Statistics

AI Learning Assistant

I can help you understand Statistics better. Ask me anything!

Summarize the main points of Statistics.

What are the most important terms to remember here?

Explain this concept like I'm five.

Give me a quick 3-question practice quiz.

CBSE Learning Objectives – Key Concepts & Skills You Must Know

Learning Objectives
- Understand how to collect and organize data into grouped frequency distributions.
- Calculate the mean from grouped data using appropriate methods.
- Interpret the results of mean calculations in the context of real-world data.
- Identify and apply the mode and median in grouped data distributions.
- Analyze the impact of extreme values on the mean, median, and mode.
- Differentiate between the use of mean, median, and mode based on the nature of the data.

CBSE Revision Notes & Quick Summary for Last-Minute Study

Statistics Notes

Measures of Central Tendency

Mean

The mean is calculated using the formula:

$x = \frac{\Sigma f_i x_i}{\Sigma f_i}$
Example: For a grouped frequency distribution, the mean can be calculated by forming class intervals and using the mid-points as representative values.

Median

The median is calculated using the formula:

$\text{Median} = l + \left( \frac{\frac{n}{2} - cf}{f} \right) \times h$
- Where:
  - $l$ : Lower limit of the median class
  - $n$ : Total number of observations
  - $cf$ : Cumulative frequency of the class preceding the median class
  - $f$ : Frequency of the median class
  - $h$ : Class width

Mode

The mode is calculated using the formula:

$\text{Mode} = l + \left( \frac{f_1 - f_0}{2f_1 - f_0 - f_2} \right) \times h$
- Where:
  - $l$ : Lower limit of the modal class
  - $f_1$ : Frequency of the modal class
  - $f_0$ : Frequency of the class preceding the modal class
  - $f_2$ : Frequency of the class succeeding the modal class
  - $h$ : Class width

Cumulative Frequency

Cumulative frequency is the sum of the frequencies of all classes preceding a given class.

Example Tables

Example 1: Marks Distribution

Marks obtained	Number of students	Cumulative frequency
0 - 10	5	5
10 - 20	3	8
20 - 30	4	12
30 - 40	3	15
40 - 50	3	18
50 - 60	4	22
60 - 70	7	29
70 - 80	9	38
80 - 90	7	45
90 - 100	8	53

Example 2: Teacher-Student Ratio

Number 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

Remarks

The mean is sensitive to extreme values, while the median is more robust in such cases.
The mode is useful for identifying the most frequent value in a dataset.
There is an empirical relationship:
$3 \text{Median} = \text{Mode} + 2 \text{Mean}$

Applications

Mean is used for comparing distributions.
Median is preferred when extreme values are present.
Mode is used to find the most popular or frequent item.

CBSE Exam Tips, Important Questions & Common Mistakes to Avoid

Common Mistakes and Exam Tips

Common Pitfalls

Misunderstanding Mean Calculation: Students often confuse the methods for calculating the mean, especially between direct, assumed mean, and step-deviation methods. Ensure you understand when to use each method.
Ignoring Class Intervals: When dealing with grouped data, failing to recognize the importance of class intervals can lead to incorrect calculations of mean, median, and mode.
Cumulative Frequency Errors: Mistakes in calculating cumulative frequencies can affect the determination of median and other statistical measures.
Assuming Midpoints: In grouped data, assuming that all values are at the midpoint can lead to inaccuracies. Always consider the range of the class interval.

Tips for Avoiding Mistakes

Double-Check Calculations: Always verify your calculations, especially when summing frequencies or products in mean calculations.
Understand the Data Structure: Familiarize yourself with the data structure (ungrouped vs grouped) and the appropriate methods for each.
Practice with Examples: Work through multiple examples to solidify your understanding of different statistical measures and their calculations.
Use Tables Effectively: Organize data in tables to visualize frequencies, cumulative frequencies, and calculations clearly.
Clarify Definitions: Ensure you understand definitions of mean, median, and mode, and how they apply to different types of data.

CBSE Quiz & Practice Test – MCQs, True/False Questions with Solutions

Multiple Choice Questions

175 units

225 units

250 units

200 units

Correct Answer: B

Solution:

To find the median, we first determine the cumulative frequencies: 5, 15, 30, 50. The total number of households is 50, so the median class is where the cumulative frequency exceeds 25. This occurs in the 200-250 units interval. Using the formula for median:

\text{Median} = l + \left(\frac{\frac{n}{2} - cf}{f}\right) \times h

, where

l = 200

cf = 15

f = 15

h = 50

. Therefore, Median =

200 + \left(\frac{25 - 15}{15}\right) \times 50 = 225

units.

3250 hours

3500 hours

3000 hours

3750 hours

Correct Answer: C

Solution:

The median class is the one where the cumulative frequency reaches 200 (half of 400). This occurs in the 2500-3000 hours class, making the median life time 3000 hours.

62.5

67.5

Correct Answer: B

Solution:

After removing the highest score (90), the scores are: 45, 50, 55, 60, 65, 70, 75, 80, 85. The median is the average of the 4th and 5th scores: (60 + 65) / 2 = 62.5.

₹2250

₹2500

₹2750

₹3000

Correct Answer: A

Solution:

To find the mean, we calculate the sum of the midpoints of each class interval multiplied by their respective frequencies, divided by the total number of families. The mean monthly expenditure is approximately ₹2250.

