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November 13th, 2017, 10:35 AM  #1 
Senior Member Joined: Jan 2013 From: Italy Posts: 154 Thanks: 7  Find standard deviation given normal probability
I have $X \sim N(0,\sigma^2), Y \sim N(1,4) \mbox{ independent}, \\ Z = XY$ find $\sigma$ such that $P(Z \le 7) = 0.84134$ what I have done is : $XY \sim N(0(1), \sigma^2+4) = (1, \sigma^2+4)$ $\begin{array}{lcl}P(Z \le 7) & = & P( \frac{Z 1}{\sqrt{\sigma^2+4}} \le \frac{71}{\sqrt{\sigma^2+4}}) \\ & = & P(W \le \frac{6}{\sqrt{\sigma^2+4}}) &= & 0.84137 \end{array}$ $W \sim N(0,1) \\ 0.84137 = \phi(1) \mbox{ therefore }$ $\begin{array}{lcl}\frac{6}{\sqrt{\sigma^2+4}} & = & 1 \\ \frac{36}{\sigma^2+4} & = & 1 \\ \frac{36(\sigma^2+4)}{\sigma^2+4} & = & \sigma^2+4 \\ 36 & = & \sigma^2 + 4 \\ \sigma^2 &= & 32 \\ \sigma & = & \sqrt{32} \end{array}$ please, can you tell me if it's correct? Thanks! 
November 14th, 2017, 12:51 AM  #2 
Math Team Joined: Jul 2013 From: काठमाडौं, नेपाल Posts: 884 Thanks: 61 Math Focus: सामान्य गणित 
seems correct


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deviation, find, normal, normal distribution, probability, standard, standard deviation 
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