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X is a normal random variable with mean μ and standard deviation σ. Then P( μ−...

X is a normal random variable with mean μ and standard deviation σ. Then P( μ− 1.6 σ ≤ X ≤ μ+ 2.6 σ) =? Answer to 4 decimal places.

math   Statistics-And-Probability   
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Answer #1

We are given that:
\\ X\sim N(\mu, \sigma^2) \\ \Rightarrow Z = \frac{X-\mu}{\sigma} \sim N(0,1) \text{, the standard normal distribution.}

Now, the required probability is given by:
\begin{align*} \boldsymbol{P(\mu - 1.6\sigma \le X \le \mu + 2.6\sigma)} &=P\left(\mu - 1.6\sigma -\mu \le X -\mu \le \mu + 2.6\sigma -\mu \right) \\ &=P\left(- 1.6\sigma \le X -\mu \le 2.6\sigma \right) \\ &=P\left(- \frac{1.6\sigma}{\sigma} \le \frac{X -\mu}{\sigma} \le \frac{2.6\sigma}{\sigma} \right) \\ &=P\left(- 1.6 \le Z \le 2.6 \right) \\ &= P(Z\le 2.6) - P(Z<-1.6) \\ &\text{Using the table for standard normal distribution, we get:} \\ &= 0.995339 - 0.054799 \\ &= 0.940540 \\ &= \bf 0.9405 \ \ \ \ \ \ \ [ANSWER \text{, Rounded to 4 decimal places}] \end{align*}

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