RD Sharma 2022 Solutions for Class 9 Maths Chapter 24 - Measures Of Central Tendency

Answer:

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

Answer:

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

Answer:

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

Answer:

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

Answer:

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

Answer:

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

Answer:

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

Answer:

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

Answer:

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

Answer:

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

Answer:

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

Answer:

Let $x_1, x_2, x_3, ..., x_{50}$ be 50 observations.

$$\begin{gathered} \therefore \frac{x_1+x_2+x_3+.....x_{50}}{50}=80.4 \\ \Rightarrow x_1+x_2+.......x_{50}=4020 .....\left(1\right) \end{gathered}$$

Let x₅ = Incorrect value = 69
$\begin{array}{rcl}{x}_5^{\text{'}}& =& \mathrm{Correct} \mathrm{value}=96\\ & =& x_5+27\end{array}$
Adding 27 on both sides in equation (1), we get

$$\begin{gathered} x_1+...+x_5+27+..... x_{50}=4020+27 \\ \Rightarrow x_1+x_2+...+x_5^{\text{'}}+..... x_{50}=4047 \end{gathered}$$
$\begin{array}{rcl}\therefore \mathrm{Actual} \mathrm{mean}& =& \frac{x_1+x_2+...+x_5^{\text{'}}+...+x_{50}}{50}\\ & =& \frac{4097}{50}\\ & =& 80.94\end{array}$

Hence, the correct mean is 80.94.

Answer:

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

Answer:

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.

Answer:

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.

Answer:

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.

Answer:

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.

Answer:

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

Answer:

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.

Answer:

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

Answer:

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

Answer:

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.

Answer:

The given distribution in tabulated form is

Prepare the following frequency table of which the first column consists of the values of the variate and the second column the corresponding frequencies. Multiply the frequency of each row with the corresponding values of variable to obtain the third column containing.

Find the sum of all entries in the second and third column to obtain N andrespectively. Therefore,

Hence the mean is

Answer:

The given distribution in tabulated form is

Prepare the following frequency table of which the first column consists of the values of the variate and the second column the corresponding frequencies. Multiply the frequency of each row with the corresponding values of variable to obtain the third column containing.

Find the sum of all entries in the second and third column to obtain N andrespectively. Therefore,

Hence the mean is

Answer:

The given distribution in tabulated form is

Prepare the following frequency table of which the first column consists of the values of the variate and the second column the corresponding frequencies. Multiply the frequency of each row with the corresponding values of variable to obtain the third column containing. Find the sum of all entries in the second and third column to obtain N andrespectively.

Find the sum of all entries in the second and third column to obtain N and respectively. Therefore,

Hence the mean is

Answer:

The given data can be tabulated in the form

The row of x denotes the number of heads per toss and the row of f denotes the number of tosses.

Prepare the following frequency table of which the first column consists of the number of heads and the second column the number of tosses (frequencies). Multiply the frequency of each row with the corresponding number of heads to obtain the third column containing.

Find the sum of all entries in the second and third column to obtain N (already given in the question) and respectively. Therefore,

Hence the mean number of heads per toss is

Answer:

(i) The given distribution in tabulated form is

We have to find the value of p using the information that the mean of the distribution is 20.6.

Prepare the following frequency table of which the first column consists of the values of the variate and the second column the corresponding frequencies. Multiply the frequency of each row with the corresponding values of variable to obtain the third column containing.

Find the sum of all entries in the second and third column to obtain N andrespectively. Therefore,

Hence, we have

(ii)

 
Mean = 20.2  (given)
$$\begin{gathered} \therefore \mathrm{Mean}=\frac{\mathrm{\Sigma }{x}_i{f}_i}{\mathrm{\Sigma }{f}_i} \\ \Rightarrow 20.2=\frac{610+20p}{30+p} \end{gathered}$$
⇒ 610 + 20p = 606 + 20.2p
⇒ 0.2p = 4
⇒ p = 20

Hence, the value of p is 20.

Answer:

(i) The given data in the tabulated form is

We have to find the value of p using the information that the mean of the data is 15.

Prepare the following frequency table of which the first column consists of the values of the variate and the second column the corresponding frequencies. Multiply the frequency of each row with the corresponding values of variable to obtain the third column containing.