Correct Answer: B

Solution:

Using the formula for mean:

\bar{x} = \frac{\Sigma f_i x_i}{\Sigma f_i}

. Given mean = 18, solve for

f

using the equation:

18 = \frac{(12 \times 7) + (14 \times 6) + (16 \times 9) + (18 \times 13) + (20 \times f) + (22 \times 5) + (24 \times 4)}{44 + f}

. Solving gives

f = 6

Correct Answer: C

Solution:

The mode is the value that appears most frequently in a data set. In this data set, the number 9 appears most frequently (five times). Therefore, the mode is 9.

The distribution is symmetric.

The distribution is positively skewed.

The distribution is negatively skewed.

The skewness cannot be determined.

Correct Answer: B

Solution:

In a positively skewed distribution, the mode is less than the mean. Here, since the mode (150 grams) is greater than the mean (145 grams), the distribution is positively skewed.

75.25

75.50

76.00

74.75

Correct Answer: A

Solution:

The total score with the incorrect entry is

40 \times 75 = 3000

. Correcting the score, the total becomes

3000 - 85 + 95 = 3010

. The correct mean is

\frac{3010}{40} = 75.25

10.5 days

12 days

15 days

9 days

Correct Answer: A

Solution:

To find the mean, multiply the mid-point of each class interval by the frequency, sum these products, and divide by the total number of students. Mean =

\frac{(3 \times 11) + (8 \times 10) + (12 \times 7) + (17 \times 4) + (24 \times 4) + (33 \times 3) + (39 \times 1)}{40} = 10.5

days.

74.5

76.5

78.5

80.5

Correct Answer: B

Solution:

Calculate the mean by multiplying the mid-point of each class interval by the number of women, summing these products, and dividing by the total number of women:

\frac{(66.5 \times 2) + (69.5 \times 4) + (72.5 \times 3) + (75.5 \times 8) + (78.5 \times 7) + (81.5 \times 4) + (84.5 \times 2)}{30} = 76.5

125-145 units

105-125 units

145-165 units

165-185 units

Correct Answer: A

Solution:

The mode is the class interval with the highest frequency. Here, the class interval 125-145 has the highest frequency of 20.

0.10 ppm

0.12 ppm

0.08 ppm

0.06 ppm

Correct Answer: C

Solution:

The mean concentration is calculated as:

\frac{(0.02 \times 4) + (0.06 \times 9) + (0.10 \times 9) + (0.14 \times 2) + (0.18 \times 4) + (0.22 \times 2)}{30} = 0.08

ppm.

₹550

₹540

₹530

₹560

Correct Answer: B

Solution:

Mean daily wage is calculated by multiplying the mid-point of each wage interval by the number of workers and dividing by the total number of workers. Mean = (510×12 + 530×14 + 550×8 + 570×6 + 590×10) / 50 = ₹540.

165 cm

160 cm

170 cm

175 cm

Correct Answer: A

Solution:

The mean height is calculated as follows:

\text{Mean} = \frac{\sum f_i x_i}{\sum f_i} = \frac{5 \times 145 + 10 \times 155 + 15 \times 165 + 10 \times 175 + 5 \times 185}{5 + 10 + 15 + 10 + 5} = 165 \text{ cm}

27.5 cm

30 cm

32.5 cm

35 cm

Correct Answer: A

Solution:

To calculate the mean, use the formula:

\bar{x} = \frac{\Sigma f_i x_i}{\Sigma f_i}

. The midpoints are 15, 25, 35, and 45. Thus,

\Sigma f_i x_i = (4 \times 15) + (6 \times 25) + (8 \times 35) + (2 \times 45) = 60 + 150 + 280 + 90 = 580

. The total frequency

\Sigma f_i = 4 + 6 + 8 + 2 = 20

. Therefore, the mean is

\bar{x} = \frac{580}{20} = 29

cm.

25-35

35-45

15-25

45-55

Correct Answer: B

Solution:

The mode is the class interval with the highest frequency. Here, the interval 35-45 has the highest frequency of 23.

10 hours

12.5 hours

15 hours

11.5 hours

Correct Answer: B

Solution:

The mean is calculated as:

\text{Mean} = \frac{5 \times 2.5 + 10 \times 7.5 + 15 \times 12.5 + 10 \times 17.5}{5 + 10 + 15 + 10} = 12.5 \text{ hours}

68%

70%

72%

74%

Correct Answer: B

Solution:

To find the mean, calculate the mid-point of each class and multiply by the frequency, then divide by the total number of cities. Mean = (503 + 6010 + 7011 + 808 + 90*3) / 35 = 70%.

₹35000

₹37500

₹40000

₹42500

Correct Answer: C

Solution:

Initially, the sum of incomes is ₹25000 + ₹30000 + ₹35000 + ₹40000 + ₹45000 = ₹175000. Adding the new income: ₹175000 + ₹50000 = ₹225000. The new mean is ₹225000 / 6 = ₹37500.

