Statistics (1 Mark) - Complete Solutions
Important Statistics Concepts:
• Mean = Sum of all observations / Total number of observations
• Median = Middle value when data is arranged in order
• Mode = Most frequently occurring value
• Assumed Mean Method: A = Assumed mean, d = x - A
• Median for grouped data: l + [(n/2 - cf)/f] × h
• Outliers affect mean more than median or mode
• Median = Middle value when data is arranged in order
• Mode = Most frequently occurring value
• Assumed Mean Method: A = Assumed mean, d = x - A
• Median for grouped data: l + [(n/2 - cf)/f] × h
• Outliers affect mean more than median or mode
1
Class Interval: 10–25, 25–40, 40–55, 55–70, 70–85, 85–100
Frequency: 2, 3, 7, 6, 6, 6
How do you find the deviation from the assumed mean for the above data? (M'15)
Frequency: 2, 3, 7, 6, 6, 6
How do you find the deviation from the assumed mean for the above data? (M'15)
Step 1: Choose assumed mean (A)
Select a convenient value from the class marks (midpoints) as assumed mean.
Usually, we choose the class mark of the middle class.
Usually, we choose the class mark of the middle class.
Step 2: Find class marks
Class mark = (Lower limit + Upper limit) / 2
For 10-25: (10+25)/2 = 17.5
For 25-40: (25+40)/2 = 32.5
For 40-55: (40+55)/2 = 47.5
For 55-70: (55+70)/2 = 62.5
For 70-85: (70+85)/2 = 77.5
For 85-100: (85+100)/2 = 92.5
For 10-25: (10+25)/2 = 17.5
For 25-40: (25+40)/2 = 32.5
For 40-55: (40+55)/2 = 47.5
For 55-70: (55+70)/2 = 62.5
For 70-85: (70+85)/2 = 77.5
For 85-100: (85+100)/2 = 92.5
Step 3: Calculate deviation (d)
Let A = 47.5 (assumed mean)
d = Class mark - A
For each class: d = x - 47.5
d = Class mark - A
For each class: d = x - 47.5
∴ Deviation from assumed mean = Class mark - Assumed mean (d = x - A)
2
Write the formula to find the median of a grouped data and explain each term.
(M'16, Aug'22, A'23)
Step 1: Median formula for grouped data
Median = l + [(n/2 - cf)/f] × h
Step 2: Explanation of terms
l = Lower limit of median class
n = Total number of observations
cf = Cumulative frequency of class preceding median class
f = Frequency of median class
h = Class width of median class
n = Total number of observations
cf = Cumulative frequency of class preceding median class
f = Frequency of median class
h = Class width of median class
∴ Median = l + [(n/2 - cf)/f] × h, where l is lower limit, n is total frequency, cf is cumulative frequency before median class, f is frequency of median class, h is class width.
3
When an observation in a data is abnormally more than or less than the remaining observations in the data, does it affect the mean or mode or median? Why?
(J'15)
Step 1: Understand the effect on mean
Mean is affected because it considers the value of every observation.
Mean = Sum of all observations / Total number of observations
An extreme value changes the sum significantly.
Mean = Sum of all observations / Total number of observations
An extreme value changes the sum significantly.
Step 2: Understand the effect on median
Median is less affected because it depends only on the position, not the values.
It is the middle value when data is arranged in order.
It is the middle value when data is arranged in order.
Step 3: Understand the effect on mode
Mode is least affected because it depends on frequency, not values.
Unless the extreme value repeats, it won't affect mode.
Unless the extreme value repeats, it won't affect mode.
∴ An abnormal observation affects the mean the most because mean considers the actual values of all observations. Median and mode are less affected as they depend on position and frequency respectively.
4
Write the formula to find the mean of a grouped data, using assumed mean method and explain each term.
(J'16)
Step 1: Assumed mean method formula
Mean = A + (Σfd/Σf)
Step 2: Explanation of terms
A = Assumed mean (conveniently chosen class mark)
d = Deviation = x - A (x is class mark)
f = Frequency of each class
Σfd = Sum of (frequency × deviation)
Σf = Total frequency = n
d = Deviation = x - A (x is class mark)
f = Frequency of each class
Σfd = Sum of (frequency × deviation)
Σf = Total frequency = n
∴ Mean = A + (Σfd/Σf), where A is assumed mean, d is deviation (x - A), f is frequency, Σfd is sum of (f × d), and Σf is total frequency.
