Question 33

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Jambmaths question: 

Find the value of t if the standard deviation of 2t, 3t, 4t, 5t, and 6t is $\sqrt{2}$

Option A: 

3

Option B: 

1

Option C: 

4

Option D: 

2

Jamb Maths Solution: 

$\begin{align}  & mean=\overline{x}=\frac{2t+3t+4t+5t+6t}{5}=\frac{20t}{5}=4t \\ & variance={{\sigma }^{2}} \\ & {{\sigma }^{2}}=\frac{{{(2t-4t)}^{2}}+{{(3t-4t)}^{2}}+{{(4t-4t)}^{2}}+{{(5t-4t)}^{2}}+{{(6t-4t)}^{2}}}{5} \\ & {{\sigma }^{2}}=\frac{4{{t}^{2}}+{{t}^{2}}+{{t}^{2}}+4{{t}^{2}}}{5} \\ & {{\sigma }^{2}}=\frac{10{{t}^{2}}}{5}=2{{t}^{2}} \\ & standard deviation=\sqrt{{{\sigma }^{2}}} \\ & \sigma =\sqrt{2{{t}^{2}}},\text{ }\sqrt{t}=t\sqrt{2},\text{  }t=1 \\\end{align}$

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