Solution:

Now,

Arranging in ascending order

42, 52, 62, 72, 82, 90, 100

Total number (N) = 7

We know that,

or, Q_{3} = 3\(\begin{pmatrix}\frac{n + 1}{4} \end{pmatrix}\)^{th}item

or, Q_{3} = 3\(\begin{pmatrix}\frac{7 + 1}{4} \end{pmatrix}\)^{th}item

or, Q_{3} = 3\(\begin{pmatrix}\frac{8}{4} \end{pmatrix}\)^{th}item

or, Q_{3} = 3 \(\times\) 2^{th}item

or, Q_{3} = 6^{th}item

\(\therefore\) The value of Q_{3}is 90.

####
Find the lower quartile and upper quartile of the following data set of scores:

18 20 23 20 23 27 24

Solution:

Arrange the values in ascending order of magnitude:

18 20 20 23 23 23 24 27 29

Here, n =7

Lower quartile = value of\(\begin{pmatrix}\frac{n+1}{4} \end{pmatrix}\)^{th}term

= value of\(\begin{pmatrix}\frac{7+1}{4} \end{pmatrix}\)^{th}term

= value of\(\begin{pmatrix}\frac{8}{4} \end{pmatrix}\)^{th}term

= value of 2^{nd}item

\(\therefore\) Lower Quartile = 20

Upper Quartile = value of\(\begin{pmatrix}\frac{3(n+1)}{4} \end{pmatrix}\)^{th}term

= value of\(\begin{pmatrix}\frac{3(7+1)}{4} \end{pmatrix}\)^{th}term

= value of\(\begin{pmatrix}\frac{3 \times 8}{4} \end{pmatrix}\)^{th}term

= value of\(\begin{pmatrix}\frac{24}{4} \end{pmatrix}\)^{th}term

= value of 6^{th}term

\(\therefore\) Upper Quartile = 27

Solution:

Here,

6, x+5, 12, 14, 17, 20, 21

Q_{1}= 8

numbers = 7

We know that,

or, Q_{1} = \(\begin{pmatrix}\frac{n + 1}{4} \end{pmatrix}\)^{th} item

or, Q_{1} = \(\begin{pmatrix}\frac{7 + 1}{4} \end{pmatrix}\)^{th} item

or, Q_{1} = \(\begin{pmatrix}\frac{8}{4} \end{pmatrix}\)^{th} item

or, Q_{1} = 2^{th} item

or, Q_{1} =x + 5

Now,

or,Q_{1}= x + 5

or, 8= x + 5

or, x = 8 - 5

\(\therefore\) x = 5

Solution:

Here,

numbers = 7

Q_{3}= 60

Now,

or, Q_{3}= 3(\(\frac{N + 1}{4}\))^{th} item

or, Q_{3}= 3(\(\frac{7 + 1}{4}\))^{th} item

or, Q_{3}= 3(\(\frac{8}{4}\))^{th} item

or, Q_{3}= 3 \(\times\) 2^{th} item

or, Q_{3}= 6^{th} item

i.e.Q_{3}= x + 25

Then,

or,Q_{3}= x + 25

or, 60 = x + 25

or, x + 60 - 25

\(\therefore\) x = 35

Solution:

Now,

Arranging in ascending order

50, 52, 56, 61, 64, 68, 72

numbers = 7

We know that,

or, Q_{1} = \(\begin{pmatrix}\frac{n + 1}{4} \end{pmatrix}\)^{th}item

or, Q_{1} = \(\begin{pmatrix}\frac{7 + 1}{4} \end{pmatrix}\)^{th}item

or, Q_{1} = \(\begin{pmatrix}\frac{8}{4} \end{pmatrix}\)^{th}item

or, Q_{1} = 2^{th}item

or,Q_{1} = 52

Again,

or, Q_{2}= \(\begin{pmatrix}\frac{n + 1}{2} \end{pmatrix}\)^{th}item

or, Q_{2}= \(\begin{pmatrix}\frac{7 + 1}{2} \end{pmatrix}\)^{th}item

or, Q_{2}= \(\begin{pmatrix}\frac{8}{2} \end{pmatrix}\)^{th}item

or, Q_{2}= 4^{th}item

or, Q_{2}= 61

Similarly,

or, Q_{3}= 3\(\begin{pmatrix}\frac{n + 1}{4} \end{pmatrix}\)^{th}item

or, Q_{3}= 3\(\begin{pmatrix}\frac{7 + 1}{4} \end{pmatrix}\)^{th}item

or, Q_{3}= 3\(\begin{pmatrix}\frac{8}{4} \end{pmatrix}\)^{th}item

or, Q_{3}= 3 \(\times\)2^{th}item

or, Q_{3}= 6^{th}item

or, Q_{3 }= 68

\(\therefore\) The value ofQ_{1}, Q_{2} and Q_{3}is 52, 61 and 68 respectively.

