The outcome of each of $30$ items was observed; $10$ items gave an outcome $\frac{1}{2} - d$ each, $10$ items gave outcome $\frac {1}{2}$ each and the remaining $10$ items gave outcome $\frac{1}{2} + d$ each. If the variance of this outcome data is $\frac {4}{3}$ then $\left| d \right|$ equals

Two sets each of 20 observations, have the same standard derivation 5. The first set has a mean 17 and the second a mean 22. Determine the standard deviation of the set obtained by combining the given two sets.

There are 60 students in a class. The following is the frequency distribution of the marks obtained by the students in a test:

$\begin{array}{|l|l|l|l|l|l|l|} \hline \text { Marks } & 0 & 1 & 2 & 3 & 4 & 5 \\ \hline \text { Frequency } & x-2 & x & x^{2} & (x+1)^{2} & 2 x & x+1 \\ \hline \end{array}$

where $x$ is a positive integer. Determine the mean and standard deviation of the marks.

Let the mean and the variance of $5$ observations $x_{1}, x_{2}, x_{3}, x_{4}, x_{5}$ be $\frac{24}{5}$ and $\frac{194}{25}$ respectively. If the mean and variance of the first $4$ observation are $\frac{7}{2}$ and $a$ respectively, then $\left(4 a+x_{5}\right)$ is equal to

Statement $1$ : The variance of first $n$ odd natural numbers is $\frac{{{n^2} - 1}}{3}$
Statement $2$ : The sum of first $n$ odd natural number is $n^2$ and the sum of square of first $n$ odd natural numbers is $\frac{{n\left( {4{n^2} + 1} \right)}}{3}$

From a lot of $12$ items containing $3$ defectives, a sample of $5$ items is drawn at random. Let the random variable $\mathrm{X}$ denote the number of defective items in the sample. Let items in the sample be drawn one by one without replacement. If variance of $X$ is $\frac{m}{n}$, where $\operatorname{gcd}(m, n)=1$, then $n-m$ is equal to..........