Exercise 7.1

1. Find the arithmetic mean of the sales per day in a fair price shop in a week. ₹10000, ₹10250, ₹10790, ₹9865, ₹15350, ₹10110

Solution:

Sum of sales = ₹ + ₹ + ₹ + ₹ + ₹ + ₹ = ₹

Number of days =

Arithmetic mean = Sum of salesNumber of days

= ₹ 665656666565

= ₹ (round off to two decimal places)

Therefore, the arithmetic mean of the sales per day is approximately ₹ 11094.17.

2. Find the mean of the data: 10.25, 9, 4.75, 8, 2.65, 12, 2.35

Solution:

Sum of data points = + + + + + + =

Number of data points =

Mean = Sum of data pointsNumber of data points

= 497

=

Therefore, the mean of the data is 7.

3. Mean of eight observations is 25. If one observation 11 is excluded, find the mean of the remaining.

Solution:

Sum of eight observations = Mean × Number of observations = × =

Sum of remaining observations = Sum of eight observations - Excluded observation

= -

=

Number of remaining observations = -

=

Mean of remaining observations = Sum of remaining observationsNumber of remaining observations

= 1897

=

Therefore, the mean of the remaining observations is 27.

4. Arithmetic mean of nine observations is calculated as 38. But in doing so, mistakenly the observation 27 is taken instead of 72. Find the actual mean of the data.

Solution:

Calculated sum of nine observations = Mean × Number of observations = × =

Difference between the correct and incorrect observation = -

=

Correct sum of nine observations = Calculated sum - Incorrect observation + Correct observation

= - +

=

Actual mean of the data = Correct sumNumber of observations

= 3879

=

Therefore, the actual mean of the data is 43.

5. Five years ago, the mean age of a family was 25 years. What is the present mean age of the family?

Solution:

Let 'n' be the number of family members.

Five years ago: Mean age = years

Sum of ages = Mean age × Number of family members

Sum of ages = ×

Present:

Each family member's age has increased by 5 years. So, the sum of their ages has increased by × n (5 years for each of the n members)

Present sum of ages = 25×n+5×n25×n5×n

= × n

Present mean age = Present sum of agesNumber of family members

= 30×nn

= years

Therefore, the present mean age of the family is 30 years.

6. Two years ago, the mean age of 40 people was 11 years. Now a person left the group, and the mean age has changed to 12 years. Find the age of the person who left the group.

Solution:

Two years ago:

Mean age of 40 people = years

Sum of ages of 40 people = 40 × 11 = years

Present:

Mean age of 40 people = 11 + 2 = years (Since 2 years have passed)

Sum of ages of 40 people = 40 × 13 = years

Number of people now = 40 - 1 =

Mean age of 39 people = years

Sum of ages of 39 people = 39 × 12 = years

Age of the person who left = Sum of ages of 40 people - Sum of ages of 39 people

= -

= years

Therefore, the age of the person who left the group is 52 years.

7. Find the sum of deviations of all observations of the data 5, 8, 10, 15, 22 from their mean.

Solution:

Calculate the mean:

Mean = 5+8+10+15+225 = /5 =

Calculate the deviations of each observation from the mean: 5 - 12 =

8 - 12 =

10 - 12 =

15 - 12 =

22 - 12 =

Sum the deviations:

-7 + (-4) + (-2) + 3 + 10 =

Therefore, the sum of deviations of all observations from their mean is 0.

8. If the sum of the 20 deviations from the mean is 100, then find the mean deviation.

Solution:

Mean deviation = Sum of absolute deviationsNumber of observations = 100/20 =

Therefore, if the sum of the ×absolute values× of the deviations is 100, the mean deviation is 5.

9. Marks of 12 students in a unit test are given as 4, 21, 13, 17, 5, 9, 10, 20, 19, 12, 20, 14. Assume a mean and calculate the arithmetic mean of the data. Assume another number as mean and calculate the arithmetic mean again. Do you get the same result? Comment.

