Mean & Median

Sum of all observations = 11 + 13 + 17 + 19 + 23 = 83

Number of observations = 5

By formula,

Mean = $\frac{\text{Sum of all observations}}{\text{Number of observations}}$

Substituting the values, we get :

Required mean = $\frac{83}{5}$ = 16.6

Hence, the mean of the given observations is 16.6.

Sum of all observations = 22 + 24 + 26 + 28 + 30 + 32 + 34 + 36

= 232

Number of observations = 8

By formula,

Mean = $\frac{\text{Sum of all observations}}{\text{Number of observations}}$

Substituting the values, we get :

Required mean = $\frac{232}{8}$ = 29

Hence, the mean of the given observations is 29.

Sum of all observations = $\frac{1}{4} + 3\frac{1}{4} + 4\frac{3}{4} + 5\frac{1}{4} + 7\frac{1}{2}$

$$\begin{aligned}= \frac{1}{4} + \frac{13}{4} + \frac{19}{4} + \frac{21}{4} + \frac{15}{2} \\[1em] = \frac{1 + 13 + 19 + 21 + 30}{4} \\[1em] = \frac{84}{4} \\[1em] = 21\end{aligned}$$

Number of observations = 5

By formula,

Mean = $\frac{\text{Sum of all observations}}{\text{Number of observations}}$

Substituting the values, we get :

Required mean = $\frac{21}{5} = 4\frac{1}{5}$

Hence, the mean of the given observations is $\mathbf{4\frac{1}{5}}$.

Sum of all observations = 6.5 + 8.2 + 9.4 + 4.6 + 7.8 + 4.9

= 41.4

Number of observations = 6

By formula,

Mean = $\frac{\text{Sum of all observations}}{\text{Number of observations}}$

Substituting the values, we get :

Required mean = $\frac{41.4}{6}$ = 6.9

Hence, the mean of the given observations is 6.9.

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

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

Number of observations = 8

By formula,

Mean = $\frac{\text{Sum of all observations}}{\text{Number of observations}}$

Substituting the values, we get :

Required mean = $\frac{36}{8}$ = 4.5

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

First five multiples of 6 are 6, 12, 18, 24 and 30.

Sum of all observations = 6 + 12 + 18 + 24 + 30 = 90

Number of observations = 5

By formula,

Mean = $\frac{\text{Sum of all observations}}{\text{Number of observations}}$

Substituting the values, we get :

Required mean = $\frac{90}{5}$ = 18

Hence, the mean of first five multiples of 6 is 18.

A prime number has exactly two factors, 1 and the number itself.

All prime numbers between 20 and 40 are 23, 29, 31 and 37.

Sum of all observations = 23 + 29 + 31 + 37

= 120

Number of observations = 4

By formula,

Mean = $\frac{\text{Sum of all observations}}{\text{Number of observations}}$

Substituting the values, we get :

Required mean = $\frac{120}{4}$ = 30

Hence, the mean of all prime numbers between 20 and 40 is 30.

Sum of all observations = 2 + 4 + 3 + 4 + 2 + 0 + 3 + 5 + 1 + 6

= 30

Number of observations = 10

By formula,

Mean = $\frac{\text{Sum of all observations}}{\text{Number of observations}}$

Substituting the values, we get :

Mean number of children per family = $\frac{30}{10}$ = 3

Hence, the mean number of children per family is 3.

Sum of all observations = 105 + 216 + 322 + 167 + 273 + 405 + 346

= 1834

Number of observations = 7

By formula,

Mean = $\frac{\text{Sum of all observations}}{\text{Number of observations}}$

Substituting the values, we get :

Mean number of books issued per day = $\frac{1834}{7}$ = 262

Hence, the mean number of books issued per day is 262.

Sum of all observations = (35.5 + 30.8 + 27.3 + 32.1 + 23.8 + 29.9)°F

= 179.4°F

Number of observations = 6

By formula,

Mean = $\frac{\text{Sum of all observations}}{\text{Number of observations}}$

Substituting the values, we get :

Mean temperature = $\frac{179.4}{6}$°F = 29.9°F

Hence, the mean temperature is 29.9°F.

