सर्विाम ककसे कहते हैं? इसके ककतिे भेद होते हैं? िाम निनिए |
Answers
Explanation:
For Mean :
frequency (fi) = 50
fi xi = 1860
Using Formula :
\longmapsto\tt\boxed{Mean=\dfrac{\sigma\:fixi}{fi}}⟼Mean=fiσfixi
Putting Values :
\longmapsto\tt{Mean=\cancel\dfrac{1860}{50}}⟼Mean=501860
\longmapsto\tt\bf{Mean=37.2}⟼Mean=37.2
For Mode :
{f}_{0}=10f0=10
{f}_{1}=12f1=12
{f}_{2}=8f2=8
l = 40
h = 10
Using Formula :
\longmapsto\tt\boxed{Mode=l+\dfrac{{f}_{1}-{f}_{0}}{{2f}_{1}-{f}_{0}-{f}_{2}}\times{h}}⟼Mode=l+2f1−f0−f2f1−f0×h
Putting Values :
\longmapsto\tt{40+\bigg(\dfrac{12-10}{24-18}\bigg)\times{10}}⟼40+(24−1812−10)×10
\longmapsto\tt{40+\dfrac{2}{6}\times{10}}⟼40+62×10
\longmapsto\tt{40+\dfrac{10}{3}}⟼40+310
\longmapsto\tt{40+3.3}⟼40+3.3
\longmapsto\tt\bf{43.3}⟼43.3
For Median :
n / 2 = 50/2 = 25
l = 40
h = 10
cf = 26
Using Formula :
\boxed{Median=l+\bigg(\dfrac{\dfrac{n}{2}-cf}{f}\bigg)\times{h}}Median=l+(f2n−cf)×h
Putting Values :
\longmapsto\tt{40=\bigg(\dfrac{25-26}{12}\bigg)\times{10}}⟼40=(1225−26)×10
\longmapsto\tt{40+\dfrac{(-1)}{12}\times{10}}⟼40+12(−1)×10
\longmapsto\tt{40-\dfrac{10}{12}}⟼40−1210
\longmapsto\tt{40-0.8}⟼40−0.8
\longmapsto\tt\bf{39.2}⟼39.2
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{f}_{0}f0 =class preceding the modal class .
{f}_{1}f1 =frequency of modal class .
{f}_{2}f2 =class secceeding the modal class .
l = lower limit
h = class size
n = number of observations
cf = cumulative frequency of class preceding the median class .
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