5
"The median of observations, –2, 5, 3, –1, 4, 6 is 3.5". Is it correct?
(M'17)
Step 1: Arrange data in ascending order
Original data: -2, 5, 3, -1, 4, 6
Arranged in order: -2, -1, 3, 4, 5, 6
Arranged in order: -2, -1, 3, 4, 5, 6
Step 2: Find median
Number of observations (n) = 6 (even)
Median = Average of (n/2)th and (n/2 + 1)th terms
= Average of 3rd and 4th terms
= (3 + 4)/2 = 7/2 = 3.5
Median = Average of (n/2)th and (n/2 + 1)th terms
= Average of 3rd and 4th terms
= (3 + 4)/2 = 7/2 = 3.5
∴ Yes, the statement is correct. The median is indeed 3.5.
6
Write the first 10 prime numbers and find their median.
(J'17)
Step 1: List first 10 prime numbers
Prime numbers: 2, 3, 5, 7, 11, 13, 17, 19, 23, 29
Step 2: Find median
Number of observations (n) = 10 (even)
Median = Average of (n/2)th and (n/2 + 1)th terms
= Average of 5th and 6th terms
= (11 + 13)/2 = 24/2 = 12
Median = Average of (n/2)th and (n/2 + 1)th terms
= Average of 5th and 6th terms
= (11 + 13)/2 = 24/2 = 12
∴ The first 10 prime numbers are 2, 3, 5, 7, 11, 13, 17, 19, 23, 29 and their median is 12.
7
Write the formula to find the median of grouped data and explain the alphabet in it.
(M'18)
Step 1: Median formula for grouped data
Median = l + [(n/2 - cf)/f] × h
Step 2: Explanation of alphabets
l = Lower limit of median class
n = Total number of observations
cf = Cumulative frequency of class preceding median class
f = Frequency of median class
h = Class width of median class
n = Total number of observations
cf = Cumulative frequency of class preceding median class
f = Frequency of median class
h = Class width of median class
∴ Median = l + [(n/2 - cf)/f] × h, where l is lower limit, n is total frequency, cf is cumulative frequency before median class, f is frequency of median class, h is class width.
8
Prathyusha stated that "the average of first 10 odd numbers is also 10". Do you agree with her? Justify your answer.
(M'18)
Step 1: List first 10 odd numbers
First 10 odd numbers: 1, 3, 5, 7, 9, 11, 13, 15, 17, 19
Step 2: Calculate average
Sum = 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 = 100
Average = Sum / 10 = 100 / 10 = 10
Average = Sum / 10 = 100 / 10 = 10
∴ Yes, I agree with Prathyusha. The average of first 10 odd numbers is indeed 10.
9
Find the median of first seven composite numbers.
(M'19)
Step 1: List first seven composite numbers
Composite numbers: Numbers with more than 2 factors
First seven composite numbers: 4, 6, 8, 9, 10, 12, 14
First seven composite numbers: 4, 6, 8, 9, 10, 12, 14
Step 2: Find median
Number of observations (n) = 7 (odd)
Median = (n+1)/2 th term = 8/2 = 4th term
4th term = 9
Median = (n+1)/2 th term = 8/2 = 4th term
4th term = 9
∴ The median of first seven composite numbers is 9.
10
Find the mode of the data 6, 8, 3, 6, 3, 7, 4, 6, 7, 3, 6.
(J'19)
Step 1: Count frequency of each number
6 appears 4 times
3 appears 3 times
7 appears 2 times
8 appears 1 time
4 appears 1 time
3 appears 3 times
7 appears 2 times
8 appears 1 time
4 appears 1 time
Step 2: Identify mode
Mode = Most frequently occurring value
6 occurs most frequently (4 times)
6 occurs most frequently (4 times)
∴ The mode of the given data is 6.
Statistics (2 Marks) - Complete Solutions
Important Statistics Concepts:
• Mean = Sum of all observations / Total number of observations
• Median = Middle value when data is arranged in order
• Mode = Most frequently occurring value
• Mean for grouped data: Σfixi / Σfi
• Mode for grouped data: l + [(f1 - f0)/(2f1 - f0 - f2)] × h
• Median = Middle value when data is arranged in order
• Mode = Most frequently occurring value
• Mean for grouped data: Σfixi / Σfi
• Mode for grouped data: l + [(f1 - f0)/(2f1 - f0 - f2)] × h
1
The heights of six members of a family are given below in the table. Find the mean height of the family members.