####
Compute Q_{1}, Q_{2} and Q_{3 }from the following data:

_{1}, Q

_{2}and Q

_{3 }from the following data:

Marks | 10 | 20 | 30 | 40 | 50 | 60 | 70 |

No. of Students | 2 | 4 | 6 | 12 | 8 | 6 | 1 |

Solution:

Marks | No. of students(f) | cumulative frequency (c.f.) |

10 | 2 | 2 |

20 | 4 | 6 |

30 | 6 | 12 |

40 | 12 | 24 |

50 | 8 | 32 |

60 | 6 | 38 |

70 | 1 | 39 |

N = 39 |

We have,

Q1= value of \(\begin{pmatrix}\frac{n + 1}{4} \end{pmatrix}\)^{th} term

= value of \(\begin{pmatrix}\frac{39 + 1}{4} \end{pmatrix}\)^{th} term

= value of 10^{th} term

In c.f. column just greater than 10 is 12 and its coressponding value is 30.

\(\therefore\) Q_{1} = 30

Q_{2 }=value of \(\begin{pmatrix}\frac{n + 1}{2} \end{pmatrix}\)^{th} term

= value of \(\begin{pmatrix}\frac{39 + 1}{2} \end{pmatrix}\)^{th} term

= value of 20^{th} term

In c.f. column just greater than 20 is 24and its coressponding value is 40.

\(\therefore\) Q_{2}=40

Q_{3 }=value of \(\begin{pmatrix}\frac{3(n + 1)}{4} \end{pmatrix}\)^{th} term

=value of \(\begin{pmatrix}\frac{3(39 + 1)}{4} \end{pmatrix}\)^{th} term

=value of \(\begin{pmatrix}\frac{3 * 40}{4} \end{pmatrix}\)^{th} term

=value of \(\begin{pmatrix}\frac{3(39 + 1)}{4} \end{pmatrix}\)^{th} term

=value of 30^{th} term

In c.f. column just greater than 30 is 32 and its coressponding value is 50.

\(\therefore\) Q_{3}= 50

####
Calculate Q_{1} and Q_{}_{3} from the following data.

_{1}and Q

_{}

_{3}from the following data.

marks | 15 | 25 | 35 | 45 | 55 |

No. of students | 5 | 10 | 16 | 7 | 5 |

Solution:

Marks | Frequency (f) | c.f |

15 | 5 | 5 |

25 | 10 | 15 |

35 | 16 | 31 |

45 | 7 | 38 |

55 | 5 | 43 |

N = 43 |

Now,

or,Q_{1}= \(\begin{pmatrix}\frac{N + 1}{4} \end{pmatrix}\)^{th} item

or,Q_{1}= \(\begin{pmatrix}\frac{43 + 1}{4} \end{pmatrix}\)^{th} item

or,Q_{1}= \(\begin{pmatrix}\frac{44}{4} \end{pmatrix}\)^{th} item

or,Q_{1}= 11^{th} item

In c.f column, c.f is just greater than 11 is 15 and its corresponding value is 25. So, Q_{1} = 25

Again,

or, Q_{3} = 3\(\begin{pmatrix}\frac{n + 1}{4} \end{pmatrix}\)

or, Q_{3} = 3\(\begin{pmatrix}\frac{43 + 1}{4} \end{pmatrix}\)

or, Q_{3} = 3\(\begin{pmatrix}\frac{44}{4} \end{pmatrix}\)

or, Q_{3} = 3 \(\times\) 11^{th}item

or, Q_{3} = 33^{th} item

In c.f column, c.f is just greater than 33 is 38 and its corresponding value is 45.

So, Q_{3} is 45.

