = 7 + (10 / 34) × 3 = 7 + 0.88 = 7.88.

Answer: Median = 8.05, Mean = 8.32, Mode = 7.88.


7. The distribution below gives the weights of 30 students of a class. Find the median weight of the students.

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:
N = 30, N/2 = 15.
Cumulative frequencies: 2, 5, 13, 19, 25, 28, 30.
Median Class is 55 - 60 (cf > 15).
l = 55, cf = 13, f = 6, h = 5.
Median = 55 + [(15 - 13) / 6] × 5
Median = 55 + (2 / 6) × 5
Median = 55 + 1.67 = 56.67 kg.
Answer: The median weight is 56.67 kg.

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Quick Review Flashcards - Click to flip and test your knowledge!
Question
In statistics, what are the three primary measures of central tendency discussed for grouped data?
Answer
The mean, median and mode.
Question
What pictorial representations of data were introduced in Class IX as a precursor to studying measures of central tendency?
Answer
Bar graphs, histograms (including those of varying widths) and frequency polygons.
Question
How is the mean ($\bar{x}$) of observations $x_1, x_2, ..., x_n$ with respective frequencies $f_1, f_2, ..., f_n$ calculated?
Answer
$\bar{x} = \frac{f_1 x_1 + f_2 x_2 + ... + f_n x_n}{f_1 + f_2 + ... + f_n}$
Question
What does the Greek letter $\Sigma$ (capital sigma) represent in statistical formulas?
Answer
Summation.
Question
The brief notation for the mean of observations is $\bar{x} = \frac{\sum f_i x_i}{\sum f_i}$. What does $i$ represent in this context?
Answer
The index of observations, varying from $1$ to $n$.
Question
Why is it often necessary to convert ungrouped data into grouped data in real-life situations?
Answer
To condense large sets of data into a meaningful form for study.
Question
When allocating frequencies to class intervals like $25-40$ and $40-55$, in which interval would an observation exactly at $40$ be placed?
Answer
The $40-55$ interval (the next class).
Question
Concept: Class mark
Answer
Definition: The mid-point of a class interval, used as the representative value for all observations in that class.
Question
What is the formula for calculating the class mark of a class interval?
Answer
Class mark $= \frac{\text{Upper class limit} + \text{Lower class limit}}{2}$
Question
In the calculation of mean for grouped data, what assumption is made regarding the frequency of each class interval?
Answer
It is assumed that the frequency is centred around the class mark (mid-point).
Question
Formula: Direct Method for the mean of grouped data
Answer
$\bar{x} = \frac{\sum f_i x_i}{\sum f_i}$ where $x_i$ is the class mark.
Question
Why might the mean calculated from grouped data differ slightly from the exact mean of the same data in ungrouped form?
Answer
Because of the mid-point assumption used for class intervals in grouped data.
Question
What is the primary reason for using the Assumed Mean Method or Step-deviation Method instead of the Direct Method?
Answer
To reduce calculation work when numerical values of $x_i$ and $f_i$ are large.
Question
In the Assumed Mean Method, what does the symbol '$a$' represent?
Answer
The assumed mean, typically chosen from the middle of the class marks ($x_i$).
Question
What is the formula for deviation ($d_i$) used in the Assumed Mean Method?
Answer
$d_i = x_i - a$
Question
Formula: Assumed Mean Method for mean
Answer
$\bar{x} = a + \frac{\sum f_i d_i}{\sum f_i}$
Question
Does the final value of the mean depend on which class mark is chosen as the assumed mean '$a$'?
Answer
No, the mean determined in each case remains the same.
Question
In the Step-deviation Method, what is the formula for calculating $u_i$?
Answer
$u_i = \frac{x_i - a}{h}$
Question
In the Step-deviation formula $u_i = \frac{x_i - a}{h}$, what does '$h$' represent?
Answer
The class size.
Question
Formula: Step-deviation Method for mean
Answer
$\bar{x} = a + h \left( \frac{\sum f_i u_i}{\sum f_i} \right)$
Question
Under what condition is the Step-deviation Method most convenient to apply?
Answer
When all deviations ($d_i$) have a common factor.
Question
How does the mean obtained by the Step-deviation Method compare to those obtained by the Direct and Assumed Mean Methods?
Answer
The mean obtained by all three methods is the same.
Question
Term: Mode (Ungrouped Data)
Answer
Definition: The value among the observations that occurs most frequently.
Question
What is the term for a data set where more than one value has the same maximum frequency?
