The random variable x has a normal distribution with $\mathbf{\mu }\mathbf{=}\mathbf{1000}$ and $\mathit{\sigma }\mathbf{=}\mathbf{10}$.

a. Find the probability that x assumes a value more than 2 standard deviations from its mean. More than 3 standard deviations from .$\mathit{\mu }$

b. Find the probability that x assumes a value within 1 standard deviation of its mean. Within 2 standard deviations of $\mathit{\mu }$.

c. Find the value of x that represents the 80th percentile of this distribution. The 10th percentile.

x is a random variable that is a normal distribution with $\mu =1000$and $\sigma =10$

a.

The probability of x assumes a value more than 2 standard deviations from its mean is,

$$\begin{gathered} p(x>\mu +2\sigma )=P\left(x>1000+\left(2\times 10\right)\right) \\ =P\left(x>1020\right) \\ =P\left(z>2\right)\left[z=\frac{1020-1000}{10}\right] \\ =1-P\left(z\le 2\right) \\ =1-P\left(z<2\right) \\ =1-0.9772 \\ =0.0228 \end{gathered}$$

Therefore, the value of x more than the 2 standard deviations from its mean is 0.0228.

Again,

The probability of x assumes a value more than 3 standard deviations from its mean is,

$$\begin{gathered} p(x>\mu +3\sigma )=P\left(x>1000+\left(3\times 10\right)\right) \\ =P\left(x>1030\right) \\ =P\left(z>3\right)\left[z=\frac{1030-1000}{10}\right] \\ =1-P\left(z\le 3\right) \\ =1-P\left(z<3\right) \\ =1-0.9987 \\ =0.0013 \end{gathered}$$

Therefore, the value of x more than the 3 standards deviation from its mean is 0.0013.

b.

The probability of x assumes a value within 1 standard deviation from its mean is,

$$\begin{gathered} p(\mu -\sigma <x<\mu +\sigma )=P\left(1000-10<x<1000+10\right) \\ =P\left(990<x<1010\right) \\ =P\left(-1<z<1\right)\left[z=\frac{x-\mu }{\sigma }\right] \\ =P\left(z<1\right)-\left(1-P\left(z<1\right)\right) \\ =0.8413-(1-0.8413) \\ =0.8413-0.1587 \\ =0.6826 \end{gathered}$$

Therefore, the probability of x within 1 standard deviation from its mean is 0.6826.

Again,

The probability of x assumes a value within 2 standard deviations from its mean is,

$$\begin{gathered} p(\mu -2\sigma <x<\mu +2\sigma )=P\left(1000-2<x<1000+20\right) \\ =P\left(980<x<1020\right) \\ =P\left(-2<z<2\right)\left[z=\frac{x-\mu }{\sigma }\right] \\ =P\left(z<2\right)-\left(1-P\left(z<2\right)\right) \\ =0.9772-(1-0.9772) \\ =0.9772-0.0228 \\ =0.9544 \end{gathered}$$

Therefore, the probability of x within 2 standard deviations from its mean is 0.9544.

c.

Assume that$x_0$ is the 80^th percentile of the distribution,

So, the value of the 80^th percentile is,

$$\begin{gathered} P(x<x_0)=0.80 \\ P\left(\frac{x-\mu }{\sigma }\le \frac{x_0-\mu }{\sigma }\right)=0.80 \\ P\left(z\le \frac{x_0-1000}{\sigma }\right)=0.80 \\ \Phi \left(\frac{x_0-1000}{\sigma }\right)=0.80 \\ \frac{x_0-1000}{\sigma }={\Phi }^{-1}\left(0.80\right) \\ {x}_0=1000+\left(10\times {\Phi }^{-1}\left(0.80\right)\right) \\ {x}_0=1000+\left(10\times 0.8416\right) \\ {x}_0=1008.416 \end{gathered}$$

Therefore, the value of$x_0$ for 80^th percentile is 1008.416.

Again, assume that$x_0$ is the 10^th percentile of the distribution.

So, the value of the 10^th percentile is,

$$\begin{gathered} P(x<x_0)=0.10 \\ P\left(\frac{x-\mu }{\sigma }\le \frac{x_0-\mu }{\sigma }\right)=0.0 \\ P\left(z\le \frac{x_0-1000}{\sigma }\right)=0.10 \\ \Phi \left(\frac{x_0-1000}{\sigma }\right)=0.10 \\ \frac{x_0-1000}{\sigma }={\Phi }^{-1}\left(0.10\right) \\ {x}_0=1000+\left(10\times {\Phi }^{-1}\left(0.10\right)\right) \\ {x}_0=1000+\left(10\times \left(-1.2816\right)\right) \\ {x}_0=987.184 \end{gathered}$$

Therefore, the value of$x_0$ for 10^th percentile is 987.184.