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  • When tested, the lives (in hours) of 5 bulbs were noted as follows: 1357, 1090, 1666, 1494, 1623 The mean deviations (in hours) from their mean is A. 178 B. 179 C. 220 D. 356

When tested, the lives (in hours) of 5 bulbs were noted as follows:

1357, 1090, 1666, 1494, 1623
The mean deviations (in hours) from their mean is
A. 178
B. 179
C. 220
D. 356

Answers (1)

\\ \text{Given~that the lives of 5 bulbs in hours is 1357, 1090, 1666, 1494 , 1623 } \\ \\ \text{ Here mean }\overline{X}=\frac{1357+1090+1666+1494+1623}{5}=\frac{7230}{7}=1446~ \\ \\ \text{ This can be written in table form as, } \\

\begin{array}{|l|l|} \hline \text { Lives (in hours }) x_{i} & \multicolumn{1}{|c|} {d_{i}=x_{i}-\bar{x}} \\ \hline 1357 & 89 \\ \hline 1090 & 356 \\ \hline 1666 & 220 \\ \hline 1494 & 48 \\ \hline 1623 & 177 \\ \hline \text { Total } & =890 \\ \hline \end{array}

Hence, mean deviation becomes  \\\\ M.D=\frac{ \Sigma d_{i}}{n}=\frac{890}{5}=178 \\

\text{M.D}=\frac{ \Sigma d_{i}}{n}=\frac{890}{5}=178 \\

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