Solution:

Let's first calculate the actual arithmetic mean of the given data:

Sum of marks = 4 + 21 + 13 + 17 + 5 + 9 + 10 + 20 + 19 + 12 + 20 + 14 =

Number of students =

Actual mean = Sum of marksNumber of students = 16412 = (upto two decimal places)

Now, let's assume a mean, say 15, and calculate the arithmetic mean using this assumed mean:

Deviations from assumed mean: , , , , , , , , , , ,

Sum of deviations = -11 + 6 + (-2) + 2 + (-10) + (-6) + (-5) + 5 + 4 + (-3) + 5 + (-1) =

Corrected mean = Assumed mean + Sum of deviationsNumber of students

= + −2012

= 15 - (upto two decimal places)

= (upto two decimal places)

Let's assume another mean, say 10:

Deviations from assumed mean: , , , , , , , , , , ,

Sum of deviations = -6 + 11 + 3 + 7 + (-5) + (-1) + 0 + 10 + 9 + 2 + 10 + 4 =

Corrected mean = Assumed mean + Sum of deviationsNumber of students

= 10 + 4512

= 10 + (upto two decimal places)

= (upto two decimal places)

We see that we don't get the same result.

10. Arithmetic mean of marks (out of 25) scored by 10 students was 15. One of the students, named Karishma, enquired the other 9 students and found the deviations from her marks are noted as -8, -6, -3, -1, 0, 2, 3, 4, 6. Find Karishma’s marks.

Solution:

Calculate the average deviation:

Sum of deviations = -8 + (-6) + (-3) + (-1) + 0 + 2 + 3 + 4 + 6 =

Average deviation = Sum of deviationsNumber of other students = −39 =

Relate to the overall mean:

Let 'K' be Karishma's marks. Let the sum of the marks of the other 9 students be 'S'.

K+S10 = 15

K + S =

S−9K9 = −1 (Since the average deviation of others from Karishma is −13)

S - 9K = -9

Calculate Karishma's marks:

Subtracting the second equation from the first:

10K =

K = (upto one decimal place)

Therefore, Karishma's marks are 15.9.

11. The sum of deviations of ‘n’ observations from 25 is 25, and the sum of deviations of the same ‘n’ observations from 35 is -25. Find the mean of the observations.

Solution:

Let the 'n' observations be x1, x2, ..., xn.

Deviations from 25:

( x1 - 25) + (x2 - 25) + ... + (xn - 25) = 25

(x1 + x2 + ... + xn) - = 25

Deviations from 35:

(x1 - 35) + (x2 - 35) + ... + (xn - 35) =

(x1 + x2 + ... + xn) - = -25

Solving for 'n' and the sum of observations (S):

Let S = x1 + x2 + ... + xn. The equations become:

S - 25n = 25

S - 35n = -25

Subtract the second equation from the first:

10n =

n =

Finding the sum of observations (S):

Substitute n = 5 into the first equation:

S - 25 × 5 = 25

S - = 25

S =

Finding the mean:

Mean = Sn

= 1505

=

Therefore, the mean of the observations is 30.

12. Find the median of the data: 3.3, 3.5, 3.1, 3.7, 3.2, 3.8

Solution:

Arrange the data in ascending order:

, , , , ,

The middle values are and .

Calculate the median: Median = 3.3+3.52

=

Therefore, the median of the data is 3.4.

13. The median of the following observations, arranged in ascending order is 15. 10, 12, 14, x - 3, x, x + 2, 25. Then find x.

Solution:

The observations are already arranged in ascending order: 10, 12, 14, x - 3, x, x + 2, 25.

There are observations (an number). The median is the value.

In this case, the middle value is .

We are given that the median is .

Therefore, x = 15.

14. Find the mode of 10, 12, 11, 10, 15, 20, 19, 21, 11, 9, 10.

Solution:

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

In the given data: 10, 12, 11, 10, 15, 20, 19, 21, 11, 9, 10

The number appears three times.

The number appears two times.

All other numbers appear only once.

Therefore, the mode is .

15. Mode of certain scores is x. If each score is decreased by 3, then find the mode of the new series.

Solution:

If the mode of a set of scores is x, and each score is decreased by 3, then the mode of the new series will be .

The mode represents the most frequent value. If every value in the dataset is shifted by the same amount (in this case, decreased by 3), the relative frequencies of the values don'tdo change.

The most frequent value in the original dataset will still be the most frequent value in the new dataset, but its value will also be shifted by .

Therefore, the mode of the new series is .

16. Find the mode of all digits used in writing the natural numbers from 1 to 100.

Solution:

Count the occurrences of each digit:

Digit 1: Appears in the ones place times (1, 11, 21,... 91).