(J'15)
Step 1: Understand the data
Height (in ft.): 5, 5.2, 5.4, 5.6
Number of Family members: 1, 2, 2, 1
Number of Family members: 1, 2, 2, 1
Step 2: Calculate total height
Total height = (5 × 1) + (5.2 × 2) + (5.4 × 2) + (5.6 × 1)
= 5 + 10.4 + 10.8 + 5.6 = 31.8 ft
= 5 + 10.4 + 10.8 + 5.6 = 31.8 ft
Step 3: Calculate total family members
Total family members = 1 + 2 + 2 + 1 = 6
Step 4: Calculate mean height
Mean height = Total height / Total family members
= 31.8 / 6 = 5.3 ft
= 31.8 / 6 = 5.3 ft
∴ The mean height of the family members is 5.3 ft
2
Find the value of Σfixi for the above data, where xi is the mid value of each class.
(J'16)
Step 1: Understand the data
Class Intervals: 10-20, 20-30, 30-40, 40-50, 50-60
Frequencies (fi): 5, 8, 10, 5, 2
Frequencies (fi): 5, 8, 10, 5, 2
Step 2: Find mid values (xi)
For 10-20: xi = (10+20)/2 = 15
For 20-30: xi = (20+30)/2 = 25
For 30-40: xi = (30+40)/2 = 35
For 40-50: xi = (40+50)/2 = 45
For 50-60: xi = (50+60)/2 = 55
For 20-30: xi = (20+30)/2 = 25
For 30-40: xi = (30+40)/2 = 35
For 40-50: xi = (40+50)/2 = 45
For 50-60: xi = (50+60)/2 = 55
Step 3: Calculate Σfixi
Σfixi = (5×15) + (8×25) + (10×35) + (5×45) + (2×55)
= 75 + 200 + 350 + 225 + 110 = 960
= 75 + 200 + 350 + 225 + 110 = 960
∴ The value of Σfixi is 960
3
Find the median of 2/3, 4/5, 1/2, 3/4, 6/5
(M'18)
Step 1: Convert fractions to decimals for easier comparison
2/3 ≈ 0.6667
4/5 = 0.8
1/2 = 0.5
3/4 = 0.75
6/5 = 1.2
4/5 = 0.8
1/2 = 0.5
3/4 = 0.75
6/5 = 1.2
Step 2: Arrange in ascending order
Original: 2/3, 4/5, 1/2, 3/4, 6/5
In order: 1/2, 2/3, 3/4, 4/5, 6/5
In order: 1/2, 2/3, 3/4, 4/5, 6/5
Step 3: Find median
Number of observations (n) = 5 (odd)
Median = (n+1)/2 th term = 6/2 = 3rd term
3rd term = 3/4
Median = (n+1)/2 th term = 6/2 = 3rd term
3rd term = 3/4
∴ The median of the given fractions is 3/4
4
Find the mean of prime numbers less than 30.
(J'18)
Step 1: List prime numbers less than 30
Prime numbers less than 30: 2, 3, 5, 7, 11, 13, 17, 19, 23, 29
Step 2: Calculate sum
Sum = 2 + 3 + 5 + 7 + 11 + 13 + 17 + 19 + 23 + 29 = 129
Step 3: Calculate mean
Number of primes = 10
Mean = Sum / Number of primes = 129 / 10 = 12.9
Mean = Sum / Number of primes = 129 / 10 = 12.9
∴ The mean of prime numbers less than 30 is 12.9
5
Write the mode formula for grouped data and explain the terms in it.
(J'15, M'17,19,'22, J'23)
Step 1: Mode formula for grouped data
Mode = l + [(f₁ - f₀)/(2f₁ - f₀ - f₂)] × h
Step 2: Explanation of terms
l = Lower limit of modal class
f₁ = Frequency of modal class
f₀ = Frequency of class preceding modal class
f₂ = Frequency of class succeeding modal class
h = Class width of modal class
f₁ = Frequency of modal class
f₀ = Frequency of class preceding modal class
f₂ = Frequency of class succeeding modal class
h = Class width of modal class
Step 3: Additional notes
• Modal class is the class with highest frequency
• This formula gives the exact mode for grouped data
• It's based on the principle that mode lies in the class with maximum frequency
• This formula gives the exact mode for grouped data
• It's based on the principle that mode lies in the class with maximum frequency
∴ Mode = l + [(f₁ - f₀)/(2f₁ - f₀ - f₂)] × h, where l is lower limit of modal class, f₁ is frequency of modal class, f₀ is frequency of preceding class, f₂ is frequency of succeeding class, and h is class width.
