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Let $X$ denote the number of scores which exceed 4 in 18 , tosses of a symmetrical die. Consider the following statements : $\quad[2014-I]$
1 . The arithmetic mean of $X$ is 6
2 . The standard deviation of $X$ is 2 . Which of the above statements is/are correct?
Options:
1 . The arithmetic mean of $X$ is 6
2 . The standard deviation of $X$ is 2 . Which of the above statements is/are correct?
Solution:
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Verified Answer
The correct answer is:
Both 1 and 2
Statement 1 $n(X)=2$
$\mathrm{p}=\frac{\mathrm{n}(\mathrm{X})}{\mathrm{n}(\mathrm{S})}=\frac{2}{6}=\frac{1}{3}$
$q=1-p=1-\frac{1}{3}=\frac{2}{3}$
arithmetric mean of $\mathrm{X}=\mathrm{n} \mathrm{p}=18 \times \frac{1}{3}=6$
Statement $2:$ Standard deviation of
$\mathrm{x}-\sqrt{\text { variance of } \mathrm{X}}-\sqrt{18 \times \frac{1}{3} \times \frac{2}{3}}-\sqrt{4}-2$
Hence, statements 1 and 2 both are correct.
Sol. $(128-130):$ \begin{array}{|c|cc|c|}
\hline Numbers (\mathrm{x}) & Frequency (\mathrm{f}) & cf. & \Sigma \mathrm{fx} \\
\hline 0 & 14 & 14 & 0 \\
1 & 21 & 35 & 21 \\
2 & 25 & 60 & 50 \\
3 & 43 & 103 & 129 \\
4 & 51 & 154 & 204 \\
5 & 40 & 194 & 200 \\
6 & 39 & 233 & 234 \\
7 & 12 & 245 & 84 \\
\hline & \mathrm{N}=245 & & \Sigma \mathrm{fx} =922 \\
\hline
\end{array}
$\mathrm{p}=\frac{\mathrm{n}(\mathrm{X})}{\mathrm{n}(\mathrm{S})}=\frac{2}{6}=\frac{1}{3}$
$q=1-p=1-\frac{1}{3}=\frac{2}{3}$
arithmetric mean of $\mathrm{X}=\mathrm{n} \mathrm{p}=18 \times \frac{1}{3}=6$
Statement $2:$ Standard deviation of
$\mathrm{x}-\sqrt{\text { variance of } \mathrm{X}}-\sqrt{18 \times \frac{1}{3} \times \frac{2}{3}}-\sqrt{4}-2$
Hence, statements 1 and 2 both are correct.
Sol. $(128-130):$ \begin{array}{|c|cc|c|}
\hline Numbers (\mathrm{x}) & Frequency (\mathrm{f}) & cf. & \Sigma \mathrm{fx} \\
\hline 0 & 14 & 14 & 0 \\
1 & 21 & 35 & 21 \\
2 & 25 & 60 & 50 \\
3 & 43 & 103 & 129 \\
4 & 51 & 154 & 204 \\
5 & 40 & 194 & 200 \\
6 & 39 & 233 & 234 \\
7 & 12 & 245 & 84 \\
\hline & \mathrm{N}=245 & & \Sigma \mathrm{fx} =922 \\
\hline
\end{array}
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