Given,

Mean of 9, 14, $x$, 16, 7, 18 = 11.5

Sum of all observations = 9 + 14 + $x$ + 16 + 7 + 18

= 64 + $x$

Number of observations = 6

By formula,

Mean = $\frac{\text{Sum of all observations}}{\text{Number of observations}}$

Substituting the values, we get :

$$\begin{aligned}\Rightarrow \frac{64 + x}{6} = 11.5 \\[1em] \Rightarrow 64 + x = 11.5 \times 6 \\[1em] \Rightarrow 64 + x = 69 \\[1em] \Rightarrow x = 69 - 64 \\[1em] \Rightarrow x = 5\end{aligned}$$

Hence, the value of $\bm{x = 5}$.

Given,

Mean of $7, 9, x + 3, 12, 2x - 1, 3$ = 9

Sum of all observations = 7 + 9 + $(x + 3)$ + 12 + $(2x - 1)$ + 3

= $3x + 33$

Number of observations = 6

By formula,

Mean = $\frac{\text{Sum of all observations}}{\text{Number of observations}}$

Substituting the values, we get :

$$\begin{aligned}\Rightarrow \frac{3x + 33}{6} = 9 \\[1em] \Rightarrow 3x + 33 = 9 \times 6 \\[1em] \Rightarrow 3x + 33 = 54 \\[1em] \Rightarrow 3x = 54 - 33 \\[1em] \Rightarrow 3x = 21 \\[1em] \Rightarrow x = \frac{21}{3} \\[1em] \Rightarrow x = 7\end{aligned}$$

Hence, the value of $\bm{x = 7}$.

On arranging the given data in ascending order of their magnitudes, we get :

3, 6, 7, 11, 12, 15, 16, 20, 23

Since, the number of terms ($n$) is 9 and 9 is odd :

$$\begin{aligned}\text{Median} = \Big(\frac{n + 1}{2}\Big)^{\text{th}}\text{ term} \\[1em] = \Big(\frac{9 + 1}{2}\Big)^{\text{th}}\text{ term} \\[1em] = 5^{\text{th}}\text{ term} \\[1em] = 12\end{aligned}$$

Hence, the median of the given data is 12.

On arranging the given data in ascending order of their magnitudes, we get :

0, 7, 9, 13, 16, 19, 25, 32, 37, 50, 51

Since, the number of terms ($n$) is 11 and 11 is odd :

$$\begin{aligned}\text{Median} = \Big(\frac{n + 1}{2}\Big)^{\text{th}}\text{ term} \\[1em] = \Big(\frac{11 + 1}{2}\Big)^{\text{th}}\text{ term} \\[1em] = 6^{\text{th}}\text{ term} \\[1em] = 19\end{aligned}$$

Hence, the median of the given data is 19.

On arranging the given data in ascending order of their magnitudes, we get :

1.8, 2.6, 3.4, 4.3, 5.6, 7.2, 9.1

Since, the number of terms ($n$) is 7 and 7 is odd :

$$\begin{aligned}\text{Median} = \Big(\frac{n + 1}{2}\Big)^{\text{th}}\text{ term} \\[1em] = \Big(\frac{7 + 1}{2}\Big)^{\text{th}}\text{ term} \\[1em] = 4^{\text{th}}\text{ term} \\[1em] = 4.3\end{aligned}$$

Hence, the median of the given data is 4.3.

