Question

Answer the question for a normal random variable x with mean u and standard deviation o specified below. (Round your answer to four decimal places.) 

μ = 1.3 and  σ = 0.16.

 Find P(1.00<x< 1.10). 

P(1.00<x< 1.10) = 

Answer the question for a normal random variable x with mean u and standard deviation o specified below. (Round your answer to four decimal places.) 

μ = 1.3 and σ = 0.16. 

Find P(x >1.35). 

P(x > 1.35) = 

Answer 1

Given that, \(\mathrm{X} \sim \mathrm{N}\left(1.3,0.16^{2}\right)\)

\(\mu=1.3\) and \(\sigma=0.16\)

6) We have to find \(P(1.00< X< 1.10)\).

$$ P(1.00< X< 1.10)=P(X< 1.10)-P(X \leq 1.00) $$

We know that if \(X \sim N\left(\mu, \sigma^{2}\right)\) then, \(Z=\frac{X-\mu}{\sigma} \sim N(0,1)\)

$$ \begin{aligned} &\therefore P(1.00< X< 1.10)=P\left(\frac{X-\mu}{\sigma}<\frac{1.10-\mu}{\sigma}\right)-P\left(\frac{X-\mu}{\sigma} \leq \frac{1.00-\mu}{\sigma}\right) \\ &\therefore P(1.00< X< 1.10)=P\left(Z<\frac{1.10-1.3}{0.16}\right)-P\left(Z \leq \frac{1.00-1.3}{0.16}\right) \\ &\therefore P(1.00< X< 1.10)=P(Z<-1.25)-P(Z \leq-1.875) \end{aligned} $$

Using "pnorm" function of \(\mathrm{R}\) we get,

\(P(Z< -1.25)=0.1056\) and \(P(Z \leq-1.875)=0.0303\)

$$ \begin{aligned} &\therefore P(1.00< X< 1.10)=0.1056-0.0303 \\ &\therefore \mathbf{P}(\mathbf{1 . 0 0}<\mathbf{X}<\mathbf{1 . 1 0})=\mathbf{0 . 0 7 5 3} \end{aligned} $$

7) We have to find \(\mathrm{P}(\mathrm{X}>1.35)\).

We know that if \(X \sim N\left(\mu, \sigma^{2}\right)\) then, \(Z=\frac{X-\mu}{\sigma} \sim N(0,1)\)

$$ \begin{aligned} &\therefore P(X>1.35)=P\left(\frac{X-\mu}{\sigma}>\frac{1.35-\mu}{\sigma}\right) \\ &\therefore P(X>1.35)=P\left(Z>\frac{1.35-1.3}{0.16}\right) \\ &\therefore P(X>1.35)=P(Z>0.3125) \end{aligned} $$

Using "pnorm" function of \(R\) we get, \(P(Z>0.3125)=0.3773\)

$$ \therefore \mathbf{P}(\mathbf{X}>\mathbf{1 . 3 5})=\mathbf{0 . 3 7 7 3} $$