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It is given that

Mean monthly salary of 20 workers = ₹ 45900

So the total monthly salary = ₹ (20 × 49500) = ₹ 918000

Mean monthly salary of 20 workers and manager = ₹ 49200

So the total monthly salary = ₹ (21 × 49200) = ₹ 1033200

We know that

Manager’s monthly salary = Total monthly salary of 20 workers and manager – Total monthly salary of 20 workers

By substituting the values

Manager’s monthly salary = 1033200 – 918000 = ₹ 115200

Therefore, the manager’s monthly salary is ₹ 115200.

It is given that

Mean weight of 36 students = 41kg

So the total weight = 41 (36) = 1476kg

It is given that if one of the students leaves the class then the mean is decreased by 200g

So the new mean = 41 – 0.2 = 40.8kg

We get the total weight of 35 students = 40.8 (35) = 1428kg

So the weight of student who left = total weight of 36 students – weight of 35 students

By substituting the values

Weight of the student who left the class = 1476 – 1428 = 48kg

Therefore, the weight of the student who left is 48kg.

It is given that

Mean weight of 25 students = 52kg

So the total weight = 25 (52) = 1300kg

Mean weight of first 13 students = 48kg

So the total weight = 13 (48) = 624kg

Mean weight of last 13 students = 55kg

So the total weight = 13 (55) = 715kg

We know that

Weight of 13^th student = total weight of the first 13 students + total weight of last 13 students – total weight of 25 students

By substituting the values

Weight of 13^th student = 624 + 715 – 1300

On further calculation

Weight of 13^th student = 39kg

Therefore, the weight of the 13^th student is 39kg.

(c) 20.5

The x- coordinate represents the median of the given data. 

Thus, median of the given data is 20.5.

According to assumed-mean method,

\(\bar{x}\) = A + \(\cfrac{\sum_if_id_i}{\sum_if_i}\)

= 25 + \(\frac{110}{50}\)

= 25 + 2.2 

= 27.2 

Thus, mean is 27.2.

The value of the middle-most observation is called the median of the data.

First arrange the given numbers in ascending order,

29, 35, 51, 55, 60, 63, 72, 82, 85, 91

Here n = 10, which is even.

Where, n is the number of the given number.

∴median = ½ {(n/ 2)^th term + ((n/2) + 1)^th term}

= ½ {(10/ 2)^th term + ((10/2) + 1)^th term}

= ½ {(5)^th term + ((5) + 1)^th term}

= ½ {(5)^th term + (6)^th term}

= ½ (60 + 63)

= ½ (123)

= (123/2)

= 61.5

Hence, the median is 61.5.

According to the question, 

4 = x/36 and 3 = y/64 

⇒ x = 4 × 36 and y = 3 × 64 

⇒ x = 144 and y = 192 

Now, x + y = 144 + 192 = 336 

And total number of observations = 36 + 64 = 100 

Thus, mean = 336/100 = 3.36.

Upper class boundary = Lowest class boundary + width × number of classes 

= 8.1 + 2.5 × 12 

= 8.1 + 30 

= 38.1 

Thus, upper class boundary of the highest class is 38.1.

Class mark = \(\frac{upper \,limit + lower \,limits}2\)

∴ class mark of 10 - 25 = \(\frac{10+25}{2}\)

= 17.5

And class mark of 35 - 55 = \(\frac{35+55}{2}\)

= 45

Upper class boundary = Lowest class boundary + width × number of classes 

= 8.1 + 2.5×12 

= 8.1 + 30 

= 38.1 

Thus, upper class boundary of the highest class is 38.1.

mode = (3 x median) – (2 x mean)

(b) centred at the class marks of the classes 

While computing the mean of the group data, we assume that the frequencies are centred at the class marks of the classes.

Correct answer is (c) centred at the lower limits of the classes 

While computing the mean of the group data, we assume that the frequencies are centred at the class marks of the classes.

