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Question: If variable takes values \(0,1,2,...,n\)with frequencies \({{q}^{n}},\dfrac{n}{1}{{q}^{n-1}}p,\dfrac...

If variable takes values 0,1,2,...,n0,1,2,...,nwith frequencies qn,n1qn−1p,n(n−1)1⋅2qn−12p2,...,pn{{q}^{n}},\dfrac{n}{1}{{q}^{n-1}}p,\dfrac{n\left( n-1 \right)}{1\cdot 2}{{q}^{n-1}}2{{p}^{2}},...,{{p}^{n}} wherep+q=1p+q=1, then the mean is $$$$
A.pq$$$$$ B.np C.$nq
D.np2n{{p}^{2}}$$$$

Explanation

Solution

We recall binomial theorem and mean of the grouped data. We use the binomial expansion (x+y)n=nC0x0yn+nC1x1yn−1+...+nCnxny0{{\left( x+y \right)}^{n}}={}^{n}{{C}_{0}}{{x}^{0}}{{y}^{n}}+{}^{n}{{C}_{1}}{{x}^{1}}{{y}^{n-1}}+...+{}^{n}{{C}_{n}}{{x}^{n}}{{y}^{0}} and the mean of the sample x1,x2,x3,....,xn{{x}_{1}},{{x}_{2}},{{x}_{3}},....,{{x}_{n}} with corresponding frequencies f1,f2,f3,....fn{{f}_{1}},{{f}_{2}},{{f}_{3}},....{{f}_{n}} then the mean is given by x‾=∑i=1nfixi∑i=1nfi\overline{x}=\dfrac{\sum\limits_{i=1}^{n}{{{f}_{i}}{{x}_{i}}}}{\sum\limits_{i=1}^{n}{{{f}_{i}}}}.$$$$

Complete step-by-step answer:
We know that binomial is the algebraic expression involving two terms and each term with distinct variable. If x,yx,y are the two terms of binomial with some positive integral power nn then the binomial expansion is given by;
(x+y)n=nC0x0yn+nC1x0yn−1+nC2x0yn−2+...+nCnxny0{{\left( x+y \right)}^{n}}={}^{n}{{C}_{0}}{{x}^{0}}{{y}^{n}}+{}^{n}{{C}_{1}}{{x}^{0}}{{y}^{n-1}}+{}^{n}{{C}_{2}}{{x}^{0}}{{y}^{n-2}}+...+{}^{n}{{C}_{n}}{{x}^{n}}{{y}^{0}}
The above expression is called binomial formula or binomial identity . We use the combinatorial formula nCr=n!r!(n−r)!=n(n−1)...(n−r+1)r(r−1)...1{}^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}=\dfrac{n\left( n-1 \right)...\left( n-r+1 \right)}{r\left( r-1 \right)...1} in the binomial formula to have;
(x+y)n=yn+nnx0yn−1+n(n−1)1.2x0yn−2+...+xn.......(1){{\left( x+y \right)}^{n}}={{y}^{n}}+{}^{n}n{{x}^{0}}{{y}^{n-1}}+\dfrac{n\left( n-1 \right)}{1.2}{{x}^{0}}{{y}^{n-2}}+...+{{x}^{n}}.......\left( 1 \right)
We can express the binomial formula in summation form as
(x+y)n=∑r=0nxryn−r.....(2){{\left( x+y \right)}^{n}}=\sum\limits_{r=0}^{n}{{{x}^{r}}}{{y}^{n-r}}.....\left( 2 \right)
We know that the mean with frequenting data is sum of product of data values and frequencies divided by sum of frequencies. If there are nn data values say x1,x2,x3...,xn{{x}_{1}},{{x}_{2}},{{x}_{3}}...,{{x}_{n}} and with corresponding frequencies f1,f2,f3...,fn{{f}_{1}},{{f}_{2}},{{f}_{3}}...,{{f}_{n}} then the mean is given by