6
Find the median of first 6 prime numbers.
(J'19)
Step 1: List first 6 prime numbers
First 6 prime numbers: 2, 3, 5, 7, 11, 13
Step 2: Find median
Number of observations (n) = 6 (even)
Median = Average of (n/2)th and (n/2 + 1)th terms
= Average of 3rd and 4th terms
= (5 + 7)/2 = 12/2 = 6
Median = Average of (n/2)th and (n/2 + 1)th terms
= Average of 3rd and 4th terms
= (5 + 7)/2 = 12/2 = 6
∴ The median of first 6 prime numbers is 6
Statistics (4 Marks) - Complete Solutions
Important Statistics Concepts:
• Mean = Σfixi / Σfi
• Mode = l + [(f₁ - f₀)/(2f₁ - f₀ - f₂)] × h
• Median = l + [(n/2 - cf)/f] × h
• Step Deviation Method: Mean = A + (Σfidi/Σfi) × h
• Ogive: Cumulative frequency curve
• Mode = l + [(f₁ - f₀)/(2f₁ - f₀ - f₂)] × h
• Median = l + [(n/2 - cf)/f] × h
• Step Deviation Method: Mean = A + (Σfidi/Σfi) × h
• Ogive: Cumulative frequency curve
1
In a village, an enumerator has surveyed for 25 households. The size of the family (number of family members) and the number of families is tabulated as follows. Find the mode of the data.
(M'15)
| Size of family | 1-3 | 3-5 | 5-7 | 7-9 | 9-11 |
|---|---|---|---|---|---|
| No. of families | 6 | 7 | 9 | 2 | 1 |
Step 1: Identify modal class
Highest frequency = 9
Modal class = 5-7
Modal class = 5-7
Step 2: Apply mode formula
Mode = l + [(f₁ - f₀)/(2f₁ - f₀ - f₂)] × h
l = 5 (lower limit of modal class)
f₁ = 9 (frequency of modal class)
f₀ = 7 (frequency of class preceding modal class)
f₂ = 2 (frequency of class succeeding modal class)
h = 2 (class width)
f₁ = 9 (frequency of modal class)
f₀ = 7 (frequency of class preceding modal class)
f₂ = 2 (frequency of class succeeding modal class)
h = 2 (class width)
Step 3: Calculate mode
Mode = 5 + [(9 - 7)/(2×9 - 7 - 2)] × 2
= 5 + [2/(18 - 9)] × 2
= 5 + (2/9) × 2
= 5 + 4/9 = 5 + 0.44 = 5.44
= 5 + [2/(18 - 9)] × 2
= 5 + (2/9) × 2
= 5 + 4/9 = 5 + 0.44 = 5.44
∴ The mode of the data is 5.44 family members
2
Daily expenditure of 25 householders is given in the following table. Draw a "less than type" cumulative frequency Ogive curve for this data.
(M'15)
| Daily expenditure (Rs) | 100-150 | 150-200 | 200-250 | 250-300 | 300-350 |
|---|---|---|---|---|---|
| No. of households | 4 | 5 | 12 | 2 | 2 |
Step 1: Create less than cumulative frequency table
| Daily expenditure (less than) | 150 | 200 | 250 | 300 | 350 |
|---|---|---|---|---|---|
| Cumulative frequency | 4 | 9 | 21 | 23 | 25 |
Step 2: Plot points for ogive
Points to plot: (150, 4), (200, 9), (250, 21), (300, 23), (350, 25)
Step 3: Draw ogive curve
Less Than Ogive Curve:
X-axis: Daily Expenditure (Rs)
Y-axis: Cumulative Frequency
Plot points: (150,4), (200,9), (250,21), (300,23), (350,25)
Join points with a smooth freehand curve
∴ The less than ogive curve is drawn by plotting points (150,4), (200,9), (250,21), (300,23), (350,25) and joining them with a smooth curve.
