RD Sharma 2020 solution class 9 chapter 24 Measures of Central Tendency Exercise 24.1

Exercise 24.1

Page-24.8

Question 1:

If the height of 5 persons are 140 cm, 150 cm, 152 cm, 158 cm and 161 cm respectively, find the mean height.

Answer 1:

Given that the heights of 5 persons are 140 cm, 150 cm, 152 cm, 158 cm and 161 cm, respectively.

We have to find the mean of their heights.

Remember the definition of mean of n values x₁, x₂… x_n is

Therefore the mean height of the 5 persons is

Question 2:

Find the mean of 994, 996, 998, 1002, 1000.

Answer 2:

The given 5 numbers are 994, 996, 998, 1002 and 1000, respectively.

We have to find their mean.

Remember the definition of mean of n values x₁, x₂… x_n is

Therefore the mean of the given 5 numbers is

Question 3:

Find the mean of first five natural numbers.

Answer 3:

The first 5 natural numbers are 1, 2, 3, 4 and 5.

We have to find their mean.

Remember the definition of mean of n values x₁, x₂… x_n is

Therefore the mean of the first 5 natural numbers is

Question 4:

Find the mean of all factors of 10.

Answer 4:

All the factors of 10 are 1, 2, 5, and 10. They are 4 in numbers.

We have to find the mean of the all factors of 10.

Remember the definition of mean of n values x₁, x₂… x_n is

Therefore the mean of the all factors of 10 is

Question 5:

Find the mean of first 10 even natural numbers.

Answer 5:

The first 10 even natural numbers are 2, 4, 6, 8, 10, 12, 14, 16, 18, and 20. They are 10 in numbers.

We have to find the mean of the first 10 even natural numbers.

Remember the definition of mean of n values x₁, x₂… x_n is

Therefore the mean of the first 10 even natural numbers is

Question 6:

Find the mean of x, x+2, x+4, x+6, x+8.

Answer 6:

We have to find the mean of, and. They are 5 in numbers.

Remember the definition of mean of n values x₁, x₂… x_n is

Therefore the required mean is

Question 7:

Find the mean of first five multiples of 3.

Answer 7:

The first five multiples of 3 are 3, 6, 9, 12, and 15. They are 5 in numbers.

We have to find the mean of first five multiples of 3.

Remember the definition of mean of n values x₁, x₂… x_n is

Therefore the mean of first five multiples of 3 is

Question 8:

Following are the weights  (in kg)  of 10 new born babies in a hospital on a particular day:
3.4, 3.6, 4.2, 4.5, 3.9, 4.1, 3.8, 4.5, 4.4, 3.6. Find the mean X.

Answer 8:

The weights (in kg) of 10 new born babies are 3.4, 3.6, 4.2, 4.5, 3.9, 4.1, 3.8, 4.5, 4.4, and 3.6. These are 10 in numbers.

We have to find the mean of their weights.

Remember the definition of mean of n values x₁, x₂… x_n is

Therefore the mean of their weights is

Question 9:

The percentage of marks obtained by students of a class in mathematics are : 64, 36, 47, 23, 0, 19, 81, 93, 72, 35, 3, 1. Find their mean.

Answer 9:

Given that the percentage of marks obtained by students of a class of mathematics are 64, 36, 47, 23, 0, 19, 81, 93, 72, 35, 3, and 1. These are 12 in numbers.

We have to find the mean of their percentage of marks.

Remember the definition of mean of n values x₁, x₂… x_n is

Therefore the mean of their weights is

Question 10:

The number of children in 10 families of a locality are:
2, 4, 3, 4, 2, 0, 3, 5, 1, 1, 5. Find the mean number of children per family.

Answer 10:

Given that the numbers of children in 10 families are 2, 4, 3, 4, 2, 3, 5, 1, 1, and 5.

We have to find the mean number of children per family.

Remember the definition of mean of n values x₁, x₂… x_n is

Therefore the mean number of children per family is

Question 11:

Explain, by taking a suitable example, how the arithmetic mean alters by (i) adding a constant k to each term, (ii) subtracting a constant k from each them, (iii) multiplying each term by a constant k and (iv) dividing each term by a non-zero constant k.

Answer 11:

Let us take n observations.

Ifbe the mean of the n observations, then we have

(i) Add a constant k to each of the observations. Then the observations becomes

Ifbe the mean of the new observations, then we have

Let us take an example to understand the above fact.

The first 5 natural numbers are 1, 2, 3, 4, and 5. Their mean is

Add 2 to each of the numbers. Then the numbers becomes 3, 4, 5, 6, and 7. The new mean is

Therefore adding a constant number to each observation the mean increased by that constant

(ii) Subtract a constant k from each of the observations.

