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Notes, Exercises, Videos, Tests and Things to Remember on Quartiles

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If three numbers divide the series into four equal parts then, these three numbers are called quartiles. Here 9, 13 and 27 divide the series into 4 equal parts. Therefore, 9, 13 and 27 are called quartiles. 9 is called the first quartile. (Q1). 25% of the items are less than Q1 and 75% of the items are more thanQ1. 13 is called the second quartile. 50% of the items are below and above it. Similarly, 27 is called the third quartile. 75% of the items are less than Q3and 25% are more than Q3.

Therefore, quartiles are also the positional value of the items.

Calculation of quartiles in continuous series
Steps:

a. Construct the cumulative frequency table.

b. Use the formula to find the quartiles class.

The first quartile, Q1 = ($$\frac{N}{4}$$)th class

The third quartile, Q3 = ($$\frac{3N}{4}$$)th class

c. Use the formula to find the exact value of quartiles.

$$Q_1 = L + \frac {\frac{N}{4} - C.f }{f} \times h$$

$$Q_3 = L + \frac {\frac{3N}{4} - C.f }{f} \times h$$

Where,
L = lower limit of the class
c.f. = cumulative frequency of preceding class
f = frequency of class
h = width of class-interval

#### Ogive (Cumulative frequency curve )

Ogive is a graphical representation of the distribution of a continuous data. While drawing ogive, points are plotted with cumulative frequency along with y-axis and its corresponding variables along with x-axis.There are two types of the ogive. They are less than ogive and more than ogive.

$$Q_1 = L + \frac {\frac{N}{4} - C.f }{f} \times h$$

$$Q_3 = L + \frac {\frac{3N}{4} - C.f }{f} \times h$$

Where,
L = lower limit of the class
c.f. = cumulative frequency of preceding class
f = frequency of class
h = width of class-interval

.

### Very Short Questions

Solution:

The given data write in ascending order: 40, 49, 50, 55, 60, 61, 70 where N = 7

\begin{align*} Position \: of\: Q_3 &= \frac{3(N + 1)^{th}}{4}term \\ &= \frac{3(7+1)^{th}}{4}\\ &= 6^{th} \: term \\ 6^{th} \: term \: &represent \: 61 \\ \therefore Upper \: quartile \: &(Q_3) = 61 \end{align*}

Solution:

The given data write in ascending order
30, 31, 37,40, 42, 43, 45, 48, where N = 8

\begin{align*} Position \: of \: Q_3 &= \frac{3(N + 1)^{th}}{4}term \\ &= \frac{3(8 + 1)^{th}}{4} term \\ &= 3 \times 2.25^{th} \: term \\ &= 6.75^{th} \: term \\ \: \\ Q_3 = 6^{th}\:term + &(7^{th}term - 6^{th} term) \times 0.75\\ Q_3 =43 + (45 - &43) \times 0.75 \\ Q_3 = 43 + 1.5 \\ \therefore \: \: Q_3 = 44.5 \: \: _{Ans} \end{align*}

Solution:

N = 7

\begin{align*} Position \: of \: Q_1 &= \frac{N + 1^{th}}{4}term \\ &= \frac{7 + 1^{th}}{4}term \\ &= 2^{nd} \: term \\ \: \\ 2^{th}\: term \: repr&esent \: 3x -1 \\ \: \\ Q_1 &= 3x -1 \\ 20 &= 3x -1 \\ or, 3x&= 20 +1 \\ or, x &= \frac{21}{3}\\ \therefore x &= 7 \: _{Ans} \end{align*}

Solution:

The given data arranging in ascending order.
11, 12, 14, 17, 18, 19, 22, 25 where N = 8.

\begin{align*} Position \: of \: first \: quartile \: (Q_1) &= \frac{N+1^{th}}{4}term \\ &= \frac{8+1}{4}term \\ &= 2.25^{th} \: term \\ \: \\ Q_1 = 2^{nd} term + (3^{rd} - &2^{nd}) term \times 0.25 \\ Q_1 = 12 + (14-12) \times &0.25 \\ Q_1 = 12 + 2 \times 0.25\\ \therefore Q_1 = 12.5 \: \: _{Ans} \end{align*}

Solution:

Calculating lower quartile (Q1)

 Marks (X) Frequency (f) Cumulative frequency (cf) 50 3 3 60 4 7 70 7 14 80 5 19 90 2 21 100 9 30 N = 30

\begin{align*}Position \: of \: lower \: quartile (Q_1) &= \frac{N + 1^{th}}{4}term \\ &= \frac{30 + 1^{th}}{4}term \\ &= 7.75^{th} term \end{align*}

7.75th term represent cf value 14.

$$\therefore$$ lower quartile (Q1) = 70 $$_{Ans}$$

Solution:

The given data write in ascending order
$$1,5,7,2x-4,x+7,2x+1,3x+2$$ where N = 7

\begin{align*} Position \: of \: Q_3 &= \frac{3(N + 1)^{th}}{4}term \\ &= \frac{3(7+1)^{th}}{4}term \\ &= 3 \times 2^{th} \: term \\ &= 6^{th} term \\ 6^{th} term \: repr&esent \: 2x + 1 \end{align*}

\begin{align*} Q_1 &= 2x+1 \\ 2x + 1 &= 15 \\ or, 2x &= 15 -1 \\ \therefore x &= \frac{14}{2} =7 \: _{Ans} \end{align*}

Solution:

Calculating the Q3 class

 Marks (x) No. of students(f) Cumulative frequenc (cf) 0 - 10 4 4 10 - 20 8 12 20 - 30 12 24 30 - 40 4 28 N = 28

\begin{align*} Position \: of \: Q_3 \: class &= \frac{3N^{th}}{4}term \\ &= \frac{3 \times 28^{th}}{4} term \\ &= 21^{th} \: term \\ Class \: of \: Q_3 &= 20 - 30 \: _{Ans}\end{align*}

solution:

Calculating the Q3 class

 Marks (x) No. of students(f) Cumulative frequenc (cf) 10 5 5 20 4 9 30 5 14 40 6 20 50 7 27 N = 27

\begin{align*} Position \: of \: upper\:quartile (Q_3) &= \frac{3(N+1)^{th}}{4}term \\ &= \frac{3(27 + 1)^{th}}{4}term \\ &= 3 \times 7^{th} \: term \\ &=21 ^{th} \: term\\ \: \\ 21^{th} \: term \: represent \: c.f \: 27 \\ \therefore Upper \: quartile \: (Q_3) &= 50 \: \: _{Ans} \end{align*}

Solution:

From graph,
Total number of boys (N) = 60

\begin{align*} Position \: of \: first \: quartile &= \frac{N^{th}}{4}term \\ &= \frac{60^{th}}{4}term \\ &= 15^{th} term \\ Class \: of \: first \: quartile &= 5 - 10 \: _{Ans} \end{align*}

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