Find the sum of all entries in the second and third column to obtain N andrespectively. Therefore,

The mean is

Hence, we have

(ii)

$$\begin{gathered} \therefore \mathrm{Mean}=\frac{\Sigma f_i{x}_i}{\Sigma f_i} \\ \Rightarrow 50=\frac{170+150a+90+1600+490a-770+1710}{12a+60} \end{gathered}$$
$$\begin{gathered} \Rightarrow \frac{640a+2800}{12a+60}=50 \\ \Rightarrow 640a+2800=600a+3000 \\ \Rightarrow a=5 \end{gathered}$$

Frequency of 30 = 5a + 3 = 5(5) + 3 = 28
Frequency of 70 = 7a – 11 = 7(5) – 11 = 24 â€‹

Answer:

The given distribution in tabulated form is

We have to find the value of p using the information that the mean of the distribution is 16.6.

Prepare the following frequency table of which the first column consists of the values of the variate and the second column the corresponding frequencies. Multiply the frequency of each row with the corresponding values of variable to obtain the third column containing.

Find the sum of all entries in the second and third column to obtain N andrespectively. Therefore,

The mean is

Hence, we have

Answer:

The given distribution in tabulated form is

We have to find the value of p using the information that the mean of the distribution is 12.58.

Prepare the following frequency table of which the first column consists of the values of the variate and the second column the corresponding frequencies. Multiply the frequency of each row with the corresponding values of variable to obtain the third column containing.

Find the sum of all entries in the second and third column to obtain N and respectively. Therefore,

The mean is

Hence, we have

Answer:

The given distribution in tabulated form is

We have to find the value of p using the information that the mean of the distribution is 7.68.

Prepare the following frequency table of which the first column consists of the values of the variate and the second column the corresponding frequencies. Multiply the frequency of each row with the corresponding values of variable to obtain the third column containing.

Find the sum of all entries in the second and third column to obtain N and respectively. Therefore,

The mean is

Hence, we have

Answer:

The given data in tabulated form is

We have to find the value of p using the information that the mean of the data is 20.

Prepare the following frequency table of which the first column consists of the values of the variate and the second column the corresponding frequencies. Multiply the frequency of each row with the corresponding values of variable to obtain the third column containing.

Find the sum of all entries in the second and third column to obtain N and respectively. Therefore,

The mean is

Hence, we have

If p is negative then the 4^th frequency becomes negative. But frequency can’t be negative. Hence the possible value of p is 1

Answer:

Let the number of candidates appeared from school III is.

Then the given data can be tabulated as

The row of x denotes the average scores and the row of f denotes the number of candidates. We have to find the value of p using the information that the average score of all the four schools is 66.

Prepare the following frequency table of which the first column consists of the average scores and the second column the number of candidate (frequencies). Multiply the frequency of each row with the corresponding average scores to obtain the third column containing.

Find the sum of all entries in the second and third column to obtain N and respectively. Therefore,

The mean is

Hence, we have

Hence the number of candidates appeared from school III is.

Answer:

The given distribution in tabulated form is

We have to find the value of missing frequencies f₁ and f₂, using the information that the mean of the distribution is 50 and the total frequency is 120.

Prepare the following frequency table of which the first column consists of the values of the variate and the second column the corresponding frequencies. Multiply the frequency of each row with the corresponding values of variable to obtain the third column containing.

Find the sum of all entries in the second and third column to obtain N and respectively. Therefore,

The mean is

Hence, we have two equations

Adding them we get

Putting the value of f₁ in the second equation we get

Answer:

The given data is 83, 37, 70, 29, 45, 63, 41, 70, 34 and 54.

Arranging the given data in ascending order, we have

29, 34, 37, 41, 45, 54, 63, 70, 70, 83

Here, the number of observation, which is an even number.

Hence, the median is

Answer:

The given data is 133, 73, 89, 108, 94, 104, 94, 85, 100 and 120.

Arranging the given data in ascending order, we have

73, 85, 89, 94, 94, 100, 104, 108, 120, 133

Here, the number of observation, which is an even number.

Hence, the median is

Answer:

The given data is 31, 38, 27, 28, 36, 25, 35 and 40.

Arranging the given data in ascending order, we have

25, 27, 28, 31, 35, 36, 38, 40

Here, the number of observation, which is an even number.

Hence, the median is

Answer:

The given data is 15, 6, 16, 8, 22, 21, 9, 18, and 25.

Arranging the given data in ascending order, we have

6, 8, 9, 15, 16, 18, 21, 22, 25

Here, the number of observation, which is an odd number.

Hence, the median is

Answer:

The given data is 41, 43, 127, 99, 71, 92, 71, 58 and 57.