3250 hours

3500 hours

3750 hours

4000 hours

Correct Answer: B

Solution:

The median is the value that divides the distribution into two equal parts. To find the median class, calculate the cumulative frequency and identify the class where the cumulative frequency exceeds half of the total frequency (200). The cumulative frequency for the class 3000-3500 is 216, which is the first to exceed 200. Using the median formula: Median = l + ((n/2 - cf)/f) * h, where l = 3000, cf = 130, f = 86, h = 500, we get Median = 3000 + ((200 - 130)/86) * 500 = 3500 hours.

₹60,000

₹65,000

₹67,500

₹70,000

Correct Answer: B

Solution:

The profits in increasing order are ₹50,000, ₹55,000, ₹60,000, ₹65,000, ₹70,000, ₹80,000. The median is the average of the 3rd and 4th values: (₹60,000 + ₹65,000)/2 = ₹62,500.

Correct Answer: B

Solution:

The mode is the interval with the highest frequency. Here, the interval 10-20 has the highest frequency of 3. Therefore, the mode is the midpoint of this interval, which is 15.

132 units

142 units

152 units

162 units

Correct Answer: B

Solution:

Calculate the midpoint of each class interval, multiply by the frequency, sum these products, and divide by the total number of consumers. The mean is approximately 142 units.

7-10

4-7

10-13

13-16

Correct Answer: A

Solution:

The median class is the one where the cumulative frequency reaches or exceeds half of the total number of observations. Here, the cumulative frequency for 7-10 letters is 76, which is the median class.

40-50

10-20

60-70

20-30

Correct Answer: A

Solution:

The mode is the class interval with the highest frequency. Here, the interval 40-50 has the highest frequency of 20.

8.5

11.5

Correct Answer: C

Solution:

To calculate the mean, we use the formula:

\bar{x} = \frac{\Sigma f_i x_i}{\Sigma f_i}

. The midpoints of the intervals are 2.5, 7.5, 12.5, and 17.5. Thus,

\Sigma f_i x_i = (3 \times 2.5) + (7 \times 7.5) + (15 \times 12.5) + (5 \times 17.5) = 7.5 + 52.5 + 187.5 + 87.5 = 335

. The total frequency

\Sigma f_i = 3 + 7 + 15 + 5 = 30

. Therefore, the mean is

\bar{x} = \frac{335}{30} = 11.5

Correct Answer: B

Solution:

To find the median, arrange the data in ascending order: 10, 12, 14, 15, 16, 18, 20. The median is the middle value, which is 15.

26.67

27.5

28.33

Correct Answer: B

Solution:

The median is calculated using the formula:

\text{Median} = l + \frac{\frac{n}{2} - cf}{f} \times h

. Here,

l = 20

n = 45

cf = 30

f = 15

, and

h = 10

. Substituting these values,

\text{Median} = 20 + \frac{22.5 - 30}{15} \times 10 = 26.67

20-30 cars

30-40 cars

40-50 cars

50-60 cars

Correct Answer: C

Solution:

The mode is the class interval with the highest frequency. Here, the class 40-50 cars has the highest frequency of 20 periods, making it the mode.

2750 hours

3250 hours

3750 hours

4250 hours

Correct Answer: B

Solution:

To find the median, calculate the cumulative frequency and locate the median class. The cumulative frequency just over 200 (half of 400) is 216 in the class 3000-3500, making it the median class. Using the formula, the median is calculated as 3250 hours.

₹2500-3000

₹1500-2000

₹3000-3500

₹2000-2500

Correct Answer: B

Solution:

The mode is the class interval with the highest frequency. Here, the class ₹1500-2000 has the highest frequency of 40 families, making it the modal class.

1-4

4-7

7-10

10-13

Correct Answer: C

Solution:

The mode is the class with the highest frequency. Here, the class 7-10 has the highest frequency of 40, making it the modal class.

550

540

530

560

Correct Answer: B

Solution:

Calculate the mid-point of each wage interval and multiply by the frequency, then divide by the total number of workers:

\text{Mean} = \frac{(510 \times 12) + (530 \times 14) + (550 \times 8) + (570 \times 6) + (590 \times 10)}{50} = 540

52.5 kg

57.5 kg

62.5 kg

67.5 kg

Correct Answer: B

Solution:

To find the median, we determine the cumulative frequency and locate the median class. The median class is 55-60 kg, as it contains the 15th student (n/2 = 15). Using the median formula, the median weight is 57.5 kg.

20-30 cars

30-40 cars

40-50 cars

50-60 cars

Correct Answer: C

Solution:

The mode is the class interval with the highest frequency. Here, the class 40-50 cars has the highest frequency of 20.

20-40

40-60

60-80

0-20

Correct Answer: B

Solution:

The mean score is calculated by finding the mid-point of each class and multiplying by the frequency, then dividing by the total number of students. The class with the highest contribution to the mean is 40-60.

4 books

3.5 books

5 books

4.5 books

Correct Answer: D

Solution:

The mean is calculated as:

\text{Mean} = \frac{3 \times 1 + 5 \times 3 + 7 \times 5 + 5 \times 7}{3 + 5 + 7 + 5} = 4.5 \text{ books}

10 books

12.5 books

15 books

7.5 books

Correct Answer: B

Solution:

To find the median, we determine the cumulative frequency and locate the median class. The cumulative frequencies are 5, 20, 40, 50. The median class is 10-15 books. Using the median formula:

\text{Median} = l + \left(\frac{\frac{n}{2} - cf}{f}\right) \times h

where

l = 10

cf = 20

f = 20

h = 5

, and

n = 50

. Substituting these values, we get

\text{Median} = 10 + \left(\frac{25 - 20}{20}\right) \times 5 = 12.5

books.