3
If the median of 60 observations given below is 28.5, then find the values of x and y.
(J'15)
| Class Interval | 0-10 | 10-20 | 20-30 | 30-40 | 40-50 | 50-60 |
|---|---|---|---|---|---|---|
| Frequency | 5 | x | 20 | 15 | y | 5 |
Step 1: Total frequency equation
Total frequency = 60
5 + x + 20 + 15 + y + 5 = 60
x + y + 45 = 60
x + y = 15 ............ (1)
5 + x + 20 + 15 + y + 5 = 60
x + y + 45 = 60
x + y = 15 ............ (1)
Step 2: Identify median class
n = 60, n/2 = 30
Median = 28.5 (given)
Median class = 20-30 (since median 28.5 lies in this class)
Median = 28.5 (given)
Median class = 20-30 (since median 28.5 lies in this class)
Step 3: Apply median formula
Median = l + [(n/2 - cf)/f] × h
l = 20, n/2 = 30, f = 20, h = 10
cf = cumulative frequency before median class = 5 + x
28.5 = 20 + [(30 - (5 + x))/20] × 10
28.5 - 20 = [(25 - x)/20] × 10
8.5 = (25 - x)/2
17 = 25 - x
x = 25 - 17 = 8
cf = cumulative frequency before median class = 5 + x
28.5 = 20 + [(30 - (5 + x))/20] × 10
28.5 - 20 = [(25 - x)/20] × 10
8.5 = (25 - x)/2
17 = 25 - x
x = 25 - 17 = 8
Step 4: Find y using equation (1)
x + y = 15
8 + y = 15
y = 15 - 8 = 7
8 + y = 15
y = 15 - 8 = 7
∴ The values are x = 8 and y = 7
4
The following distribution gives the daily profits (in rupees) earned by 50 shops in a locality. Convert the above distribution to a 'less than type' cumulative frequency distribution and draw its Ogive.
(J'15)
| Daily Profits (Rs.) | 0-50 | 50-100 | 100-150 | 150-200 | 200-250 | 250-300 |
|---|---|---|---|---|---|---|
| No. of shops | 6 | 9 | 13 | 10 | 8 | 4 |
Step 1: Create less than cumulative frequency table
| Daily profits (less than) | 50 | 100 | 150 | 200 | 250 | 300 |
|---|---|---|---|---|---|---|
| Cumulative frequency | 6 | 15 | 28 | 38 | 46 | 50 |
Step 2: Plot points for ogive
Points to plot: (50, 6), (100, 15), (150, 28), (200, 38), (250, 46), (300, 50)
Step 3: Draw ogive curve
Less Than Ogive Curve:
X-axis: Daily Profits (Rs)
Y-axis: Cumulative Frequency
Plot points: (50,6), (100,15), (150,28), (200,38), (250,46), (300,50)
Join points with a smooth freehand curve
∴ The less than ogive is drawn by plotting points (50,6), (100,15), (150,28), (200,38), (250,46), (300,50) and joining them with a smooth curve.
5
Consider the following distribution of daily wages of 50 workers of a factory. Find the mean daily wages of the workers in the factory by using step-deviation method.
(M'16)
| Daily wages (Rs.) | 200-250 | 250-300 | 300-350 | 350-400 | 400-450 |
|---|---|---|---|---|---|
| No. of workers | 6 | 8 | 14 | 10 | 12 |
Step 1: Create calculation table
| Class | Frequency (f) | Midpoint (x) | d = (x - A)/h A=325, h=50 |
fd |
|---|---|---|---|---|
| 200-250 | 6 | 225 | -2 | -12 |
| 250-300 | 8 | 275 | -1 | -8 |
| 300-350 | 14 | 325 | 0 | 0 |
| 350-400 | 10 | 375 | 1 | 10 |
| 400-450 | 12 | 425 | 2 | 24 |
| Total | 50 | 14 |
Step 2: Apply step-deviation formula
Mean = A + (Σfd/Σf) × h
A = 325, h = 50, Σfd = 14, Σf = 50
Mean = 325 + (14/50) × 50
= 325 + 14 = 339
Mean = 325 + (14/50) × 50
= 325 + 14 = 339
∴ The mean daily wage of the workers is Rs. 339