On arranging the given data in ascending order of their magnitudes, we get :

108, 109, 118, 122, 125, 127

As the number of terms ($n$) is 6 and 6 is even :

$$\begin{aligned}\Rightarrow \text{Median} = \frac{1}{2}\Big[\Big(\frac{n}{2}\Big)^{\text{th}}\text{ term} + \Big(\frac{n}{2} + 1\Big)^{\text{th}}\text{ term}\Big] \\[1em] = \frac{1}{2}\Big[\Big(\frac{6}{2}\Big)^{\text{th}}\text{ term} + \Big(\frac{6}{2} + 1\Big)^{\text{th}}\text{ term}\Big] \\[1em] = \frac{1}{2}[3^{\text{rd}}\text{ term} + 4^{\text{th}}\text{ term}] \\[1em] = \frac{1}{2}(118 + 122) \\[1em] = \frac{1}{2} \times 240 \\[1em] = 120\end{aligned}$$

Hence, the median of the given data is 120.

On arranging the given data in ascending order of their magnitudes, we get :

17, 19, 24, 25, 26, 31, 40

Since, the number of terms ($n$) is 7 and 7 is odd :

$$\begin{aligned}\text{Median} = \Big(\frac{n + 1}{2}\Big)^{\text{th}}\text{ term} \\[1em] = \Big(\frac{7 + 1}{2}\Big)^{\text{th}}\text{ term} \\[1em] = 4^{\text{th}}\text{ term} \\[1em] = 25\end{aligned}$$

Hence, the median score is 25.

On arranging the given data in ascending order of their magnitudes, we get :

0, 13, 15, 20, 27, 29, 31, 34, 43, 50, 56

Since, the number of terms ($n$) is 11 and 11 is odd :

$$\begin{aligned}\text{Median} = \Big(\frac{n + 1}{2}\Big)^{\text{th}}\text{ term} \\[1em] = \Big(\frac{11 + 1}{2}\Big)^{\text{th}}\text{ term} \\[1em] = 6^{\text{th}}\text{ term} \\[1em] = 29\end{aligned}$$

Hence, the median score is 29.

On arranging the given data in ascending order of their magnitudes, we get :

9.8, 10.6, 12.7, 13.4, 14.3, 15, 16.5, 17.2

As the number of terms ($n$) is 8 and 8 is even :

$$\begin{aligned}\Rightarrow \text{Median} = \frac{1}{2}\Big[\Big(\frac{n}{2}\Big)^{\text{th}}\text{ term} + \Big(\frac{n}{2} + 1\Big)^{\text{th}}\text{ term}\Big] \\[1em] = \frac{1}{2}\Big[\Big(\frac{8}{2}\Big)^{\text{th}}\text{ term} + \Big(\frac{8}{2} + 1\Big)^{\text{th}}\text{ term}\Big] \\[1em] = \frac{1}{2}[4^{\text{th}}\text{ term} + 5^{\text{th}}\text{ term}] \\[1em] = \frac{1}{2}(13.4 + 14.3) \\[1em] = \frac{1}{2} \times 27.7 \\[1em] = 13.85\end{aligned}$$

Hence, the median weight is 13.85 kg.

On arranging the given data in ascending order of their magnitudes, we get :

143.7, 144.2, 145, 146.5, 147.3, 148.5, 149.6, 150, 152.1

Since, the number of terms ($n$) is 9 and 9 is odd :

$$\begin{aligned}\text{Median} = \Big(\frac{n + 1}{2}\Big)^{\text{th}}\text{ term} \\[1em] = \Big(\frac{9 + 1}{2}\Big)^{\text{th}}\text{ term} \\[1em] = 5^{\text{th}}\text{ term} \\[1em] = 147.3\end{aligned}$$

Hence, the median height is 147.3 cm.

Sum of all observations = 5 + 11 + 17 + 23 + 31 + 36

= 123

Number of observations = 6

By formula,

Mean = $\frac{\text{Sum of all observations}}{\text{Number of observations}}$

Substituting the values, we get :

Required mean = $\frac{123}{6}$ = 20.5

Hence, option 2 is the correct option.

Given,

Mean of 8 numbers = 136.5

Number of observations = 8

By formula,

Mean = $\frac{\text{Sum of all observations}}{\text{Number of observations}}$

⇒ Sum of all observations = Mean × Number of observations

Substituting the values, we get :

Sum of these numbers = 136.5 × 8 = 1092

Hence, option 3 is the correct option.