Correct answer is (b)0

We know that \(\bar{x}\) = \(\cfrac{\sum f_ix_i}{\sum f_i}\)

⇒ \(\bar{x}\) Σ f_i = Σ f_ix_i … (i) 

Now, Σ f_i (xi − \(\bar{x}\)) = Σ f_ix_i − \(\bar{x}\)Σ f_i 

⇒ Σ f_i (x − \(\bar{x}\)) = Σ f_ix_i − Σ f_ix_i [Using (i)] 

⇒ Σ f_i (x_i − \(\bar{x}\)) = 0

The d_i′ s are the deviations from A of midpoints of the classes.

Correct answer is (b) Median

The abscissa of the point of intersection of the ‘less than type’ and that of the ‘more than type’ cumulative frequency curves of a grouped data gives its median.

(b) Median

The cumulative frequency table is useful in determining the median.

(b) Median

The abscissa of the point of intersection of the ‘less than type’ and that of the ‘more than type’ cumulative frequency curves of a grouped data gives its median.

Since, 8 is less than 30 and 32 is more than 30, so the middle value remains unchanged Thus, the median of 21 observations taken together is 30.

We know that

Number of observations = 7

It is given that mean = 18

It can be written as

(3 + 21 + 25 + 17 + x + 3 + 19 + x – 4)/7 = 18

On further calculation

2x + 84 = 126

By subtraction

2x = 42

By division

x = 21

By substituting the value of x

(x + 3) = 21 + 3 = 24

(x – 4) = 21 – 4 = 17

So we get

3, 21, 25, 17, 24, 19, 17

We know that 17 occurs maximum number of times so the mode is 17.

Since, 8 is less than 30 and 32 is more than 30, so the middle value remains unchanged Thus, the median of 21 observations taken together is 30.

We know that

Number of observations = 9

It is given that median = 45

So we get

Median = [(n + 1)/2]th value

By substituting the values

Median = [(9 + 1)/2]th value

We get

Median = 5^th value = 2x + 3

It is given that median = 45

We get

2x + 3 = 45

On further calculation

2x = 42

By division

x = 21

By substituting the value of x

2x + 3 = 2(21) + 3 = 45

We get

42, 43, 44, 44, 45, 45, 46, 47

We know that 45 occurs maximum number of times so the mode is 45.

We know that

Number of observations = 9

By arranging the numbers in ascending order

We get

52, 53, 54, 54, (2x + 1), 55, 55, 56, 57

So we get

Median = [(n + 1)/2]th value

By substituting the values

Median = [(9 + 1)/2]th value

We get

Median = 5^th value = 2x + 1

It is given that median = 55

We can write it as

2x + 1 = 55

On further calculation

2x = 54

By division

x = 27

By substituting the value of x

2x + 1 = 2(27) + 1 = 55

So we get

52, 53, 54, 54, 55, 55, 55, 56, 57

We know that 55 occurs the maximum number of times so the mode is 55.

By arranging the numbers in ascending order

We get

9, 19, 27, 28, 30, 32, 35, 50, 50, 50, 50, 60

From the table we know that 50 occurs maximum number of times so the mode is 50.

By arranging the numbers in ascending order

We get

10, 13, 15, 18, x + 1, x + 3, 30, 32, 35, 41

We know that n = 10 which is even

So we get

Median = ½ {(n/2)th term + (n/2 + 1)th term}

It can be written as

Median = ½ {5^th term + 6^th term)

By substituting the values

Median = ½ (x + 1 + x + 3)

On further calculation

Median = ½ (2x + 4)

We get

Median = x + 2

It is given that median = 24

So we get

x + 2 = 24

On further calculation

x = 24 – 2 = 22

Therefore, the value of x is 22.

By arranging the numbers in ascending order

We get

32, 34, 36, 37, 4, 44, 47, 50, 53, 54

We know that n = 10 which is even

So we get

Median = ½ {(n/2)th term + (n/2 + 1)th term}

It can be written as

Median = ½ {5^th term + 6^th term)

By substituting the values

Median = ½ (40 + 44)

On further calculation

Median = ½ (84)

We get

Median = 42

Therefore, the median age is 42 years.

Mode is the value of the variable which occurs most frequently.