Appears in the tens place times (10-19).

Appears in 100.

Total: 10 + 10 + 1 = times.

Digits 2 through 9:

Each of these digits appears times in the ones place and times in the tens place.

Total: 10 + 10 = times each.

Digit 0: Appears times in the ones place (10, 20,... 90) and in 100.

Total: 9 + 1 = times.

Identify the most frequent digit:

0: times

1: times

2: times

3: times

4: times

5: times

6: times

7: times

8: times

9: times

The digit appears most frequently.

Therefore, the mode is 1.

17. Observations of a raw data are 5, 28, 15, 10, 15, 8, 24. Add four more numbers so that the mean and median of the data remain the same, but the mode increases by 1.

Solution:

Current Data: 5, 28, 15, 10, 15, 8, 24

Current Mean: 5+8+10+15+15+24+287 = 10571157 =

Current Median: Arranged data: , , , , , , .

The median is .

Current Mode: (appears twice)

Add four numbers so that:

Mean remains 15 (so the sum of all 11 numbers must be 15 × = ).

Median remains 15.

Mode becomes .

Adding the numbers:

To make the mode 16, we need at least 16s.

We have - = to add in total.

If we use three 16s, we have - ( × 16) = - = left for the last number.

Adding 12, 16, 16, and 16 gives us a total sum of 165, which means the mean will be .

Checking the median:

The ordered list becomes , , , , , , , , , , .

The median (the middle value) is still .

Therefore, the four numbers that need to be added are , , , and .

18. If the mean of a set of observations x₁, x₂, ..., x₁₀ is 20. Find the mean of x1 + 4, x2 + 8, x3 + 12, ..., x10 + 40.

Solution:

Given that the mean of x1, x2, ..., x10 is 20, we can write:

(x1 + x2 + ... + x10)/10 =

Therefore, (x1 + x2 + ... + x10) = 20 × 10 =

We need to find the mean of the new set of observations: x1 + 4, x2 + 8, x3 + 12, ..., x10 + 40.

The sum of the new observations is:

(x1 + 4) + (x2 + 8) + (x3 + 12) + ... + (x10 + 40)

= (x1 + x2 + ... + x10) + (4 + 8 + 12 + ... + 40)

We already know that (x1 + x2 + ... + x10) = .

The series 4, 8, 12, ..., 40 is an arithmetic progression with a first term of , a common difference of and terms.

The sum of this arithmetic progression is:

n2 × (2a + (n-1)d) = 102 × (2×4 + (10-1)×4) = 5 × (8 + 36) = 5 × =

Therefore, the sum of the new observations is 200 + 220 = .

The mean of the new observations is:

Sum of new observations10 = 42010 =

Therefore, the mean of x₁ + 4, x₂ + 8, x₃ + 12, ..., x₁₀ + 40 is 42.

19. Six numbers from a list of nine integers are 7, 8, 3, 5, 9 and 5. Find the largest possible value of the median of all nine numbers in this list.

Solution:

Given Numbers: 3, 5, 5, 7, 8, 9

We need to add 3 numbers. The median of the resulting 9 numbers will be the th number in the ordered list.

To maximize the median, we want the 5th number to be as large as possible.

Consider adding numbers greater than or equal to 8: If we add three numbers >= 8 (e.g., 8, 9, 10), the ordered list will be , , , , , , , , .

The median is .

Consider adding numbers greater than or equal to 9: If we add three numbers >= 9 (e.g., 9, 10, 11), the ordered list will be , , , , , , , , .

The median is .

Key Insight: No matter what three numbers we add that are greater than or equal to 8, the 5th number in the ordered list will still be 8. We cannot "push" a larger number into the 5th position because we only have three additional numbers to insert.

Therefore, the largest possible value of the median is 8.

20. The median of a set of 9 distinct observations is 20. If each of the largest 4 observations of the set is increased by 2, find the median of the resulting set.

Solution:

We have 9 distinct observations. The median is the middle value when the observations are arranged in order. Since there are 9 observations, the median is the th observation.

The median is given as . This means the 5th observation in the ordered list is .

The largest 4 observations are increased by . This means the 6th, 7th, 8th, and 9th observations are .

The question asks for the median of the resulting set. The median is the th value in the ordered list.

Since only the largest 4 values were changed, the 5th value which is the median remains .

Therefore, the median of the resulting set is still .