Arranging the given data in ascending order, we have

41, 43, 57, 58, 71, 71, 92, 99, 127

Here, the number of observation, which is an odd number.

Hence, the median is

Answer:

The given data is 25, 34, 31, 23, 22, 26, 35, 29, 20 and 32.

Arranging the given data in ascending order, we have

20, 22, 23, 25, 26, 29, 31, 32, 34, 35

Here, the number of observation, which is an even number.

Hence, the median is

Answer:

The given data is 12, 17, 3, 14, 5, 8, 7 and 15.

Arranging the given data in ascending order, we have

3, 5, 7, 8, 12, 14, 15, 17

Here, the number of observation, which is an even number.

Hence, the median is

Answer:

The given data is 92, 35, 67, 85, 72, 81, 56, 51, 42 and 69.

Arranging the given data in ascending order, we have

35, 42, 51, 56, 67, 69, 72, 81, 85, 92

Here, the number of observation, which is an even number.

Hence, the median is

Answer:

The given data is 46, 64, 87, 41, 58, 77, 35, 90, 55, 92 and 33.

Arranging the given data in ascending order, we have

33, 35, 41, 46, 55, 58, 64, 77, 87, 90, 92

Here, the number of observation, which is an odd number.

Hence, the median is

If 92 and 41 are replaced by 99 and 43 respectively, then the given data becomes 46, 64, 87, 43, 58, 77, 35, 90, 55, 99 and 33.

Arranging the new data in ascending order, we have

33, 35, 43, 46, 55, 58, 64, 77, 87, 90, 99

Here, the number of observation, which is an odd number.

Hence, the new median is

Answer:

The given data is 41, 43, 127, 99, 61, 92, 71, 58 and 57.

Arranging the given data in ascending order, we have

41, 43, 57, 58, 61, 71, 92, 99, 127

Here, the number of observation, which is an odd number.

Hence, the median is

Ifis replaced by 85, then the given data become 41, 43, 127, 99, 61, 92, 71, 85 and 57.

Arranging the new data in ascending order, we have

41, 43, 57, 61, 71, 85, 92, 99, 127

Here, the number of observation, which is an odd number.

Hence, the new median is

Answer:

The weights (in kg) of 15 students are 31, 35, 27, 29, 32, 43, 37, 41, 34, 28, 36, 44, 45, 42 and 30.

Arranging them in ascending order, we have

27, 28, 29, 30, 31, 32, 34, 35, 36, 37, 41, 42, 43, 44, 45

Here, the number of observation, which is an odd number.

Hence, the median is

If the weights 44kg and 27kg are replaced by 46kg and 25kg respectively, then the given data becomes 31, 35, 25, 29, 32, 43, 37, 41, 34, 28, 36, 46, 45, 42 and 30.

Arranging the new data in ascending order, we have

25, 28, 29, 30, 31, 32, 34, 35, 36, 37, 41, 42, 43, 45, 46

Here, the number of observation, which is an odd number.

Hence, the new median is

Answer:

The given data in descending order is

50, 42, 35, 2x+10, 2x-8, 12, 11, 8

Here, the number of observation, which is an even number.

Hence, the median is

But, it is given that the median is 25. Hence, we have

Answer:

The given data in ascending order is

29, 32, 48, 50, x, x+2, 72, 78, 84, 95

Here, the number of observation, which is an even number.

Hence, the median is

But, it is given that the median is 63. Hence, we have

Answer:

Arrange the data is ascending order, 6, 14, 15, 17, x + 1, 2x – 13, 30, 32, 34, 43.
Total observation, n = 10
$\begin{array}{rcl}\therefore \mathrm{Median}& =& \frac{{\left({\displaystyle \frac{n}{2}}\right)}^{\mathrm{th}} \mathrm{observation} + {\left({\displaystyle \frac{n}{2}}+1\right)}^{\mathrm{th}} \mathrm{observation}}{2}\\ & =& \frac{5^{\mathrm{th}} \mathrm{observation} + 6^{\mathrm{th}} \mathrm{observation}}{2}\\ & =& \frac{\left(x+1\right)+\left(2x-13\right)}{2}\\ & =& \frac{3x-12}{2}\end{array}$
But median = 24 (given)
$$\begin{gathered} \Rightarrow \frac{3x-12}{2}=24 \\ \Rightarrow 3x-12=48 \\ \Rightarrow 3x=60 \\ \Rightarrow x=20 \end{gathered}$$

Hence, the value of x is 20.