Then the observations becomes

Ifbe the mean of the new observations, then we have

Let us take an example to understand the above fact.

The first 5 even natural numbers are 2, 4, 6, 8 and 10. Their mean is

Subtract 1 from each of the numbers. Then the numbers becomes 1, 3, 5, 7, and 9. The new mean is

Therefore subtracting a constant number to each observation the mean decreased by that constant

(iii) Multiply a constant k to each of the observations. Then the observations becomes

Ifbe the mean of the new observations, then we have

Let us take an example to understand the above fact.

The first 5 even natural numbers are 2, 4, 6, 8 and 10. Their mean is

Multiply 2 to each of the numbers. Then the numbers becomes 4, 8, 12, 16, and 20. The new mean is

Therefore multiplying by a constant number to each observation the mean becomes the constant multiplied by the old mean

(vi) Divide a nonzero constant k to each of the observations. Then the observations becomes

Ifbe the mean of the new observations, then we have

Let us take an example to understand the above fact.

The first 5 even natural numbers are 2, 4, 6, 8 and 10. Their mean is

Divide 2 to each of the numbers. Then the numbers becomes 1, 2, 3, 4, and 5. The new mean is

Therefore dividing by a constant number to each observation the mean becomes the old mean divided by the constant

Question 12:

The mean of marks scored by 100 students was found to be 40. Later on it was discovered that a score of 53 was misread as 83. Find the correct mean.

Answer 12:

Given that the mean of marks scored by 100 students is 40. Let us denote the marks of 100 students by.

Ifbe the mean of the n observations, then we have

Hence we have

This sum is incorrect. We have to find the correct sum

It was found that a score 53 was misread as 83. So, 83 must be replaced by 53.

To get the correct sum at first we have to remove the incorrect entry and then add the correct entry.

Therefore the correct sum is

Thus, the correct mean is

Question 13:

The  traffic police recorded the speed (in km/hr) of 10 motorists as 47, 53, 49, 60, 39, 42, 55, 57, 52, 48. Later on an error in recording instrument was found. Find the overage speed of the motorists if the instrument  recorded 5 km/hr less in each case.

Answer 13:

Given that the recorded speed (in km/hr) of 10 motorists are 47, 53, 49, 60, 39, 42, 55, 57, 52, and 48. Let us denote the speeds of 10 motorists by.

Ifbe the mean of the n observations, then we have

Hence we have

This mean is incorrect. We have to find the correct mean.

It was found that the instrument recorded 5 km/hr less in each case.

To get the correct we have to add 5 with each entry. Then the correct entries are

Recall that ifbe the mean of the n observationsand if we add a constant k with each of the observations, then the new mean becomes

Therefore the correct mean speed of the motorists is

Page-24.9

Question 14:

The mean of five numbers is 27. If one number is excluded, their mean is 25. Find the excluded number.

Answer 14:

Given that the mean of 5 numbers is 27. Let us denote the numbers by.

Ifbe the mean of the n observations, then we have

Hence the sum of 5 numbers is

If one number is excluded then the mean becomes 25 and the total numbers becomes 4.

Let the number excluded is x.

After excluding one number the sum becomesand then the mean is

But it is given that after excluding one number the mean becomes 25.

Hence we have

Thus the excluded number is.

Question 15:

The mean weight per student in a group of 7 students is 55 kg. The individual weights of 6 of them (in kg) are 52, 54, 55, 53, 56 and 54. Find the weight of the seventh student.

Answer 15:

Given that the mean weight (in kg) of 7 students is 55 kg. Let us denote their weights by.

Ifbe the mean of the n observations, then we have

Hence the sum of the weights of 7 students is

The individual weights (in kg) of 6 of them are 52, 54, 55, 53, 56 and 54.

Let the weight of the remaining student is x.

According to this information the sum of their weights is

Hence we have

Thus the weight of the seventh student is.

Question 16:

The mean weight of 8 numbers is 15. If each number is multiplied by 2, what will be the new mean?

Answer 16:

Let us denote the 8 numbers by. Their mean is 15.

Ifbe the mean of the n numbers, then we have

Therefore the sum of the numbers is

If each numbers is multiplied by 2, then the numbers becomes.

The mean of the new numbers is

Note that the new mean is equal to the old mean multiplied by 2.

Question 17:

The mean of 5 numbers is 18. If one number is excluded, their mean is 16. Find the excluded number.

Answer 17:

Given that the mean of 5 numbers is 18. Let us denote the numbers by.

Ifbe the mean of the n observations, then we have

Hence the sum of 5 numbers is

If one number is excluded then the mean becomes 16 and the total numbers becomes 4.

Let the number excluded be x.

After excluding one number the sum becomesand then the mean is

But it is given that after excluding one number the mean becomes 16.