Correct Answer: A

Solution:

The mean is calculated as the total number of plants divided by the number of houses. Total plants = (0 \times 1) + (2 \times 2) + (4 \times 1) + (6 \times 5) + (8 \times 6) + (10 \times 2) + (12 \times 3) = 80. Number of houses = 20. Mean = 80/20 = 4.

₹540

₹550

₹560

₹570

Correct Answer: B

Solution:

The mean is calculated using the formula:

\text{Mean} = \frac{\Sigma f_i x_i}{\Sigma f_i}

. Here, the mean is calculated as:

\frac{(510 \times 12) + (530 \times 14) + (550 \times 8) + (570 \times 6) + (590 \times 10)}{50} = 550

1-4 letters

4-7 letters

7-10 letters

10-13 letters

Correct Answer: C

Solution:

The mode is the value that appears most frequently. Here, the highest frequency is 40, corresponding to the 7-10 letters interval. Therefore, the modal size is 7-10 letters.

400 hours

450 hours

475 hours

500 hours

Correct Answer: B

Solution:

To find the median, we determine the cumulative frequency and locate the median class. The cumulative frequencies are 10, 40, 80, 95, 100. The median class is 400-500 hours. Using the median formula:

\text{Median} = l + \left(\frac{\frac{n}{2} - cf}{f}\right) \times h

where

l = 400

cf = 40

f = 40

h = 100

, and

n = 100

. Substituting these values, we get

\text{Median} = 400 + \left(\frac{50 - 40}{40}\right) \times 100 = 450

hours.

55 kg

57.5 kg

60 kg

62.5 kg

Correct Answer: B

Solution:

The median class is the one where the cumulative frequency reaches 15.5 (half of 30). This occurs in the 55-60 kg class, making the median weight 57.5 kg.

Correct Answer: D

Solution:

The mode is the value that appears most frequently. In this grouped data, the class interval with the highest frequency is 40-60 with 20 students. Therefore, the mode is the midpoint of this interval, which is 50.

65%

70%

75%

80%

Correct Answer: B

Solution:

Calculate the mean by multiplying the mid-point of each class interval by the number of cities, summing these products, and dividing by the total number of cities:

\frac{(50 \times 3) + (60 \times 10) + (70 \times 11) + (80 \times 8) + (90 \times 3)}{35} = 70

7.5

8.5

6.5

5.5

Correct Answer: A

Solution:

To find the mean, calculate the mid-point of each interval and multiply by the corresponding frequency, then divide by the total number of houses:

\text{Mean} = \frac{(1 \times 1) + (3 \times 2) + (5 \times 1) + (7 \times 5) + (9 \times 6) + (11 \times 2) + (13 \times 3)}{20} = 7.5

1.5

2.0

2.5

3.0

Correct Answer: B

Solution:

The z-score is calculated using the formula:

z = \frac{x - \mu}{\sigma}

, where

x

is the student's height,

\mu

is the mean, and

\sigma

is the standard deviation. Substituting the given values,

z = \frac{180 - 160}{10} = 2.0

25-35 years

35-45 years

45-55 years

55-65 years

Correct Answer: B

Solution:

The mode is the class interval with the highest frequency, which is 35-45 years with 23 patients.

55.5 mangoes

56.5 mangoes

57.5 mangoes

58.5 mangoes

Correct Answer: C

Solution:

Mean number of mangoes = (5115 + 54110 + 57135 + 60115 + 63*25) / 400 = 57.5 mangoes.

10.5 hours

12.5 hours

8.5 hours

11.5 hours

Correct Answer: A

Solution:

To find the mean, we use the formula:

\text{Mean} = \frac{\sum f_i x_i}{\sum f_i}

. The midpoints are 2.5, 7.5, 12.5, and 17.5. The mean is

\frac{(3 \times 2.5) + (7 \times 7.5) + (15 \times 12.5) + (5 \times 17.5)}{3 + 7 + 15 + 5} = 10.5

hours.

40 years

35 years

45 years

50 years

Correct Answer: A

Solution:

The mean age is calculated as:

\text{Mean} = \frac{10 \times 25 + 15 \times 35 + 20 \times 45 + 5 \times 55}{10 + 15 + 20 + 5} = 40 \text{ years}

Approximately 68%

Approximately 95%

Approximately 99.7%

Approximately 50%

Correct Answer: A

Solution:

In a normal distribution, approximately 68% of the data falls within one standard deviation of the mean. Since 2850 to 3150 hours is one standard deviation from the mean (3000 ± 150), the probability is approximately 68%.

Correct Answer: A

Solution:

The frequency of an interval can be found by subtracting the cumulative frequency of the previous interval from the cumulative frequency of the current interval. Therefore, the frequency of the interval 40-50 is 120 - 90 = 30.

165 cm

167 cm

162 cm

160 cm

Correct Answer: C

Solution:

To find the mean, we use the formula:

\bar{x} = \frac{\Sigma f_i x_i}{\Sigma f_i}

where

f_i

is the frequency and

x_i

is the midpoint of each class interval. The midpoints are 145, 155, 165, 175, and 185. Calculating the sum of

f_i x_i

gives 8100, and the sum of frequencies is 50. Thus,

\bar{x} = \frac{8100}{50} = 162

cm.

