Question

Hey Mates...

Please answer these 2 question given in the attachment with full explanation...

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Answer 1

hope it helps you ✔️✅ ✔️✅✔️✅ ✔️

Answer 2

Q 4. Solution :-

$\begin{tabular}{|c|c|}\cline{1-2}Class Interval&Frequency\\\cline{1-2} 3-6&2\\\cline{1-2}6-9&5\\\cline{1-2}9-12&10\\\cline{1-2}12-15&23\\\cline{1-2}15-18&21\\\cline{1-2}18-21&12\\\cline{1-2}21-24&3\\\cline{1-2}\end{tabular}$

The modal class is 12-15, because it has the highest frequency.

• Lower limit of the class, l = 12
• Frequency of the class, f₁ = 23
• Frequency of the previous class, f₀ = 10
• Frequency of the next class, f₂ = 21
• Class width, h = 3

We know,

$\sf{Mode=l+\bigg(\dfrac{f_1-f_0}{2f_1-f_0-f_2} \bigg) \times h}\\\\\sf{\implies Mode=12+ \bigg(\dfrac{23-10}{2\times23-10-21} \bigg) \times 3}\\\\\sf{\implies Mode=12+\bigg(\dfrac{13}{46-10-21}\bigg) \times 3 }\\\\\sf{\implies Mode= 12+\bigg(\dfrac{13}{15}\bigg) \times 3 }\\\\\sf{\implies Mode= 12+\dfrac{13}{5} }\\\\\sf{\implies Mode= \dfrac{60+13}{5} }\\\\\sf{\implies Mode= \dfrac{73}{5} }\\\\\boxed{\boxed{\sf{\implies Mode= 14.6 }}}$

$\rule{170}2$

Q 5. Solution :-

$\begin{tabular}{|c|c|}\cline{1-2}Weekly wages&No. of days\\\cline{1-2} 60-69&5\\\cline{1-2}70-79&15\\\cline{1-2}80-89&20\\\cline{1-2}90-99&30\\\cline{1-2}100-109&20\\\cline{1-2}110-119&8\\\cline{1-2}\end{tabular}$

Difference between the upper limit of a class and the lower limit of the succeeding class = 1.

→ 1 ÷ 2 = 0.5

Subtracting 0.5 from the lower limit and adding 0.5 to the upper limit and then calculating the cumulative frequency(c.f.) →

$\begin{tabular}{|c|c|c|}\cline{1-3}Weekly wages&No. of days&C.F.\\\cline{1-3} 59.5-69.5&5&5\\\cline{1-3}69.5-79.5&15&15+5=20\\\cline{1-3}79.5-89.5&20&20+20=40\\\cline{1-3}89.5-99.5&30&40+30=70\\\cline{1-3}99.5-109.5&20&70+20=90\\\cline{1-3}109.5-119.5&8&90+8=98\\\cline{1-3}&\sum f=98& \\\cline{1-3}\end{tabular}$

The median is in the class where the cumulative frequency reaches half the sum of the absolute frequencies. n/2 = 98/2 = 49

So, the median class is 89.5-99.5.

• Lower limit of the class, l = 89.5
• n = sum of the frequencies = 98
• Frequency of the median class, f = 30
• C.F. of the previous class = 40
• Width of each class, h = 10

We know,

$\displaystyle{\sf{Median = l+\frac{\frac{n}{2}-C.F. }{f} \times h }}\\\\\displaystyle{\sf{\implies Median =89.5+\frac{49-40}{30} \times 10}$

$\displaystyle{\sf{\implies Median =89.5+\frac{9}{30}\times10 }$

$\displaystyle{\sf{\implies Median = 89.5+\frac{9}{3} }$

$\displaystyle{\sf{\implies Median = 89.5+3 }$

$\boxed{\boxed{\sf{\implies Median= 92.5 }}}$