Given,

Mean of the marks of 10 students = 21

Sum of observations = 24 + 30 + 26 + 15 + 22 + 20 + 12 + 25 + 21 + $x$

= 195 + $x$

Number of observations = 10

By formula,

Mean = $\frac{\text{Sum of all observations}}{\text{Number of observations}}$

Substituting the values, we get :

$$\begin{aligned}\Rightarrow \frac{195 + x}{10} = 21 \\[1em] \Rightarrow 195 + x = 21 \times 10 \\[1em] \Rightarrow 195 + x = 210 \\[1em] \Rightarrow x = 210 - 195 \\[1em] \Rightarrow x = 15\end{aligned}$$

Hence, option 4 is the correct option.

First five whole numbers are 0, 1, 2, 3 and 4.

Sum of all observations = 0 + 1 + 2 + 3 + 4 = 10

Number of observations = 5

By formula,

Mean = $\frac{\text{Sum of all observations}}{\text{Number of observations}}$

Substituting the values, we get :

Required mean = $\frac{10}{5}$ = 2

Hence, option 1 is the correct option.

Given,

Mean of 3, 5, 8, $p$ = 7

Sum of all observations = 3 + 5 + 8 + $p$

= 16 + $p$

Number of observations = 4

By formula,

Mean = $\frac{\text{Sum of all observations}}{\text{Number of observations}}$

Substituting the values, we get :

$$\begin{aligned}\Rightarrow \frac{16 + p}{4} = 7 \\[1em] \Rightarrow 16 + p = 7 \times 4 \\[1em] \Rightarrow 16 + p = 28 \\[1em] \Rightarrow p = 28 - 16 \\[1em] \Rightarrow p = 12\end{aligned}$$

Hence, option 4 is the correct option.

First five prime numbers are 2, 3, 5, 7 and 11.

Sum of all observations = 2 + 3 + 5 + 7 + 11

= 28

Number of observations = 5

By formula,

Mean = $\frac{\text{Sum of all observations}}{\text{Number of observations}}$

Substituting the values, we get :

Required mean = $\frac{28}{5}$ = 5.6

Hence, option 3 is the correct option.

First five multiples of 3 are 3, 6, 9, 12 and 15.

Sum of all observations = 3 + 6 + 9 + 12 + 15 = 45

Number of observations = 5

By formula,

Mean = $\frac{\text{Sum of all observations}}{\text{Number of observations}}$

Substituting the values, we get :

Required mean = $\frac{45}{5}$ = 9

Hence, option 2 is the correct option.

Given,

Mean of 10 observations = 18

By formula,

Mean = $\frac{\text{Sum of all observations}}{\text{Number of observations}}$

⇒ Sum of all observations = Mean × Number of observations

⇒ Sum of 10 observations (with the misread value) = 18 × 10 = 180

Since 12 was misread as 21, the wrong value 21 was included in this sum in place of the correct value 12. So, to get the correct sum, we add 12 and subtract 21.

Correct sum = 180 + 12 − 21 = 171

∴ Correct mean = $\frac{171}{10}$ = 17.1

Hence, option 2 is the correct option.

Given,

Mean of $10, 6, 13, x + 3, 2, 2x - 1 = 9$

Sum of all observations = 10 + 6 + 13 + $(x + 3)$ + 2 + $(2x - 1)$

= 3x + 33

Number of observations = 6

By formula,

Mean = $\frac{\text{Sum of all observations}}{\text{Number of observations}}$

Substituting the values, we get :

$$\begin{aligned}\Rightarrow \frac{3x + 33}{6} = 9 \\[1em] \Rightarrow 3x + 33 = 9 \times 6 \\[1em] \Rightarrow 3x + 33 = 54 \\[1em] \Rightarrow 3x = 54 - 33 \\[1em] \Rightarrow 3x = 21 \\[1em] \Rightarrow x = \frac{21}{3} \\[1em] \Rightarrow x = 7\end{aligned}$$

Hence, option 1 is the correct option.

