Statistics Test...

$$5$$ students of a class have an average height $$150\ cm$$ and variance $$18\ cm^{2}$$. A new student, whose height is $$156\ cm$$, joined them. The variance (in $$cm^{2})$$ of the height of these six students is

If the mean of the data : $$7, 8, 9, 7, 8, 7, \lambda, 8$$ is $$8$$, then the variance of this data is

Let the observations $$x_i(1\leq i \leq 10)$$ satisfy the equations, $$\displaystyle\sum^{10}_{i=1}(x_i-5)=10$$ and $$\displaystyle\sum^{10}_{i=1}(x_i-5)^2=40$$. If $$\mu$$ and $$\lambda$$ are the mean and the variance of the observations, $$x_1-3, x_2-3, ....., x_{10}-3$$, then the ordered pair $$(\mu, \lambda)$$ is equal to?

A student scores the following marks in five test: $$45, 54, 41, 57, 43$$. His score is not known for the sixth test. If the mean score is 48 in the six tests, then the standard deviation of the marks in six test is 

The mean and variance of $$20$$ observations are found to be $$10$$ and $$4$$, respectively. On rechecking, it was found that an observation $$9$$ was incorrect and the correct observation was $$11$$. Then the correct variabce is:

If the data $$x_1, x_2, .., x_{10}$$ is such that the mean of first four of these is $$11$$, the mean of the remaining six is $$16$$ and the sum of square of all of these is $$2,000$$; then the standard deviation of this data is?

For two data sets, each of size $$ 5$$, the variances are given to be $$4$$ and $$5$$  and the corresponding means are given to be $$2$$ and $$4,$$ respectively. The variance of the combined data set is 

The mean and the standard deviation (s.d) of 10 observations are 20 and 2 respectively. Each of these 10 observations is multiplied by $$p$$ and then reduced by $$q$$, where $$p \neq 0$$ and $$q  \neq 0$$. If the new mean and new s.d. become half of their original values, then $$q$$ is equal to:

If both the mean and the standard deviation of $$50$$ observations $${ x }_{ 1 },{ x }_{ 2 },......,{ x }_{ 50 }$$ are equal to $$16$$, then the mean of $${ \left( { x }_{ 1 }-4 \right)  }^{ 2 },{ \left( { x }_{ 2 }-4 \right)  }^{ 2 },.....{ \left( { x }_{ 50 }-4 \right)  }^{ 2 }$$ is:

If $$\displaystyle \sum_{i = 1}^{9}(x_{i} - 5) = 9$$ and $$\displaystyle \sum_{i = 1}^{9}(x_{i} - 5)^{2} = 45$$, then the standard deviation of the $$9$$ items $$x_{1}, x_{2}, ...., x_{9}$$ is