Question

Find mean, median and mode from the following
data
X (Less then)
​

Answer 1

Answer:

Classesxi fi fixi c.f  0−20106 60  6 20−40308 240  14 40−605010 500 24 60−807012840 36 80−10090 6 540  42 100−1201105 550  47 120−140130 3 390  50   ∑fi=50∑fixi=3120  ⇒  Mean=∑fi∑fixi=503120=62.4

Answer 2

Answer:

• Mean= 28.73
• Median= 29
• Mode= 32.77

Explanation:

Kindly, refer to the attachment first!

(1) Median

$\to\sf{\bar{x} = \dfrac{\sum fx}{\sum f}}$

$\to\sf{\bar{x} = \dfrac{1925}{67}}$

$\to\boxed{\sf{\bf{\bar{x} = 28.73 }}}$

(2) Median

N =67

N/2=33.5

33.5 falls under less than 30 class hence,20-30will be the median class.

• cf = 20 (cumulative frequency of the class preceding the median class)
• f = 15 (frequency of the median class)
• l1=20 (lower class boundary of median class)
• l2 =30 (upper class boundary of median class)

$\to \sf{Median = l_1 +\dfrac{\bigg(\dfrac{N}{2} - cf\bigg)}{f} \times (l_2 - l_1)}$

$\to \sf{Median = 20+\dfrac{33.5 - 20}{15} \times (30-20)}$

$\to \sf{Median = 20+\dfrac{13.5}{15} \times 10}$

$\to\boxed{ \sf{\bf{Median = 29}}}$

(3) Mode

The highest frequency is 20

So,30-40 will be the modal class.

• f1 =15  (frequency of the class preceding the modal class)
• f2 =7  (frequency of the class succeeding the modal class)
• f = 20 (frequency of the modal class)
• l1=30 (lower class boundary of modal class)
• l2 =40 (upper class boundary of modal class)

$\to \sf{Mode = l_1 +\dfrac{(f-f_1)}{(f-f_1)+(f-f_2)}\times (l_2 - l_1)}$

$\to \sf{Mode = 30 +\dfrac{(20-15)}{(20-15)+(20-7)}\times (40-30)}$

$\to \sf{Mode = 30 +\dfrac{5}{5+13}\times 10}$

$\to \boxed{\sf{\bf{Mode = 32.77}}}$