On arranging the given data in ascending order of their magnitudes, we get :

20, 27, 41, 45, 50, 55, 56

Since, the number of terms ($n$) is 7 and 7 is odd :

$$\begin{aligned}\text{Median} = \Big(\frac{n + 1}{2}\Big)^{\text{th}}\text{ term} \\[1em] = \Big(\frac{7 + 1}{2}\Big)^{\text{th}}\text{ term} \\[1em] = 4^{\text{th}}\text{ term} \\[1em] = 45\end{aligned}$$

Hence, option 3 is the correct option.

On arranging the given data in ascending order of their magnitudes, we get :

8, 14, 16, 23, 26, 31, 35, 40

As the number of terms ($n$) is 8 and 8 is even :

$$\begin{aligned}\Rightarrow \text{Median} = \frac{1}{2}\Big[\Big(\frac{n}{2}\Big)^{\text{th}}\text{ term} + \Big(\frac{n}{2} + 1\Big)^{\text{th}}\text{ term}\Big] \\[1em] = \frac{1}{2}\Big[\Big(\frac{8}{2}\Big)^{\text{th}}\text{ term} + \Big(\frac{8}{2} + 1\Big)^{\text{th}}\text{ term}\Big] \\[1em] = \frac{1}{2}[4^{\text{th}}\text{ term} + 5^{\text{th}}\text{ term}] \\[1em] = \frac{1}{2}(23 + 26) \\[1em] = \frac{1}{2} \times 49 \\[1em] = 24.5\end{aligned}$$

Hence, option 4 is the correct option.

(i) Given,

Mean of 8 numbers = 18

Number of observations = 8

By formula,

Mean = $\frac{\text{Sum of all observations}}{\text{Number of observations}}$

⇒ Sum of these numbers = Mean × Number of observations

= 18 × 8

= 144

Hence, 8 numbers have mean 18. The sum of these numbers is 144.

(ii) Prime numbers between 20 and 30 are 23 and 29.

Sum of all observations = 23 + 29 = 52

Number of observations = 2

By formula,

Mean = $\frac{\text{Sum of all observations}}{\text{Number of observations}}$

Substituting the values, we get :

Required mean = $\frac{52}{2}$ = 26

Hence, the mean of all prime numbers between 20 and 30 is 26.

(iii) Sum of all observations

$$\begin{aligned}= x + 2x + 3 + (5x + 1) \\[1em] = 8x + 4\end{aligned}$$

Number of observations = 4

By formula,

Mean = $\frac{\text{Sum of all observations}}{\text{Number of observations}}$

Substituting the values, we get :

$$\begin{aligned}\text{Required mean} = \frac{8x + 4}{4} \\[1em] = \frac{4(2x + 1)}{4} \\[1em] = 2x + 1\end{aligned}$$

Hence, the mean of the given numbers is $\bm{2x + 1}$.

(iv) On arranging the given data in ascending order of their magnitudes, we get :

0, 2, 5, 6, 7, 8, 9

Since, the number of terms ($n$) is 7 and 7 is odd :

$$\begin{aligned}\text{Median} = \Big(\frac{n + 1}{2}\Big)^{\text{th}}\text{ term} \\[1em] = \Big(\frac{7 + 1}{2}\Big)^{\text{th}}\text{ term} \\[1em] = 4^{\text{th}}\text{ term} \\[1em] = 6\end{aligned}$$

Hence, the median of 5, 2, 6, 8, 0, 9, 7 is 6.