####
Calculate the first quartile (lower quartile) from the given data.

Class | 0 -10 | 10 -20 | 20 - 30 | 30 - 40 | 40 - 50 |

frequency | 6 | 8 | 10 | 7 | 5 |

Solution:

x | f | c.f |

0 -10 | 6 | 6 |

10 - 20 | 8 | 14 |

20 - 30 | 10 | 24 |

30 - 40 | 7 | 31 |

40 - 50 | 5 | 36 |

N = 36 |

Now,

or, Q_{1} = \(\begin{pmatrix}\frac{N}{4} \end{pmatrix}\)^{th} item

or, Q_{1} = \(\begin{pmatrix}\frac{36}{4} \end{pmatrix}\)^{th} item

or, Q_{1} = 9^{th} item

In c.f column, c.f is just greater than 9 is 14 and its corresponding value is 10 - 20.

So, Q_{1} lies in class 10 - 20

Then,

l = 10, c.f = 6, f = 8, and i = 10

or, Q_{1} =L + \(\frac{\frac{N}{4}-C.F}{f}\) × i

or,Q_{1} = 10 + \(\frac{\frac{36}{4}-6}{8}\) × 10

or,Q_{1} = 10 + \(\frac{9 - 6}{8}\) × 10

or,Q_{1} = 10 + \(\frac{3}{8}\) × 10

or,Q_{1} = 10 + \(\frac{3}{4}\) × 5

or,Q_{1} = 10 + 0.75× 5

or,Q_{1} = 10 +3.75

or,Q_{1} = 13.75

\(\therefore\) The value of Q_{1}is 13.75.

####
Calculate the first quartile and third quartile from the given data:

Class | 0-5 | 5-10 | 10-15 | 15-20 | 20-25 | 25-30 |

Frequency | 6 | 8 | 10 | 7 | 5 | 3 |

Solution:

Class | Frequency(f) | c.f. |

0-5 | 6 | 6 |

5-10 | 8 | 14 |

10-15 | 10 | 24 |

15-20 | 7 | 31 |

20-25 | 5 | 36 |

25-30 | 4 | 40 |

N =40 |

Now,

Q1 Class = value of (\(\frac{N}{4}\))^{th} item

= value of ( \(\frac{40}{4}\))^{th} item

= value of 10^{th} item

In c.f. column, c.f. just greater than 10 is 14 and its corresponding class is (5-10).

\(\therefore\) Q_{1} lies in the class 5-10

Here,

\(\frac{N}{4}\) = 10,l= 5, c.f. = 6, f = 8 and i = 5

\(\therefore\)Q_{1}= l + \(\frac{\frac{N}{4}− c.f}{f}\) × i

= 5 + \(\frac{10− 6}{8}\) ×5

=5 + \(\frac{4}{8}\) ×5

= 5 + 2.5

= 7.5

Again,

Q_{3} Class = value of(\(\frac{3N}{4}\))^{th} item

= value of(\(\frac{3 * 40}{4}\))^{th} item

=value of30^{th} item

In c.f. column, c.f.just greater than 30 is 31 and its corresponding class is 15-20.

Here,

\(\frac{3N}{4}\) = 30,l= 5, c.f. = 24, f = 7and i = 5

\(\therefore\)Q_{3}= l + \(\frac{\frac{3N}{4}− c.f}{f}\) × i

= 5 + \(\frac{30− 24}{7}\) ×5

= 5 + 4.28

= 9.28

####
Calculate the third quartile (upper quartile) from the following data.

Class | 5 -10 | 10 -15 | 15 - 20 | 20 - 25 | 25 - 30 | 30 - 35 | 35 - 40 |

Frequency | 4 | 6 | 10 | 15 | 10 | 6 | 5 |

Solution:

Class | frequency | c.f |

5 - 10 | 4 | 4 |

10 - 15 | 6 | 10 |

15 - 20 | 10 | 20 |

20 - 25 | 15 | 35 |

25 - 30 | 10 | 45 |

30 - 35 | 6 | 51 |

35 - 40 | 5 | 56 |

N = 56 |

Now,

or, Q_{3}= 3(\(\frac{N}{4}\))^{th} item

or, Q_{3}= 3(\(\frac{56}{4}\))^{th} item

or, Q_{3}= 3 \(\times\) 14^{th} item

or, Q_{3}= 42^{th} item

In c.f column, c.f is just greater than 42 is 45 and its corresponding value is 25 - 30.