Answer:

Given that the marks obtained by 15 students are 4, 6, 5, 7, 9, 8, 10, 4, 7, 6, 5, 9, 8, 7 and 7.

Make the following frequency table.

Since the value 7 occurs maximum number of times, that is, 4. Hence, the mode value is 7.

Answer:

The given data is 125, 175, 225, 125, 225, 175, 325, 125, 375, 225, and 125.

Make the following frequency table.

Since the value 125 occurs maximum number of times, that is, 4. Hence, the mode value is

Answer:

The given series is 7.5, 7.3, 7.2, 7.2, 7.4, 7.7, 7.7, 7.5, 7.3, 7.2, 7.6 and 7.2.

Make the following frequency table.

Since the value 7.2 occurs maximum number of times, that is, 4. Hence, the modal value is.

Answer:

(i) The given data is 14, 25, 14, 28, 18, 17, 18, 14, 23, 22, 14 and 18.

Make the following frequency table.

Here occurs maximum number of times (that is 4). Hence, the modal value is 14.

(ii) The given data is 7, 9, 12, 13, 7, 12, 15, 7, 12, 7, 25, 18 and 7.

Make the following frequency table.

Since the value 7 occurs maximum number of times, that is, 5. Hence, the modal value is 7.

Answer:

The demand of different shirt sizes is given as

Here x denotes the size of the shirts and f denotes the number of peoples (frequency) wearing it.

The value 39 occur maximum number of times, that is, 39. Hence, the modal shirt size is 39.

Answer:

Arranging in ascending order,
54, 65, 67, 68, 69, 70, 71, 73, 73, 75, 76, 77, 77, 78, 80, 81, 81, 83, 85, 85, 85, 85, 87
Total = 25
Sum of total numbers = 1891
$\begin{array}{rcl}\therefore \mathrm{Mean}& =& \frac{\mathrm{Sum} \mathrm{of} \mathrm{the} \mathrm{numbers}}{\mathrm{Total} \mathrm{numbers}}=\frac{1891}{25}=75.64\\ \mathrm{Median}& =& {\left(\frac{25+1}{2}\right)}^{\mathrm{th}} \mathrm{observation}\\ & =& {13}^{\mathrm{th}} \mathrm{observation}\\ & =& 77\\ & & \end{array}$​

Mode of a data set is the data with the highest frequency.

∴ Mode = 85

Answer:

Arranging in ascending order,
2, 5, 7, 7, 8, 10, 10, 10, 14, 17, 18, 24, 25, 27, 28, 48

Total = 16

$\begin{array}{rcl}\because \mathrm{Median}& =& \frac{1}{2}\left[{\left(\frac{\mathrm{n}}{2}\right)}^{\mathrm{th}}\mathrm{observation}+{\left(\frac{\mathrm{n}}{2}+1\right)}^{\mathrm{th}}\mathrm{observation}\right]\\ & =& \frac{1}{2}\left[8^{\mathrm{th}} \mathrm{observation}+9^{\mathrm{th}} \mathrm{observation}\right]\\ & =& \frac{1}{2}\left[10+14\right]\\ & =& 12\end{array}$

Mode of a data set is the data with the highest frequency.

∴ Mode = 10

Answer:

Given that the ratio of mode and median of a certain data is 6:5. That is,

We know that

Answer:

The given data is x+2, 2x+3, 3x+4, 4x+5. They are four in numbers.

The mean is

But, it is given that the mean is x+2. Hence, we have

Answer:

Given that the median of the scores, whereis 6. The number of scores n is 5, which is an odd number. We have to find

Note that the scores are in descending order. Hence the median is

But, it is given that the median is 6. Hence, we have

Answer:

The given data is. They are 6 in numbers.

The mean is

But, it is given that the mean is 5. Hence, we have

Answer:

The given data is.

The mode is the value which occur maximum number of times, that is, the mode has maximum frequency. If the maximum frequency occurs for more than 1 value, then the number of mode is more than 1 and is not unique.

Here it is given that the mode is 4. So, x must be 4, otherwise it contradicts that the mode is 4. Hence

Answer:

The given data is. The total number of values is, is an even number. Hence the median depends on theobservation andobservation.

Since we have to find the maximum possible value of x.So we must put it in the 4^th position when ordering in ascending order.

Arranging the data in ascending order, we have

Hence, the median is

Here it is given that the median is 29. So, we have

Answer:

The given data is 1, 2, x, 4 and 5. Since, the given data is already in ascending order.