(v) On arranging the given data in ascending order of their magnitudes, we get :

9, 10, 12, 13, 14, 15, 16, 17

As the number of terms ($n$) is 8 and 8 is even :

$$\begin{aligned}\Rightarrow \text{Median} = \frac{1}{2}\Big[\Big(\frac{n}{2}\Big)^{\text{th}}\text{ term} + \Big(\frac{n}{2} + 1\Big)^{\text{th}}\text{ term}\Big] \\[1em] = \frac{1}{2}\Big[\Big(\frac{8}{2}\Big)^{\text{th}}\text{ term} + \Big(\frac{8}{2} + 1\Big)^{\text{th}}\text{ term}\Big] \\[1em] = \frac{1}{2}[4^{\text{th}}\text{ term} + 5^{\text{th}}\text{ term}] \\[1em] = \frac{1}{2}(13 + 14) \\[1em] = \frac{1}{2} \times 27 \\[1em] = 13.5\end{aligned}$$

Hence, the median of 13, 10, 12, 17, 14, 15, 16, 9 is 13.5.

(i) Given,

Mean of 9, 14, 7, 16, $x$, 18 = 11.5

Sum of all observations = 9 + 14 + 7 + 16 + $x$ + 18

= 64 + x

Number of observations = 6

By formula,

Mean = $\frac{\text{Sum of all observations}}{\text{Number of observations}}$

Substituting the values, we get :

$$\begin{aligned}\Rightarrow \frac{64 + x}{6} = 11.5 \\[1em] \Rightarrow 64 + x = 11.5 \times 6 \\[1em] \Rightarrow 64 + x = 69 \\[1em] \Rightarrow x = 69 - 64 \\[1em] \Rightarrow x = 5\end{aligned}$$

So, the value of $x$ is 5, as given in the statement.

Hence, the given statement is true (T).

(ii) First four multiples of 3 are 3, 6, 9 and 12.

Sum of all observations = 3 + 6 + 9 + 12 = 30

Number of observations = 4

By formula,

Mean = $\frac{\text{Sum of all observations}}{\text{Number of observations}}$

Substituting the values, we get :

Required mean = $\frac{30}{4}$ = 7.5

So, the mean of first four multiples of 3 is 7.5, as given in the statement.

Hence, the given statement is true (T).

(iii) When the number of observations ($n$) is odd, the median is the middle term of the data arranged in ascending order.

By formula,

$$\begin{aligned}\text{Median} = \Big(\frac{n + 1}{2}\Big)^{\text{th}}\text{ term} \\[1em] = \Big[\frac{1}{2}(n + 1)\Big]^{\text{th}}\text{ term}\end{aligned}$$

This is the same as the median given in the statement.

Hence, the given statement is true (T).

(iv) First 8 natural numbers are 1, 2, 3, 4, 5, 6, 7 and 8.

These numbers are already arranged in ascending order.

As the number of terms ($n$) is 8 and 8 is even :

$$\begin{aligned}\Rightarrow \text{Median} = \frac{1}{2}\Big[\Big(\frac{n}{2}\Big)^{\text{th}}\text{ term} + \Big(\frac{n}{2} + 1\Big)^{\text{th}}\text{ term}\Big] \\[1em] = \frac{1}{2}\Big[\Big(\frac{8}{2}\Big)^{\text{th}}\text{ term} + \Big(\frac{8}{2} + 1\Big)^{\text{th}}\text{ term}\Big] \\[1em] = \frac{1}{2}[4^{\text{th}}\text{ term} + 5^{\text{th}}\text{ term}] \\[1em] = \frac{1}{2}(4 + 5) \\[1em] = \frac{1}{2} \times 9 \\[1em] = 4.5\end{aligned}$$

So, the median of first 8 natural numbers is 4.5, as given in the statement.

Hence, the given statement is true (T).

(v) The median is the middle term of the data arranged in ascending (or descending) order.

The observation which occurs most frequently in a data is called the mode of the data, not the median.

Hence, the given statement is false (F).

1. Temperatures recorded on the first 3 days (Monday, Tuesday and Wednesday) are 78.2°F, 66.2°F and 70.1°F.

Sum of all observations = (78.2 + 66.2 + 70.1)°F

= 214.5°F

Number of observations = 3

By formula,

Mean = $\frac{\text{Sum of all observations}}{\text{Number of observations}}$

Substituting the values, we get :

Average temperature = $\frac{214.5}{3}$°F = 71.5°F

Hence, option (a) is the correct option.

