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Hint: Find the probability distribution for X and then find the mean of the distribution. Using the mean, you can calculate the variance and standard deviation of X.

Complete step-by-step answer:

It is given that X denotes the sum of the numbers obtained when two fair dice are rolled.

X can take values from 2(1 + 1) through 12 (6 + 6). Hence, X can take values 2, 3, 4, 5, 6, 7, 8, 9, 10, 11 and 12.

We know that the total outcome when two dice are rolled is 6 \[ \times \] 6, that is 36.

Let us find the probability distribution of X.

Now, we find the mean of the distribution.

Mean, \[\bar X = \sum {XP(X)} \]

\[\bar X = \dfrac{1}{{36}}\left[ {2 \times 1 + 3 \times 2 + 4 \times 3 + 5 \times 4 + 6 \times 5 + 7 \times 6 + 8 \times 5 + 9 \times 4 + 10 \times 3 + 11 \times 2 + 12 \times 1} \right]\]

\[\bar X = \dfrac{{252}}{{36}}\]

\[\bar X = 7\]

We now find the variance of the distribution as follows:

Variance, \[{\sigma ^2} = \sum {{X^2}P(X) - {{\bar X}^2}} \]

\[{\sigma ^2} = \dfrac{1}{{36}}\left[ {{2^2} \times 1 + {3^2} \times 2 + {4^2} \times 3 + {5^2} \times 4 + {6^2} \times 5 + {7^2} \times 6 + {8^2} \times 5 + {9^2} \times 4 + {{10}^2} \times 3 + {{11}^2} \times 2 + {{12}^2} \times 1} \right] - {7^2}\]

\[{\sigma ^2} = \dfrac{{1974}}{{36}} - 49\]

\[{\sigma ^2} = \dfrac{{1974 - 1764}}{{36}}\]

\[{\sigma ^2} = \dfrac{{210}}{{36}}\]

\[{\sigma ^2} = \dfrac{{35}}{6}\]

We can calculate the standard deviation by taking the square root of the variance.

\[\sigma = \sqrt {\dfrac{{35}}{6}} \]

Hence, the value of variation is \[\dfrac{{35}}{6}\] and the value of standard deviation is \[\sqrt {\dfrac{{35}}{6}} \].

Note: Mean can also be found by multiplying X with the number of possible outcomes and adding them and dividing by the total number of outcomes in that case, it is not necessary to find the probability of each outcome.

Complete step-by-step answer:

It is given that X denotes the sum of the numbers obtained when two fair dice are rolled.

X can take values from 2(1 + 1) through 12 (6 + 6). Hence, X can take values 2, 3, 4, 5, 6, 7, 8, 9, 10, 11 and 12.

We know that the total outcome when two dice are rolled is 6 \[ \times \] 6, that is 36.

Let us find the probability distribution of X.

X | Possible Outcomes | P(X) |

2 | (1, 1) | \[\dfrac{1}{{36}}\] |

3 | (2, 1) (1, 2) | \[\dfrac{2}{{36}}\] |

4 | (3, 1) (2, 2) (1, 3) | \[\dfrac{3}{{36}}\] |

5 | (4, 1) (3, 2) (2, 3) (1, 4) | \[\dfrac{4}{{36}}\] |

6 | (5, 1) (4, 2) (3, 3) (2, 4) (1, 5) | \[\dfrac{5}{{36}}\] |

7 | (6, 1) (5, 2) (4, 3) (3, 4) (2, 5) (1, 6) | \[\dfrac{6}{{36}}\] |

8 | (6, 2) (5, 3) (4, 4) (3, 5) (2, 6) | \[\dfrac{5}{{36}}\] |

9 | (6, 3) (5, 4) (4, 5) (3, 6) | \[\dfrac{4}{{36}}\] |

10 | (6, 4) (5, 5) (4, 6) | \[\dfrac{3}{{36}}\] |

11 | (6, 5) (5, 4) | \[\dfrac{2}{{36}}\] |

12 | (6, 6) | \[\dfrac{1}{{36}}\] |

Now, we find the mean of the distribution.

Mean, \[\bar X = \sum {XP(X)} \]

\[\bar X = \dfrac{1}{{36}}\left[ {2 \times 1 + 3 \times 2 + 4 \times 3 + 5 \times 4 + 6 \times 5 + 7 \times 6 + 8 \times 5 + 9 \times 4 + 10 \times 3 + 11 \times 2 + 12 \times 1} \right]\]

\[\bar X = \dfrac{{252}}{{36}}\]

\[\bar X = 7\]

We now find the variance of the distribution as follows:

Variance, \[{\sigma ^2} = \sum {{X^2}P(X) - {{\bar X}^2}} \]

\[{\sigma ^2} = \dfrac{1}{{36}}\left[ {{2^2} \times 1 + {3^2} \times 2 + {4^2} \times 3 + {5^2} \times 4 + {6^2} \times 5 + {7^2} \times 6 + {8^2} \times 5 + {9^2} \times 4 + {{10}^2} \times 3 + {{11}^2} \times 2 + {{12}^2} \times 1} \right] - {7^2}\]

\[{\sigma ^2} = \dfrac{{1974}}{{36}} - 49\]

\[{\sigma ^2} = \dfrac{{1974 - 1764}}{{36}}\]

\[{\sigma ^2} = \dfrac{{210}}{{36}}\]

\[{\sigma ^2} = \dfrac{{35}}{6}\]

We can calculate the standard deviation by taking the square root of the variance.

\[\sigma = \sqrt {\dfrac{{35}}{6}} \]

Hence, the value of variation is \[\dfrac{{35}}{6}\] and the value of standard deviation is \[\sqrt {\dfrac{{35}}{6}} \].

Note: Mean can also be found by multiplying X with the number of possible outcomes and adding them and dividing by the total number of outcomes in that case, it is not necessary to find the probability of each outcome.

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