Find the mean of the given data-Whine Class interval - Frequency f - i - Class mark x - i - d - i - x - i - a - u - i -frac d - i h Whine0-100 - 40 - 50 -200 -211Whine100-200 - 39 - 150 -100 -1 Whine200-300 - 34 - 250 - 0 - 0 1Whine 300-400 - 30 - 350 - 100 - 111Whine400-500 - 45 - 450 - 200 - 211Wline array A- 250B- 1-188C- 250-52D- 0

The correct option is B 1/188Let A =250 [ Class interval amp; Frequency (f_i) amp; Class mark (x_i) amp; d_i = x_i A amp; u_i = d_i/h ...

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Byju's Answer

Standard VI

Mathematics

Relation between mean of Step- Deviations (u) and mean

Find the mean...

Question

Find the mean of the given data:
$\begin{matrix} C l a s s i n t e r v a l & F r e q u e n c y (f_{i}) & C l a s s m a r k (x_{i}) & d_{i} = x_{i} - a & u_{i} = \frac{d_{i}}{h} 0 - 100 & 40 & 50 & - 200 & - 2 100 - 200 & 39 & 150 & - 100 & - 1 200 - 300 & 34 & 250 & 0 & 0 300 - 400 & 30 & 350 & 100 & 1 400 - 500 & 45 & 450 & 200 & 2 \end{matrix}$

0

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\frac{1}{188}

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250

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250.52

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Solution

The correct option is B $\frac{1}{188}$
Let A =250
$\begin{matrix} C l a s s i n t e r v a l & F r e q u e n c y (f_{i}) & C l a s s m a r k (x_{i}) & d_{i} = x_{i} - A & u_{i} = \frac{d_{i}}{h} & f_{i} u_{i} 0 - 100 & 40 & 50 & - 200 & - 2 & - 80 100 - 200 & 39 & 150 & - 100 & - 1 & - 39 200 - 300 & 34 & 250 & 0 & 0 & 0 300 - 400 & 30 & 350 & 100 & 1 & 30 400 - 500 & 45 & 450 & 200 & 2 & 90 T o t a l & \sum f_{i} = 188 & \sum f_{i} u_{i} = 1 \end{matrix}$
Mean $= A + \frac{\sum f_{i} u_{i}}{\sum f_{i}} \times h = 200 + (\frac{1}{188} \times 100)$
Mean = 250.53

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Similar questions

Q. Find the mean of the step deviations, if the assumed mean "a"=250:

\begin{matrix} C l a s s i n t e r v a l & F r e q u e n c y (f_{i}) & C l a s s m a r k (x_{i}) & d_{i} = x_{i} - a & u_{i} = \frac{d_{i}}{h} 0 - 100 & 40 & 50 & - 200 & - 2 100 - 200 & 39 & 150 & - 100 & - 1 200 - 300 & 34 & 250 & 0 & 0 300 - 400 & 30 & 350 & 100 & 1 400 - 500 & 45 & 450 & 200 & 2 \end{matrix}

Q. Find the mean of the step deviations:

\begin{matrix} C l a s s & F r e q u e n c y & C l a s s m a r k & d_{i} & u_{i} = \frac{d_{i}}{h} i n t e r v a l & (f_{i}) & (x_{i}) & (x_{i} - a) 0 - 100 & 40 & 50 & - 200 & - 2 100 - 200 & 39 & 150 & - 100 & - 1 200 - 300 & 34 & 250 & 0 & 0 300 - 400 & 30 & 350 & 100 & 1 400 - 500 & 45 & 450 & 200 & 2 \end{matrix}

Find the mean of the step deviations:
$\begin{matrix} Class interval & Frequency (f_{i}) & Class mark (x_{i}) & d_{i} = x_{i} - a & u_{i} = \frac{d_{i}}{h} 0 - 100 & 40 & 50 & - 200 & - 2 100 - 200 & 39 & 150 & - 100 & - 1 200 - 300 & 34 & 250 & 0 & 0 300 - 400 & 30 & 350 & 100 & 1 400 - 500 & 45 & 450 & 200 & 2 \end{matrix}$

Find the mean of the step deviations:
$\begin{matrix} C l a s s i n t e r v a l & F r e q u e n c y (f_{i}) & C l a s s m a r k (x_{i}) & d_{i} = x_{i} - a & u_{i} = \frac{d_{i}}{h} 0 - 100 & 40 & 50 & - 200 & - 2 100 - 200 & 39 & 150 & - 100 & - 1 200 - 300 & 34 & 250 & 0 & 0 300 - 400 & 30 & 350 & 100 & 1 400 - 500 & 45 & 450 & 200 & 2 \end{matrix}$

Q. Calculate

\frac{\sum f_{i} u_{i}}{\sum f_{i}}

for the given data using step-deviation method.

\begin{matrix} Class interval & Frequency (f_{i}) & Class mark (x_{i}) & d_{i} = x_{i} - a & u_{i} = \frac{d_{i}}{h} 0 - 100 & 40 & 50 & - 200 & - 2 100 - 200 & 39 & 150 & - 100 & - 1 200 - 300 & 34 & 250 & 0 & 0 300 - 400 & 30 & 350 & 100 & 1 400 - 500 & 45 & 450 & 200 & 2 \end{matrix}

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Find the mean of the given data: Class intervalFrequency(fi)Class mark(xi)di=xi−aui=dih0−1004050−200−2100−20039150−100−1200−3003425000300−400303501001400−500454502002

The correct option is B 1188Let A =250 Class intervalFrequency(fi)Class mark(xi)di=xi−Aui=dihfiui0−1004050−200−2−80100−20039150−100−1−39200−30034250000300−40030350100130400−50045450200290Total∑fi=188∑fiui=1 Mean =A+∑fiui∑fi×h=200+(1188 ×100) Mean = 250.53