Find the mean of the step deviations-Whine Class - Frequency - Class mark - d - i - u - i -frac d - i h 1interval - f - i - x - i - x - i - a -Whine0-100 - 40 - 50 -200 -211Whe100-200 - 39 - 150 -100 -1 1Whine200-300 - 34 - 250 - 0 - 0 1Whine 300-400 - 30 - 350 - 100 - 111Whine400-500 - 45 - 450 - 200 - 2 1Whelend array A- 0B- 1-188C- 250D- 250-52

The correct option is B 1/188[ Class amp; Frequency amp; Class mark amp; d_i amp; u_i = d_i/h amp; f_i u_i; interval am ...

Byju's Answer

Standard XI

Statistics

Mean of Continuous Series

Find the mean...

Question

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

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

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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 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}

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}

Find the mean of the step deviations: ClassFrequencyClass markdiui=dihinterval(fi)(xi)(xi−a)0−1004050−200−2100−20039150−100−1200−3003425000300−400303501001400−500454502002

The correct option is B 1188ClassFrequencyClass markdiui=dihfiuiinterval(fi)(xi)(xi−a)0−1004050−200−2−80100−20039150−100−1−39200−30034250000300−40030350100130400−50045450200290Total1881i Mean of step deviations =∑fiui∑fi=1188