Now,

l = 25, c.f = 35, f = 10, and i = 5

Then,

or, Q_{3}= L + \(\frac{\frac{3N}{4}-c.f}{f}\)× i

or, Q_{3}= 25 + \(\frac{42 - 35}{10}\)× 5

or, Q_{3}= 25 + \(\frac{7}{2}\)

or, Q_{3}= 25 +3.5

or, Q_{3}= 28.5

\(\therefore\) The value of Q_{3} is 28.5.

####
Calculate first, second and third quartile from the following data:

Marks | 30-40 | 40-50 | 50-60 | 60-70 | 70-80 | 80-90 |

Students | 5 | 12 | 25 | 15 | 8 | 3 |

Solution:

Marks | Students (f) | c.f. |

30-40 | 5 | 5 |

40-50 | 12 | 17 |

50-60 | 25 | 42 |

60-70 | 15 | 57 |

70-80 | 8 | 65 |

80-90 | 3 | 68 |

N =68 |

Now,

Q_{1} Class = value of (\(\frac{N}{4}\))^{th} item

= value of ( \(\frac{68}{4}\))^{th} item

= value of 17^{th} item

In c.f. column, c.f. just greater than 17 is 42 and its corresponding class is (50-60).

\(\therefore\) Q_{1} lies in the class 50-60

Here,

\(\frac{N}{4}\) = 17, l= 50, c.f. = 17, f = 25 and i =10

\(\therefore\)Q_{1}= l + \(\frac{\frac{N}{4}− c.f}{f}\) × i

= 50 + \(\frac{17− 17}{25}\) ×10

=50 + \(\frac{0}{25}\) ×10

= 50 +0

= 50

Again,

Q_{2}Class = value of (\(\frac{N}{2}\))^{th} item

= value of ( \(\frac{68}{2}\))^{th} item

= value of 34^{th} item

In c.f. column, c.f. just greater than 34 is 42 and its corresponding class is (50-60).

\(\therefore\) Q_{2}lies in the class 50-60

Here,

\(\frac{N}{2}\) = 34, l = 50, c.f. = 17, f = 25 and i =10

\(\therefore\)Q_{1}= l + \(\frac{\frac{N}{2}− c.f}{f}\) × i

= 50 + \(\frac{34− 17}{25}\) ×10

=50 + \(\frac{17}{25}\) ×10

= 50 +6.8

= 56.8

Q_{3} Class = value of(\(\frac{3N}{4}\))^{th} item

= value of(\(\frac{3 * 68}{4}\))^{th} item

=value of 51^{th} item

In c.f. column, c.f.just greater than 51 is 57 and its corresponding class is 60-70.

Here,

\(\frac{3N}{4}\) = 51, l= 60, c.f. = 42, f = 15 and i = 10

\(\therefore\)Q_{3}= l + \(\frac{\frac{3N}{4}− c.f}{f}\) × i

= 60 + \(\frac{51− 42}{15}\) × 10

= 60 + 6

= 66

####
Find the Q_{1} and Q_{3} of the given data.

_{1}and Q

_{3}of the given data.

Weight (in kg) | 50 | 60 | 70 | 80 | 90 |

No. of teachers | 8 | 16 | 12 | 4 | 5 |

Solution:

x | f | c.f |

50 | 8 | 8 |

60 | 16 | 24 |

70 | 12 | 36 |

80 | 4 | 40 |

90 | 3 | 43 |

N = 43 |

Now,

or, Q_{1} =(\(\frac{N+1}{4}\))^{th} item

or, Q_{1} =(\(\frac{43+1}{4}\))^{th} item

or, Q_{1} =(\(\frac{44}{4}\))^{th} item

or, Q_{1} =11^{th} item

In c.f column, c.f is just greater than 11 is 24 and its corresponding value is 60.

So,Q_{1} = 60

Again,

or, Q_{3} =3(\(\frac{N+1}{4}\))^{th} item

or, Q_{3} =3(\(\frac{43+1}{4}\))^{th} item

or, Q_{3} =3(\(\frac{44}{4}\))^{th} item

or, Q_{3} =3 \(\times\) 11^{th} item

or, Q_{3} =33^{th} item

In c.f column, c.f is just greater than 33 is 36 and its corresponding value is 70.

i.e.Q_{3} = 70