5.5

6.5

7.5

8.5

Correct Answer: A

Solution:

The mean number of plants is calculated by multiplying the mid-point of each interval by the number of houses and dividing by the total number of houses. Mean = (1×1 + 3×2 + 5×1 + 7×5 + 9×6 + 11×2 + 13×3) / 20 = 5.5.

10.5 days

12.5 days

14.5 days

16.5 days

Correct Answer: C

Solution:

Mean number of days absent is calculated by multiplying the mid-point of each interval by the number of students and dividing by the total number of students. Mean = (3×11 + 8×10 + 12×7 + 17×4 + 24×4 + 33×3 + 39×1) / 40 = 14.5 days.

162.5 cm

163.0 cm

163.5 cm

164.0 cm

Correct Answer: B

Solution:

To find the mean, we use the formula:

\bar{x} = \frac{\Sigma f_i x_i}{\Sigma f_i}

. The midpoints of the class intervals are 152.5, 157.5, 162.5, 167.5, and 172.5. Calculating the sum of products of frequencies and midpoints:

5 \times 152.5 + 10 \times 157.5 + 15 \times 162.5 + 10 \times 167.5 + 10 \times 172.5 = 8152.5

. The total frequency is 50. Therefore, the mean height is

\frac{8152.5}{50} = 163.0

cm.

₹200

₹220

₹240

₹260

Correct Answer: A

Solution:

Using the direct method, the mean is calculated as follows:

\bar{x} = \frac{\Sigma f_i x_i}{\Sigma f_i}

. The mean daily expenditure is ₹200.

100-150 widgets

150-200 widgets

200-250 widgets

250-300 widgets

Correct Answer: C

Solution:

The mode is the class interval with the highest frequency. Here, the class 200-250 widgets has the highest frequency of 4 days, making it the mode.

₹2000-2500

₹2500-3000

₹3000-3500

₹1500-2000

Correct Answer: D

Solution:

The modal class is the one with the highest frequency. Here, the class interval with the highest frequency is 1500-2000 with 40 families.

4 hours

5 hours

6 hours

4.5 hours

Correct Answer: D

Solution:

The mean is calculated as:

\text{Mean} = \frac{8 \times 1 + 12 \times 3 + 10 \times 5 + 10 \times 7}{8 + 12 + 10 + 10} = 4.5 \text{ hours}

65%

70%

75%

80%

Correct Answer: B

Solution:

To find the mean literacy rate, calculate the sum of the midpoints of each class interval multiplied by their respective frequencies, divided by the total number of cities. The mean literacy rate is approximately 70%.

0.1126

0.1251

0.0996

0.1353

Correct Answer: B

Solution:

The probability of observing

k

events in a Poisson distribution is given by

P(k; \lambda) = \frac{e^{-\lambda} \lambda^k}{k!}

. For

\lambda = 10

and

k = 8

P(8; 10) = \frac{e^{-10} 10^8}{8!} \approx 0.1251

-1

-0.5

0.5

Correct Answer: A

Solution:

The standard deviation is the square root of the variance, which is

\sqrt{4} = 2

. The deviation from the mean in standard deviation units is

\frac{3 - 5}{2} = -1

0-20 hours

40-60 hours

60-80 hours

80-100 hours

Correct Answer: C

Solution:

The mode is the class interval with the highest frequency. Here, the class 60-80 hours has the highest frequency of 61.

56.5 mangoes

58.5 mangoes

60.5 mangoes

62.5 mangoes

Correct Answer: B

Solution:

To find the mean, multiply the mid-point of each class interval by the frequency, sum these products, and divide by the total number of boxes. Mean =

\frac{(51 \times 15) + (54 \times 110) + (57 \times 135) + (60 \times 115) + (63 \times 25)}{400} = 58.5

mangoes.

₹200

₹220

₹240

₹260

Correct Answer: B

Solution:

The mean daily expenditure is calculated as

\frac{(125 \times 4) + (175 \times 5) + (225 \times 12) + (275 \times 2) + (325 \times 2)}{25} = ₹220

₹540

₹550

₹560

₹520

Correct Answer: A

Solution:

The formula for the median is:

\text{Median} = l + \left( \frac{\frac{n}{2} - cf}{f} \right) \times h

. Here, the median class is ₹520-540. Thus,

l = 520, n = 50, cf = 12, f = 14, h = 20

. Substituting these values, we get the median as ₹540.

32.5 years

37.5 years

42.5 years

47.5 years

Correct Answer: B

Solution:

The median is the age at which the cumulative frequency is greater than or equal to 50 (since 100/2 = 50). The cumulative frequency just greater than 50 is 78, which corresponds to the class 35-40. Using the median formula, we find the median to be approximately 37.5 years.

150-160 cm

160-170 cm

170-180 cm

140-150 cm

Correct Answer: B

Solution:

The total number of students is 3 + 5 + 12 + 8 + 2 = 30. The median class is the one where the cumulative frequency is greater than or equal to 30/2 = 15. The cumulative frequency reaches 20 at the class 160-170 cm, making it the median class.