Here, the number of observation, which is an odd number.

Hence, the median is

Here, it is given that the median is 3. Hence, we have.

The mean is

Answer:

Given that the ratio of mean and median of a certain data is 2:3. That is,

We know that

Answer:

Given that the arithmetic mean and mode of a data are 24 and 12 respectively. That is,

We have to find median

We know that

Answer:

Given that the difference of mode and median of a data is 24. That is,

We have to find the difference between median and mean

We know that

Answer:

It is given that, the mean of n observations x₁, x₂,..., x_n is $\overline{X}$.

$\therefore \overline{X}=\frac{x_1+x_2+...+x_n}{n}$   
   
$\Rightarrow x_1+x_2+...+x_n=n\overline{X}$        .....(1)

Now,

$\sum _{i=1}^n\left(x_i-\overline{X}\right)$

$=\left(x_1-\overline{X}\right)+\left(x_2-\overline{X}\right)+...+\left(x_n-\overline{X}\right)$

$=x_1+x_2+...+x_n-\left(\underset{n \mathrm{times}}{\underbrace{\overline{X}+\overline{X}+...+\overline{X}}}\right)$

$=n\overline{X}-n\overline{X}$          [From (1)]

$=0$

Thus, the value of $\sum _{i=1}^n\left(x_i-\overline{X}\right)$ is 0.

If $\overline{\mathrm{X}}$ represents the mean of n observations x₁,x₂,...,x_n, then the value of  $\sum _{i=1}^n \left(x_i - \overline{\mathrm{X}}\right)$ is ____0____.

Answer:

Let the mean of n observations x₁, x₂,..., x_n be $\overline{X}$.

$\therefore \overline{X}=\frac{x_1+x_2+...+x_n}{n}$              $\left(\mathrm{Mean}=\frac{\mathrm{Sum} \mathrm{of} \mathrm{observations}}{\mathrm{Number} \mathrm{of} \mathrm{observations}}\right)$
   
$\Rightarrow x_1+x_2+...+x_n=n\overline{X}$        .....(1)

If each observation is increased by 5, then the new observations are x₁ + 5, x₂ + 5,..., x_n + 5.

∴ Mean of new observations

$=\frac{\left(x_1+5\right)+\left(x_2+5\right)+...+\left(x_n+5\right)}{n}$

$=\frac{x_1+x_2+...+x_n+\underset{n \mathrm{times}}{\underbrace{5+5+...+5}}}{n}$

$=\frac{n\overline{X}+5n}{n}$

$=\frac{n\left(\overline{X}+5\right)}{n}$

$=\overline{X}+5$

Thus, the mean of new observations is increased by 5.

If each observation of the data is increased by 5, then their mean is ___increased___ by 5.

Answer:

It is given that, the mean of x₁, x₂,..., x_n  is $\overline{X}$.

$\therefore \overline{X}=\frac{x_1+x_2+...+x_n}{n}$   
   
$\Rightarrow x_1+x_2+...+x_n=n\overline{X}$        .....(1)

Let $\overline{X\text{'}}$ be the mean of data ax₁ + b, ax₂ + b,..., ax_n+ b.

$\therefore \overline{X\text{'}}=\frac{\left(a{x}_1+b\right)+\left(a{x}_2+b\right)+...+\left(a{x}_n+b\right)}{n}$

$\Rightarrow \overline{X\text{'}}=\frac{\left(a{x}_1+a{x}_2+...+a{x}_n\right)+\left(\underset{n \mathrm{times}}{\underbrace{b+b+...+b}}\right)}{n}$

$\Rightarrow \overline{X\text{'}}=\frac{a\left(x_1+x_2+...+x_n\right)+nb}{n}$

$\Rightarrow \overline{X\text{'}}=\frac{an\overline{X}+nb}{n}$          [Using (1)]

$\Rightarrow \overline{X\text{'}}=\frac{n\left(a\overline{X}+b\right)}{n}$

$\Rightarrow \overline{X\text{'}}=a\overline{X}+b$

Thus, the mean of data ax₁ + b, ax₂ + b,..., ax_n+ b is $a\overline{X}+b$.

If the mean of the data x₁, x₂,..., x_n is $\overline{X}$, then the mean of ax₁ + b, ax₂ + b,..., ax_n+ b is $\underline{ a\overline{X}+b }$ .