Answer
Multimodal.
Question
Concept: Modal class
Answer
Definition: The class interval in a grouped frequency distribution that has the maximum frequency.
Question
Is it possible to determine the exact mode of grouped data simply by looking at the frequencies?
Answer
No, one can only locate the modal class; the mode must then be calculated using a formula.
Question
Formula: Mode for grouped data
Answer
$\text{Mode} = l + \left( \frac{f_1 - f_0}{2f_1 - f_0 - f_2} \right) \times h$
Question
In the mode formula for grouped data, what does '$l$' represent?
Answer
The lower limit of the modal class.
Question
In the mode formula for grouped data, what does '$f_1$' represent?
Answer
The frequency of the modal class.
Question
In the mode formula for grouped data, what does '$f_0$' represent?
Answer
The frequency of the class preceding the modal class.
Question
In the mode formula for grouped data, what does '$f_2$' represent?
Answer
The frequency of the class succeeding the modal class.
Question
In the mode formula for grouped data, what does '$h$' represent?
Answer
The size of the class interval (assuming all class sizes are equal).
Question
If a situation requires finding the 'average of the marks obtained by most students', which measure of central tendency should be used?
Answer
The mode.
Question
Term: Median
Answer
Definition: A measure of central tendency that gives the value of the middle-most observation in the data.
Question
What is the first step in finding the median of ungrouped data?
Answer
Arrange the data values in ascending order.
Question
If the number of observations ($n$) in ungrouped data is odd, how is the median identified?
Answer
It is the $(\frac{n + 1}{2})^{th}$ observation.
Question
If the number of observations ($n$) in ungrouped data is even, how is the median calculated?
Answer
It is the average of the $(\frac{n}{2})^{th}$ and $(\frac{n}{2} + 1)^{th}$ observations.
Question
Concept: Cumulative Frequency Table
Answer
Definition: A table created by adding the frequency of each class to the sum of the frequencies of all previous classes.
Question
What does a median marks of $28.5$ signify regarding the student population?
Answer
About $50\%$ of students scored less than $28.5$ and about $50\%$ scored more than $28.5$.
Question
In a 'less than type' cumulative frequency distribution, what values from the class intervals are used as the boundaries?
Answer
The upper limits of the class intervals.
Question
In a 'more than or equal to type' cumulative frequency distribution, what values from the class intervals are used as the boundaries?
Answer
The lower limits of the class intervals.
Question
How is the 'more than or equal to' cumulative frequency for the first class interval calculated?
Answer
It is equal to the total number of observations ($n$).
Question
How is the 'less than' cumulative frequency for the final class interval related to the total number of observations?
Answer
It is equal to the total number of observations ($n$).
Question
The cumulative frequency curves drawn to represent data are popularly called _____.
Answer
Ogives
Question
If the class sizes of grouped data are unequal but class marks are large, which method for finding the mean can still be applied?
Answer
The Step-deviation Method, by taking '$h$' to be a suitable divisor for all $d_i$ values.
Question
What does the 'mean number of wickets' signify in the context of cricket matches?
Answer
The average number of wickets taken by a bowler per match.
Question
Can the mode be calculated for grouped data with unequal class sizes?
Answer
Yes, although the provided material notes it will not be discussed in detail.
Question
When finding the median of grouped data, what is the 'middle observation' used to locate the median class?
Answer
The observation at the $\frac{n}{2}$ position.
Question
Concept: Median Class
Answer
Definition: The class whose cumulative frequency is greater than (and nearest to) $\frac{n}{2}$.
Question
In Example 1, marks obtained by $30$ students resulted in a mean of $59.3$. Is this value considered an exact or approximate mean?
Answer
Exact mean (since it was calculated from ungrouped data).
Question
Process: Converting 'less than type' to standard grouped data
Answer
Subtract the cumulative frequency of the previous class from the current class to find the individual class frequency.