Correct Answer: B

Solution:

The cumulative frequency for the class 50-60 is obtained by adding the frequencies of all preceding classes: 5 + 3 + 4 + 3 + 3 = 18. Adding the frequency of the 50-60 class, we get 18 + 4 = 22.

2750 hours

3250 hours

3750 hours

4250 hours

Correct Answer: B

Solution:

To find the median, we need to determine the cumulative frequency and locate the median class. The median class is 3000-3500 hours, as it contains the 200th lamp (n/2 = 200). Using the formula for median, the median life time is 3250 hours.

₹1750

₹2000

₹2250

₹2500

Correct Answer: B

Solution:

The mean is calculated as: (10 * 1250 + 15 * 1750 + 20 * 2250 + 5 * 2750) / (10 + 15 + 20 + 5) = ₹2000.

Correct Answer: A

Solution:

Using the mean formula: Mean = (Σfᵢxᵢ) / Σfᵢ, and given that the mean is ₹18, solve for f. Calculate the total sum of allowances with the midpoints and solve for f to get f = 5.

₹550

₹540

₹560

₹570

Correct Answer: B

Solution:

The mean is calculated using the formula:

\bar{x} = \frac{\Sigma f_i x_i}{\Sigma f_i}

. Here,

\Sigma f_i x_i = (500 \times 12) + (520 \times 14) + (540 \times 8) + (560 \times 6) + (580 \times 10) = 27000

and

\Sigma f_i = 50

. Therefore, mean = 27000/50 = ₹540.

145.5 mm

146.5 mm

147.5 mm

148.5 mm

Correct Answer: B

Solution:

The median class is 145-153 mm. Using the median formula, the median length is

146.5

mm.

Direct method

Assumed mean method

Step deviation method

Convert to continuous classes first, then use the median formula

Correct Answer: D

Solution:

To accurately calculate the median for grouped data, the class intervals must be continuous. Therefore, converting to continuous classes first is necessary before applying the median formula.

30 years

35 years

40 years

45 years

Correct Answer: B

Solution:

The median is the age at which half the policyholders are younger and half are older. With 100 policyholders, the median age is the age at which the cumulative frequency first exceeds 50. The cumulative frequency exceeds 50 at the 'Below 35' category, making the median age 35 years.

32.5 years

35 years

37.5 years

40 years

Correct Answer: C

Solution:

The median is the age at which half the policy holders are younger and half are older. With 100 policy holders, the median is the 50th policy holder. The cumulative frequency just above 50 is 78, which corresponds to the age group 35-40. Therefore, the median age is approximately 37.5 years.

₹20,833

₹21,000

₹22,500

₹23,000

Correct Answer: A

Solution:

The mean is calculated using:

\bar{x} = \frac{\Sigma f_i x_i}{\Sigma f_i}

where

x_i

is the midpoint of each class interval. The midpoints are ₹12,500, ₹17,500, ₹22,500, ₹27,500, and ₹32,500. Calculating the sum of

f_i x_i

gives ₹1,250,000, and the sum of frequencies is 60. Thus,

\bar{x} = \frac{1,250,000}{60} = ₹20,833

Correct Answer: A

Solution:

The total number of students is 30. The median is the score at the 15.5th position. The cumulative frequency table is: 0-20: 4, 20-40: 10, 40-60: 20, 60-80: 28, 80-100: 30. The median class is 40-60. Using the median formula:

\text{Median} = l + \left(\frac{\frac{n}{2} - cf}{f}\right) \times h

, where

l = 40

cf = 10

f = 10

h = 20

. Therefore, Median =

40 + \left(\frac{15 - 10}{10}\right) \times 20 = 50

4000-5000 runs

5000-6000 runs

6000-7000 runs

7000-8000 runs

Correct Answer: A

Solution:

The mode is the class interval with the highest frequency. Here, the interval 4000-5000 runs has the highest frequency of 18.

True or False

Correct Answer: True

Solution:

Continuous class intervals are necessary for applying the median formula to grouped data.

Correct Answer: True

Solution:

The mode is defined as the value that appears most frequently in a dataset.

Correct Answer: False

Solution:

For calculating the median, class intervals must be continuous. If they are not, adjustments need to be made.

Correct Answer: True

Solution:

To find the median class, we locate the class whose cumulative frequency is greater than and nearest to half the total number of observations.

Correct Answer: True

Solution:

The median class is determined by finding the class whose cumulative frequency is greater than and nearest to half the total number of observations.

Correct Answer: True

Solution:

The formula for calculating the median assumes that the class intervals are continuous.

Correct Answer: True

Solution:

The median is a better measure of central tendency in the presence of extreme values as it is not influenced by them.

Correct Answer: True

Solution:

The mean takes into account all the observations and lies between the extremes, making it a widely used measure of central tendency.

Correct Answer: True

Solution:

The cumulative frequency of a class in a cumulative frequency distribution is indeed the sum of frequencies of all classes up to and including that class.

Correct Answer: True

Solution:

The median is less affected by extreme values, making it a better measure of central tendency in such cases.

Correct Answer: True

Solution:

For calculating the median of grouped data, it is necessary that the class intervals are continuous before applying the formula.

Correct Answer: True

Solution:

The cumulative frequency is calculated by summing the frequencies of all classes up to and including the given class.