Answer:

Mode of the data = 3

Median of the data = 3

Now,

Mode = 3Median − 2Mean

⇒ 3 = 3 × 3 − 2Mean

⇒ 2Mean = 9 − 3 = 6

⇒ Mean = $\frac{6}{2}$ = 3

Thus, the mean of the data is 3.

If mode and median of the certain data are 3 and 3 respectively, then mean is ___3___.

Answer:

It is given that,

Mode − Mean = 36

We know

Mode = 3Median − 2Mean

⇒ Mode − Mean = 3Median − 3Mean

⇒ Mode − Mean = 3(Median − Mean)

⇒ 3(Median − Mean) = 36           (Given)

⇒ Median − Mean = $\frac{36}{3}$ = 12

Thus, the value of Median − Mean is 12.

If Mode − Mean = 36, then Median − Mean = ____12____.

Answer:

Mean of the data = 24

Mode of the data = 12

We know

Mode = 3Median − 2Mean

⇒ 12 = 3Median − 2 × 24

⇒ 3Median = 48 + 12 = 60

⇒ Median = $\frac{60}{3}$ = 20

Thus, the median of data is 20.

The mean and mode of a data are 24 and 12 respectively, then median of the data is ___20___.

Answer:

We know

Mean = $\frac{\mathrm{Sum} \mathrm{of} \mathrm{observations}}{\mathrm{Number} \mathrm{of} \mathrm{observations}}$

⇒ Sum of observations = Mean × Number of observations

Mean of 12 observations = 15      (Given)

∴ Sum of 12 observations = 15 × 12 = 180

If two observations 20 and 25 are removed, then

Sum of remaining 10 observations = Sum of 12 observations − 20 − 25

⇒ Sum of remaining 10 observations = 180 − 20 − 25 = 135

∴ Mean of remaining 10 observations = $\frac{\mathrm{Sum} \mathrm{of} \mathrm{remaining} 10 \mathrm{observation}}{10}=\frac{135}{10}=13.5$

Thus, the mean of remaining observations is 13.5.

If the mean of 12 observations is 15. If two observations 20 and 25 are removed, then the mean of remaining observations is ___13.5___.

Answer:

Mean : Median = 2 : 3          (Given)

Let mean = 2x and median = 3x, where x is constant           .....(1)

We know

Mode = 3Median − 2Mean

⇒ Mode = 3 × 3x − 2 × 2x            [From (1)]

⇒ Mode = 9x − 4x = 5x

∴ Mode : Mean = 5x : 2x = 5 : 2

Thus, the ratio of mode to mean is 5 : 2.

If Mean : Median = 2 : 3, then Mode : Mean = ___5 : 2___.

Answer:

We know

Mean = $\frac{\mathrm{Sum} \mathrm{of} \mathrm{observations}}{\mathrm{Number} \mathrm{of} \mathrm{observations}}$

⇒ Sum of observations = Mean × Number of observations

Mean of a set of 12 observations = 10                 (Given)

∴ Sum of a set of 12 observations = 10 × 12 = 120                   .....(1)

Mean of another set of 8 observations = 12         (Given)

∴ Sum of another set of 8 observations = 12 × 8 = 96              .....(2)

Now,

Sum of all 20 observations 

= Sum of a set of 12 observations + Sum of another set of 8 observations

= 120 + 96                    [From (1) and (2)]

= 216

∴ Mean of all 20 observations = $\frac{\mathrm{Sum} \mathrm{of} \mathrm{all} 20 \mathrm{observations}}{20}=\frac{216}{20}$ = 10.8

Thus, the mean of all 20 observations is 10.8.

The mean of a set of 12 observations is 10 and another set of 8 observations is 12. The mean of all 20 observations is ___10.8___.

Answer:

We know

Mean = $\frac{{\displaystyle \underset{i=1 }{\overset{n }{\sum f_i{x}_i}}}}{\underset{i=1 }{\overset{n }{\sum f_i}}}$

The mean of given distribution is 5.

$\therefore 5=\frac{{\displaystyle 3\times 2+5\times a+7\times 5+4\times b}}{2+a+5+b}$

$\Rightarrow 5=\frac{{\displaystyle 6+5a+35+4b}}{7+a+b}$

$\Rightarrow 5\left(7+a+b\right)=41+5a+4b$

$\Rightarrow 35+5a+5b=41+5a+4b$

$\Rightarrow 5b-4b=41-35$

$\Rightarrow b=6$

Thus, the value of b is 6.


The mean of the following distribution 

x :    3    5    7    4
y :    2    a    5    b

is 5. Then the value of b is ____6____.