Arranging the given numbers in ascending order, we get

28, 31, 32, 32, 32, 32, 34, 36, 38, 40, 41

Clearly, 32 occurs maximum number of times.

Hence, mode of the ages (in years) of 11 cricket players is 32 years

Correct answer is (d) Standard Deviation

The standard deviation is a measure of dispersion. It is the action or process of distributing thing over a wide area (nothing about central location).

Correct answer is (a)

l + \(\{h\times\cfrac{(\frac{N}{2}-cf)}f\}\)

The first six odd natural number are 1, 3, 5, 7, 9 and 11.

∴mean = (sum of the given numbers)/ (number of the given numbers)

Then,

Sum of the six odd natural numbers,

= 1 + 3 + 5 + 7 + 9 + 11

= 36

Number of the given number = 6

Now,

mean = 36 / 6

= 6

Hence, the mean of the six odd natural numbers is 6.

We know that

First eight natural numbers = 1, 2, 3, 4, 5, 6, 7 and 8

So we get

Mean = sum of numbers/ total numbers

By substituting the values

Mean = (1 + 2 + 3 + 4 + 5 + 6 + 7 + 8)/8

On further calculation

Mean = 36/8

By division

Mean = 4.5

Therefore, the mean of the first eight natural numbers is 4.5.

It is given that the number of books issued are 105, 216, 322, 167, 273, 405 and 346.

We know that

Mean = sum of numbers/ total numbers

By substituting the values

Mean = (105 + 216 + 322 + 167 + 273 + 405 + 346)/ 7

On further calculation

Mean = 1834/7

By division

Mean = 262

Therefore, the average number of books issued per day is 262.

Mode is the value of the variable which occurs most frequently.

Arranging the given numbers in ascending order, we get

4, 6, 7, 8, 8, 8, 8, 10, 11, 15

Clearly, 8 occurs maximum number of times.

Hence, mode of the given number is 8

The first 50 whole numbers are 0, 1, 2, 3, 4, …, 49.

Here n = 50, which is even.

Where, n is the number of the given number.

∴median = ½ {(n/ 2)^th term + ((n/2) + 1)^th term}

= ½ {(50/ 2)^th term + ((50/2) + 1)^th term}

= ½ {(25)^th term + ((25) + 1)^th term}

= ½ {(25)^th term + (26)^th term}

= ½ (24 + 25)

= ½ (49)

= (49/2)

= 24.5

Hence, the median of first 50 whole numbers is 24.5.

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

Here n = 10, which is even.

Where, n is the number of the given number.

∴median = ½ {(n/ 2)^th term + ((n/2) + 1)^th term}

= ½ {(10/ 2)^th term + ((10/2) + 1)^th term}

= ½ {(5)^th term + ((5) + 1)^th term}

= ½ {(5)^th term + (6)^th term}

= ½ (10 + 12)

= ½ (22)

= (22/2)

= 11

Hence, the median of first 10 even numbers is 11.

(i) The first eight natural numbers are 1, 2, 3, 4, 5, 6, 7, 8

∴ Sum of these observations =1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 = 36

and, number of their observations = 8

∴ Required mean = 36/8 = 4.5

(ii) The first six even natural numbers are 1 = 2, 4, 6, 8, 10, 12

∴Sum of these observations = 2, 4, 6, 8, 10, 12 = 42

and, number of their observations = 6

∴ Required mean = 42/6 = 7

(iii) The first five odd natural numbers are = 1, 3, 5, 7, 9

∴Sum of these observations =1 + 3 + 5 + 7 + 9 = 25

and, number of their observations = 5

∴Required mean = 25/5 = 5

(iv) The all prime numbers upto 30 are 2, 3, 5, 7, 11, 13, 17, 19, 23, 29

∴Sum of these observations = 2 + 3 +5 + 7+ 11 + 13 + 17 + 19 + 23 +29 = 129

and, number of their observations = 10

∴ Required mean = 129/10= 12.9

(v) All prime numbers between 20 and 40 are 23, 29, 31, 37

Sum of these observations = 23 + 29 + 31 + 37 = 120 .

and, number of their observations = 4

∴ Required mean = 120/4 = 30