Correct Answer: False

Solution:

For calculating the mode of grouped data, it is necessary that the class intervals are continuous to apply the formula correctly.

Correct Answer: True

Solution:

To apply the median formula for grouped data, class intervals must be continuous.

Correct Answer: False

Solution:

The mean is not always the best measure of central tendency when there are extreme values, as it can be affected by those values. In such cases, the median is often a better measure.

Correct Answer: True

Solution:

Cumulative frequency distribution helps in determining the count of observations below a certain threshold.

Correct Answer: True

Solution:

Cumulative frequency distributions can be constructed as 'less than' or 'more than' types, depending on whether they accumulate frequencies from the lower or upper ends of the data.

Correct Answer: True

Solution:

The mean for grouped data can indeed be calculated using three methods: the direct method, the assumed mean method, and the step deviation method.

Correct Answer: True

Solution:

The mode for grouped data can be found by using a formula, which takes into account the class interval and frequencies.

Correct Answer: True

Solution:

The median is the value that divides the data into two equal halves.

Correct Answer: False

Solution:

The mean is not always the best measure of central tendency when there are extreme values, as it can be heavily influenced by them. In such cases, the median is often a better measure.

Correct Answer: True

Solution:

The mode can be calculated for grouped data with unequal class sizes, although it is not discussed in detail.

Correct Answer: False

Solution:

The mean is not always a better measure of central tendency than the mode. The mode is preferred in situations requiring the most frequent value or most popular item, such as finding the most popular TV program.

Correct Answer: True

Solution:

The median is often more appropriate than the mean in situations where extreme values may skew the data, as it is not affected by outliers.

Correct Answer: True

Solution:

This is the correct formula for calculating the mode of grouped data, where

l

is the lower limit of the modal class,

f_m

is the frequency of the modal class,

f_1

and

f_2

are the frequencies of the classes before and after the modal class, and

h

is the class width.

Correct Answer: True

Solution:

The modal class is defined as the class interval with the highest frequency in a frequency distribution.

Correct Answer: False

Solution:

Cumulative frequency distributions can be of both 'less than' and 'more than' types.

Correct Answer: False

Solution:

The mean is not always the best measure of central tendency, especially in the presence of extreme values. In such cases, the median might be more appropriate.

Correct Answer: True

Solution:

The cumulative frequency of a class is obtained by adding the frequencies of all the classes preceding the given class.

Correct Answer: False

Solution:

For calculating the median of grouped data, class intervals must be continuous to apply the median formula correctly.

Correct Answer: True

Solution:

The cumulative frequency of a class is indeed the sum of the frequencies of all classes preceding it.

Correct Answer: True

Solution:

The mode identifies the most frequently occurring value, making it ideal for such analyses.

Correct Answer: True

Solution:

The median is less affected by extreme values and provides a better measure of central tendency in such cases.

Correct Answer: False

Solution:

Cumulative frequency distributions can be constructed for both equal and unequal class intervals.

Correct Answer: True

Solution:

The mean for grouped data can indeed be calculated using the direct method, assumed mean method, and step deviation method, as each method provides a way to handle the data effectively.

Correct Answer: True

Solution:

The formula for finding the median of grouped data is correctly given as

\text{Median} = l + \left( \frac{\frac{n}{2} - cf}{f} \right) \times h

, where each symbol has a specific meaning.

Correct Answer: True

Solution:

The mode is used to find the most frequent or popular item in a dataset, making it the best measure for such purposes.

Correct Answer: True

Solution:

For calculating the median of grouped data, it is essential that the class intervals are continuous to apply the formula correctly.

Correct Answer: True

Solution:

The median is more appropriate than the mean when data has extreme values because it is not affected by them and provides a 'typical' observation.

Correct Answer: False

Solution:

For calculating the median for grouped data, it is necessary to ensure that the class intervals are continuous before applying the formula.

Correct Answer: True

Solution:

Ogives are graphical representations of cumulative frequency distributions.

Correct Answer: True

Solution:

Extreme values can skew the mean, making it a poor representative of the data in such cases.

Correct Answer: True

Solution:

The mean considers all data points, making it a comprehensive measure of central tendency.

Correct Answer: False

Solution:

The mean is the most frequently used measure of central tendency because it considers all observations. The mode is used to find the most frequent value.

Statistics

AI Learning Assistant

CBSE Learning Objectives – Key Concepts & Skills You Must Know

CBSE Revision Notes & Quick Summary for Last-Minute Study

Statistics Notes

Measures of Central Tendency

Mean

Median

Mode

Cumulative Frequency

Example Tables

Example 1: Marks Distribution

Example 2: Teacher-Student Ratio

Remarks

Applications

CBSE Exam Tips, Important Questions & Common Mistakes to Avoid

Common Mistakes and Exam Tips

Common Pitfalls

Tips for Avoiding Mistakes

Multiple Choice Questions

Q1.In a survey of household electricity consumption, the data was grouped as follows: 100-150 units (5 households), 150-200 units (10 households), 200-250 units (15 households), and 250-300 units (20 households). What is the median consumption?

Correct Answer: B

Correct Answer: C

Q3.A class of students has the following scores in a test: 45, 50, 55, 60, 65, 70, 75, 80, 85, 90. If the highest score is removed, what will be the new median score?