Hence we have

Thus the excluded number is.

Question 18:

The mean of 200 items was 50, Later on, it was discovered that the two items were misread as 92 and 8 instead of 192 and 88, Find the correct mean.

Answer 18:

Given that the mean of 200 items is 50. Let us denote the 200 items by.

Ifbe the mean of the n observations, then we have

Hence we have

This sum is incorrect. We have to find the correct sum

It was discovered that the two items were misread as 92 and 8 instead of 192 and 88. So, 92 and 8 are must be replaced by 192 and 8 respectively.

To get the correct sum at first we have to remove the incorrect items and then add the correct items.

Therefore the correct sum is

Thus, the correct mean is

Question 19:

If M is the mean of x₁, x₂, x₃, x₄, x₅ and x₆, prove that
(x₁ − M) + (x₂ − M) (x₃ − M) (x₄ − M) + (x₅ − M) + (x₆ − M) = 0.

Answer 19:

Given that the mean of x₁, x₂, x₃, x₄, x₅, and x₆ is M.

We have to prove that

Remember the definition of mean of n values x₁, x₂… x_n is

Therefore, we have

Now,

Hence the proof is complete.

Question 20:

Durations of sunshine (in hours) in Amritsar to first 10 days of August 1997 as reported by the Meteorological Department are given below:
9.6, 5.2, 3.5, 1.5, 1.6, 2.4, 2.6, 8.4, 10.3, 10.9

(i) Find the mean X.

(ii) Verify that $\sum _{i=1}^{10} (x_i-\overline{X}) = 0$

Answer 20:

Given that the durations of sunshine (in hours) in Amritsar for 10 days are 9.6, 5.2, 3.5, 1.5, 1.6, 2.4, 2.6, 8.4, 10.3, and 10.9.

(i) We have to find their mean.

Remember the definition of mean of n values x₁, x₂… x_n is

Therefore the mean duration of sunshine is

(ii) We have to prove that

We have

Hence the proof is complete.

Question 21:

Find the values of n and X in each of the following cases:

(i) $\sum _{t=1}^n(x_i-12)=-10$ and $\sum _{t=1}^n(x_i-3)=62$

(ii) $\sum _{t=1}^n(x_i-10)=30$ and $\sum _{t=1}^n(x_i-6)=150$

Answer 21:

Ifbe the mean of the n observations, then we have

(i) The given two equations are

The equations can be rewritten as

We have to solve the above equations forand.

Subtracting the first equation from the second equation, we have

Substitute the value of n in the first equation, we have

(ii) The given two equations are

The equations can be rewritten as

We have to solve the above equations forand.

Subtracting the first equation from the second equation, we have

Substitute the value of n in the first equation, we have

Question 22:

The sum of the deviations of a set of n values x₁, x₂,...., x_n, measured from 15 and −3 are −90 and 54 respectively. Find the value of n and mean.

Answer 22:

Ifbe the mean of the n observations, then we have

Given that the sums of the deviations of a set of n valuesmeasured from 15 and -3 are -90 and 54 respectively. So that we get two equations

These equations can be rewritten as

We have to solve the above equations forandunknowns.

Subtracting the first equation from the second equation, we have

Substitute the value of n in the first equation, we have

Question 23:

Find the sum of the deviations of the variate values 3, 4, 6, 7, 8, 14 from their mean.

Answer 23:

Ifbe the mean of the n observations, then we have

The given 6 variate values are 3, 4, 6, 7, 8 and 14. Let us denote the values by $X_1, X_2, ..., X_6$. Their mean value is given by

$$\begin{gathered} \overline{X}=\frac{1}{6}\sum _{i=1}^6 X_i \\ =\frac{1}{6}\left(3+4+6+7+8+14\right) \\ =\frac{1}{6}\times 42=7 \end{gathered}$$

The sum of the deviations of the variate values from their mean is

$$\begin{gathered} \sum _{i=1}^6\left(X_i-\overline{X}\right) \\ =\sum _{i=1}^6\left(X_i-7\right) \\ =\sum _{i=1}^6{X}_i-\sum _{i=1}^{6}7 \\ =6\overline{X}-6\times 7 \\ =6\times 7-6\times 7 \\ =0 \end{gathered}$$

Hence the proof is complete.

Question 24:

If $\overline{X}$ is the mean of the ten natural numbers x₁, x₂, x₃,..., x₁₀, show that
$(x_1-\overline{X})+(x_2-\overline{X})+...+(x_{10}-\overline{X}) = 0$.
 

Answer 24:

Ifbe the mean of the n observations, then we have

Given thatis the mean of the 10 natural numbers. So, we have

We have to show that

That is to show that the sum of the deviations of the 10 natural numbers from their mean is 0.

We have

Hence the proof is complete.