2. Temperatures recorded on the last 3 days (Friday, Saturday and Sunday) are 68.4°F, 72.8°F and 74.2°F.

Sum of all observations = (68.4 + 72.8 + 74.2)°F = 215.4°F

Number of observations = 3

By formula,

Mean = $\frac{\text{Sum of all observations}}{\text{Number of observations}}$

Substituting the values, we get :

Average temperature = $\frac{215.4}{3}$°F = 71.8°F

Hence, option (b) is the correct option.

3. Given, average temperature over the entire week (7 days) = 72.9°F

By formula,

Mean = $\frac{\text{Sum of all observations}}{\text{Number of observations}}$

⇒ Sum of all observations = Mean × Number of observations

∴ Sum of temperatures of the 7 days = (72.9 × 7)°F

= 510.3°F

Sum of temperatures of the remaining 6 days = (78.2 + 66.2 + 70.1 + 68.4 + 72.8 + 74.2)°F

= 429.9°F

The temperature on Thursday is the difference between the sum of temperatures of all 7 days and the sum of temperatures of the remaining 6 days.

∴ Temperature recorded on Thursday ($x$) = (510.3 − 429.9)°F

= 80.4°F

Hence, option (d) is the correct option.

4. The temperatures (in °F) of the entire week, taking Thursday's temperature as 80.4°F (found in part 3), are 78.2, 66.2, 70.1, 80.4, 68.4, 72.8 and 74.2.

On arranging the given data in ascending order of their magnitudes, we get :

66.2, 68.4, 70.1, 72.8, 74.2, 78.2, 80.4

Since, the number of terms ($n$) is 7 and 7 is odd :

$$\begin{aligned}\text{Median} = \Big(\frac{n + 1}{2}\Big)^{\text{th}}\text{ term} \\[1em] = \Big(\frac{7 + 1}{2}\Big)^{\text{th}}\text{ term} \\[1em] = 4^{\text{th}}\text{ term} \\[1em] = 72.8^{\circ}\text{F}\end{aligned}$$

So, the median (medium) temperature over the entire week is 72.8°F.

Hence, option (c) is the correct option.

On arranging the given data in ascending order of their magnitudes, we get :

40, 50, 60, 68, 94, 98, 99

Since, the number of terms ($n$) is 7 and 7 is odd :

$$\begin{aligned}\text{Median} = \Big(\frac{n + 1}{2}\Big)^{\text{th}}\text{ term} \\[1em] = \Big(\frac{7 + 1}{2}\Big)^{\text{th}}\text{ term} \\[1em] = 4^{\text{th}}\text{ term} \\[1em] = 68\end{aligned}$$

So, Assertion (A) is true.

When the number of observations is odd, the middle term of the data arranged in ascending order is the median. So, Reason (R) is true.

The median 68 in Assertion (A) is found as the middle (4^th) term of the 7 observations.

So, Reason (R) is true.

∴ Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).

Hence, option 1 is the correct option.

The mean of a data need not be one of the given observations. For example, the mean of 1 and 2 is $\frac{1 + 2}{2}$ = 1.5, which is not in the given data.

So, Assertion (A) is false.

Mean = $\frac{\text{Sum of observations}}{\text{Number of observations}}$ is the correct formula for the mean.

So, Reason (R) is true.

∴ Assertion (A) is false but Reason (R) is true.

Hence, option 4 is the correct option.

Given,

Mean of 25, 29, 25, 32, 24, $x$ = 27

Sum of all observations = 25 + 29 + 25 + 32 + 24 + $x$

= 135 + $x$

Number of observations = 6

By formula,

Mean = $\frac{\text{Sum of all observations}}{\text{Number of observations}}$

Substituting the values, we get :

$$\begin{aligned}\Rightarrow \frac{135 + x}{6} = 27 \\[1em] \Rightarrow 135 + x = 27 \times 6 \\[1em] \Rightarrow 135 + x = 162 \\[1em] \Rightarrow x = 162 - 135 \\[1em] \Rightarrow x = 27\end{aligned}$$

So, the data is 25, 29, 25, 32, 24 and 27.