Correct Answer: B

Q4.The following data gives the distribution of total monthly household expenditure of 200 families. What is the mean monthly expenditure?

Correct Answer: A

Q5.The following distribution shows the daily pocket allowance of children of a locality. If the mean pocket allowance is Rs 18, find the missing frequency fff.

Correct Answer: B

Q6.In a study of plant growth, the following data represents the number of leaves on 20 plants: 5, 7, 9, 6, 8, 10, 11, 9, 8, 7, 10, 12, 11, 9, 8, 7, 6, 5, 8, 9. What is the mode of the number of leaves?

Correct Answer: C

Q7.A researcher is analyzing the distribution of weights of a sample of apples. The mode of the distribution is found to be 150 grams. If the mean weight is 145 grams, what can be inferred about the skewness of the distribution?

Correct Answer: B

Q8.In a class of 40 students, the mean score on a mathematics test is 75. If one student's score is mistakenly recorded as 85 instead of 95, what is the correct mean score?

Correct Answer: A

Q9.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.

Correct Answer: A

Q10.In a hospital, the number of heartbeats per minute of 30 women were recorded as follows: 65-68 (2 women), 68-71 (4 women), 71-74 (3 women), 74-77 (8 women), 77-80 (7 women), 80-83 (4 women), 83-86 (2 women). What is the mean number of heartbeats per minute?

Correct Answer: B

Q11.The following frequency distribution gives the monthly consumption of electricity of 68 consumers of a locality. Find the mode of the data.

Correct Answer: A

Correct Answer: C

Q13.The following table shows the daily wages of 50 workers in a factory: ₹500-520 (12 workers), ₹520-540 (14 workers), ₹540-560 (8 workers), ₹560-580 (6 workers), and ₹580-600 (10 workers). What is the mean daily wage?

Correct Answer: B

Correct Answer: A

Q15.In a study of plant growth, the heights of plants were recorded and grouped into intervals: 10-20 cm (4 plants), 20-30 cm (6 plants), 30-40 cm (8 plants), 40-50 cm (2 plants). What is the mean height of the plants?

Correct Answer: A

Q16.The following table shows the ages of patients admitted in a hospital during a year. What is the mode of the data?

Correct Answer: B

Q17.The following table shows the number of hours spent by students on homework per week: 0-5 hours (5 students), 5-10 hours (10 students), 10-15 hours (15 students), 15-20 hours (10 students). What is the mean number of hours spent on homework?

Correct Answer: B

Q18.In a study of the literacy rate of 35 cities, the rates are grouped as follows: 45-55% (3 cities), 55-65% (10 cities), 65-75% (11 cities), 75-85% (8 cities), 85-95% (3 cities). What is the mean literacy rate?

Correct Answer: B

Q19.The following data represents the monthly income (in ₹) of 5 families: ₹25000, ₹30000, ₹35000, ₹40000, ₹45000. If a new family with an income of ₹50000 is added, what will be the new mean monthly income?

Correct Answer: C

Correct Answer: B

Q21.A company's profit over the last five years is given as follows: Year 1: ₹50,000, Year 2: ₹60,000, Year 3: ₹55,000, Year 4: ₹65,000, Year 5: ₹70,000. If the profit for Year 6 is ₹80,000, what is the new median profit?

Correct Answer: B

Q22.A factory records the number of defective items produced in a week as follows: 0-10 (2 days), 10-20 (3 days), 20-30 (1 day), 30-40 (1 day). What is the mode of the number of defective items?

Correct Answer: B

Correct Answer: B

Correct Answer: A

Q25.The table below shows the number of cars passing through a spot on a road for 100 periods of 3 minutes each. Which class interval has the mode? | 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 |

Correct Answer: A

Q26.A group of students collected data on the number of books read by their classmates over a month. The data was grouped into intervals: 0-5, 5-10, 10-15, and 15-20 books. The frequencies for these intervals were 3, 7, 15, and 5 respectively. What is the mean number of books read by the students?

Correct Answer: C

Q27.A hospital records the number of patients visiting in a week as follows: Monday (10), Tuesday (12), Wednesday (15), Thursday (14), Friday (16), Saturday (20), Sunday (18). What is the median number of patients visiting per day?

Correct Answer: B

Q28.In a frequency distribution, the median class is identified as 20-30 with a cumulative frequency of 45. The cumulative frequency of the class preceding the median class is 30, and the frequency of the median class is 15. If the class width is 10, what is the median?

Correct Answer: B

Correct Answer: C

Correct Answer: B

Correct Answer: B

Q32.Given the following frequency distribution of the number of letters in surnames: 1-4 (6 surnames), 4-7 (30 surnames), 7-10 (40 surnames), 10-13 (16 surnames), 13-16 (4 surnames), 16-19 (4 surnames). What is the modal number of letters in the surnames?

Correct Answer: C

Q33.In a factory, the daily wages of workers are distributed as follows: 500-520 for 12 workers, 520-540 for 14 workers, 540-560 for 8 workers, 560-580 for 6 workers, and 580-600 for 10 workers. What is the mean daily wage of the workers?

Correct Answer: B

Q34.The distribution below gives the weights of 30 students of a class. What is the median weight of the students?

Correct Answer: B

Q35.A student noted the number of cars passing through a spot on a road for 100 periods each of 3 minutes. What is the mode of the data?