x‾=∑i=1nxifi∑i=1nfi=∑xf∑f\overline{x}=\dfrac{\sum\limits_{i=1}^{n}{{{x}_{i}}{{f}_{i}}}}{\sum\limits_{i=1}^{n}{{{f}_{i}}}}=\dfrac{\sum{xf}}{\sum{f}}
We are give in the question that the variable takes values 0,1,2,...,n0,1,2,...,nwith frequencies qn,n1qn−1p,n(n−1)1⋅2qn−12p2,...,pn{{q}^{n}},\dfrac{n}{1}{{q}^{n-1}}p,\dfrac{n\left( n-1 \right)}{1\cdot 2}{{q}^{n-1}}2{{p}^{2}},...,{{p}^{n}} wherep+q=1p+q=1, So let us find the sum of frequencies
∑f=qn+n1qn−1p+n(n−1)1⋅2qn−12p2+...+pn\sum{f}={{q}^{n}}+\dfrac{n}{1}{{q}^{n-1}}p+\dfrac{n\left( n-1 \right)}{1\cdot 2}{{q}^{n-1}}2{{p}^{2}}+...+{{p}^{n}}
We put the binomial formula (1) of (x+y)n{{\left( x+y \right)}^{n}} with x=p,y=qx=p,y=q to have
∑f=qn+n1qn−1p+n(n−1)1⋅2qn−12p2+...+pn=(p+q)n\sum{f}={{q}^{n}}+\dfrac{n}{1}{{q}^{n-1}}p+\dfrac{n\left( n-1 \right)}{1\cdot 2}{{q}^{n-1}}2{{p}^{2}}+...+{{p}^{n}}={{\left( p+q \right)}^{n}}
We put the given p+q=1p+q=1 in the above step to have
∑f=qn+n1qn−1p+n(n−1)1⋅2qn−12p2+...+pn=(p+q)n=1\sum{f}={{q}^{n}}+\dfrac{n}{1}{{q}^{n-1}}p+\dfrac{n\left( n-1 \right)}{1\cdot 2}{{q}^{n-1}}2{{p}^{2}}+...+{{p}^{n}}={{\left( p+q \right)}^{n}}=1
Let us find the sum of product of data values and frequencies.

& \sum{xf}=1\cdot {{q}^{n}}+2\cdot \dfrac{n}{1}{{q}^{n-1}}p+3\cdot \dfrac{n\left( n-1 \right)}{1\cdot 2}{{q}^{n-1}}2{{p}^{2}}+...+n\cdot {{p}^{n}} \\\ & \Rightarrow \sum{xf}={}^{n}{{C}_{0}}{{p}^{0}}{{q}^{n}}+2\cdot {}^{n}{{C}_{1}}{{p}^{1}}{{q}^{n-1}}p+3\cdot {}^{n}{{C}_{2}}{{p}^{2}}{{q}^{n-1}}+...+n\cdot {{p}^{n}} \\\ \end{aligned}$$ We can write the summation in the left hand side as; $$\Rightarrow \sum{xf}=\sum\limits_{r=1}^{n}{r\cdot {}^{n}{{C}_{r}}{{p}^{r}}{{q}^{n-r}}}$$ We use the combinatorial identity ${}^{n}{{C}_{r}}=\dfrac{n}{r}{}^{n-1}{{C}_{r-1}}$ in the above step to have; $$\begin{aligned} & \Rightarrow \sum{xf}=\sum\limits_{r=1}^{n}{r\cdot \dfrac{n}{r}\cdot {}^{n-1}{{C}_{r-1}}{{p}^{r}}{{q}^{n-r}}} \\\ & \Rightarrow \sum{xf}=\sum\limits_{r=1}^{n}{n\cdot {}^{n-1}{{C}_{r}}p\cdot {{p}^{r-1}}{{q}^{n-1+1-r}}} \\\ & \Rightarrow \sum{xf}=np\sum\limits_{r=1}^{n}{{}^{n-1}{{C}_{r-1}}{{p}^{r-1}}{{q}^{\left( n-1 \right)-\left( r-1 \right)}}} \\\ \end{aligned}$$ We use the binomial formula ${{\left( x+y \right)}^{n}}=\sum\limits_{r=0}^{n}{{{x}^{r}}}{{y}^{n-r}}$ for $x=p,y=q,n=n-1,r=r-1$ in tha above step to have $$\begin{aligned} & \Rightarrow \sum{xf}=np{{\left( p+q \right)}^{n-1}} \\\ & \Rightarrow \sum{xf}=np{{\left( 1 \right)}^{n-1}}\left( \because p+q=1 \right) \\\ & \Rightarrow \sum{xf}=np \\\ \end{aligned}$$ So the required mean is $$\overline{x}=\dfrac{\sum{xf}}{\sum{f}}=\dfrac{np}{1}=np$$ **So, the correct answer is “Option B”.** **Note:** We note that the mean is one of the measures of central tendency along with mode and median. We also note that the variable that takes values is also called random variable and the values taken are called data values. We can also show $nx{{\left( x+y \right)}^{n-1}}=\sum\limits_{r=0}^{n}{r\cdot {}^{n}{{C}_{r}}{{x}^{r}}{{y}^{n-r}}}$by differentiating the binomial formula with respect to $x$ and then multiplying$x$. The terms ${}^{n}{{C}_{o}},{}^{n}{{C}_{1}},...,{}^{n}{{C}_{n}}$ are called binomial coefficients whose sum is ${{2}^{n}}$.