Question
The mean and variance of $n$ observations are 8 and 16 , respectively. If the sum of the first ( $n - 1$ ) observations is 48 and the sum of squares of the first ( $n - 1$ ) observations is 496 , then the value of n is :
Options
Solution
Sol. [Key Concept: Use $\Sigma\ x$ and $\Sigma\ x^{2}$ with mean $8$, variance $16$ to form a quadratic in $n$.]
$$x_{1} + x_{2} + \ldots + x_{n - 1} + x_{n} = 8n48 + x_{n} = 8n$$
$$\begin{array}{r} \Rightarrow x_{n} = 8n - 48\#(1) \end{array}$$
$${\Rightarrow 16 = \frac{496 + x_{n}^{2}}{n} - (8)^{2} }{\Rightarrow 80n = 496 + x_{n}^{2}}$$
$$\begin{array}{r} \Rightarrow x_{n}^{2} = 80n - 496\#(2) \end{array}$$
$${\Rightarrow (8n - 48)^{2} = 80n - 496 }{\Rightarrow 64(n - 6)^{2} = 8(10n - 62) }{\Rightarrow 8(n - 6)^{2} = 10n - 62 }{\Rightarrow 4(n - 6)^{2} = 5n - 31 }{\Rightarrow 4\left( n^{2} - 36 - 12n \right) = 5n - 31 }{\Rightarrow 4n^{2} + 144 - 48n = 5n - 31 }{\Rightarrow 4n^{2} - 53n + 175 = 0 }{\Rightarrow 4n^{2} - 28n - 25n + 175 = 0 }{\Rightarrow 4n(n - 7) - 25(n - 7) = 0 }{n = 7 }$$
Create a free account to view solution
View Solution Free