On arranging the given data in ascending order of their magnitudes, we get :

24, 25, 25, 27, 29, 32

As the number of terms ($n$) is 6 and 6 is even :

$$\begin{aligned}\Rightarrow \text{Median} = \frac{1}{2}\Big[\Big(\frac{n}{2}\Big)^{\text{th}}\text{ term} + \Big(\frac{n}{2} + 1\Big)^{\text{th}}\text{ term}\Big] \\[1em] = \frac{1}{2}\Big[\Big(\frac{6}{2}\Big)^{\text{th}}\text{ term} + \Big(\frac{6}{2} + 1\Big)^{\text{th}}\text{ term}\Big] \\[1em] = \frac{1}{2}[3^{\text{rd}}\text{ term} + 4^{\text{th}}\text{ term}] \\[1em] = \frac{1}{2}(25 + 27) \\[1em] = \frac{1}{2} \times 52 \\[1em] = 26\end{aligned}$$

Hence, option 3 is the correct option.

The given observations in ascending order are 8, 11, 12, $(x + 6)$, 17, 18 and 23.

Since, the number of terms ($n$) is 7 and 7 is odd :

$$\begin{aligned}\text{Median} = \Big(\frac{n + 1}{2}\Big)^{\text{th}}\text{ term} \\[1em] = \Big(\frac{7 + 1}{2}\Big)^{\text{th}}\text{ term} \\[1em] = 4^{\text{th}}\text{ term} \\[1em] = x + 6\end{aligned}$$

Given, median = 15

$$\begin{aligned}\Rightarrow x + 6 = 15 \\[1em] \Rightarrow x = 15 - 6 \\[1em] \Rightarrow x = 9\end{aligned}$$

Hence, option 2 is the correct option.

Given,

Mean of the marks of 25 students = 30

By formula,

Mean = $\frac{\text{Sum of all observations}}{\text{Number of observations}}$

⇒ Sum of all observations = Mean × Number of observations

∴ Sum of the marks = 30 × 25 = 750

On adding 15 marks to every student, the sum increases by 15 × 25 = 375.

∴ New sum of the marks = 750 + 375 = 1125

∴ New mean = $\frac{1125}{25}$ = 45

So, if 15 marks are added to every student, the mean also increases by 15, i.e. new mean = 30 + 15 = 45.

Hence, option 3 is the correct option.

In any data, every observation is greater than or equal to the smallest observation. So, the sum of all observations is greater than or equal to (number of observations × smallest observation). Dividing both by the number of observations, the mean is greater than or equal to the smallest observation.

Similarly, every observation is less than or equal to the largest observation. So, the sum of all observations is less than or equal to (number of observations × largest observation), and hence the mean is less than or equal to the largest observation.

Thus, the mean of a data always lies between the smallest and the largest observations of the data (both inclusive).

So, the mean can neither be smaller than the smallest number nor bigger than the largest number of the data.

∴ Both the statements (i) and (ii) are wrong.

Hence, option 4 is the correct option.

Total number of observations = 5 + 2 + 8 = 15

On arranging the given data in ascending order of their magnitudes, we get :

10 (1^st to 5^th terms), 11 (6^th and 7^th terms) and 12 (8^th to 15^th terms).

Since, the number of terms ($n$) is 15 and 15 is odd :

$$\begin{aligned}\text{Median} = \Big(\frac{n + 1}{2}\Big)^{\text{th}}\text{ term} \\[1em] = \Big(\frac{15 + 1}{2}\Big)^{\text{th}}\text{ term} \\[1em] = 8^{\text{th}}\text{ term} \\[1em] = 12\end{aligned}$$

Hence, option 4 is the correct option.