Question

The random variable xhas a normal distribution with$\mathit{\mu }\mathbf{=}\mathbf{75}$and $\mathit{\sigma }\mathbf{=}\mathbf{10}$. Find the following probabilities:

a.$\mathit{P}\left(x⩽80\right)$

b.$\mathit{P}\left(x⩾85\right)$

c.$\mathit{P}\left(70⩽x⩽75\right)$

d.$\mathit{P}\left(x>80\right)$

e.$\mathit{P}\left(x=78\right)$

f.$\mathit{P}\left(x⩽110\right)$

Short answer

a.$P\left(x⩽80\right)=$$0.6915$

b.$P\left(x⩾85\right)=$$0.1587$

c.$P\left(70⩽x⩽75\right)=$$0.1915$

d.$P\left(x>80\right)=$$0.3085$

e.$P\left(x=78\right)=0$

f.$P\left(x⩽110\right)=$$0.9998$

Step-by-step solution

Given information

X is anormalrandom variable.

$$\begin{gathered} \mu =75 \\ \sigma =10 \end{gathered}$$

Define the probability density function

The p.d.f of X follows a normal distribution which is given by is:

$f\left(x\right)=\frac{1}{\sigma \sqrt{2\pi }}{e}^{-\frac{1}{2}{\left[\left(x-\mu \right)/\sigma \right]}^2}$

(a) Calculate Px⩽80

$\begin{array}{rcl}P\left(x⩽80\right)& =& P\left(\frac{x-\mu }{\sigma }⩽\frac{80-75}{10}\right)\\ & =& P\left(Z⩽0.5\right)\\ & =& 0.6915\\ & & \end{array}$

Hence,$P\left(x⩽80\right)=$$0.6915$

(b) Calculate Px⩾85

$\begin{array}{rcl}P\left(x⩾85\right)& =& P\left(\frac{x-\mu }{\sigma }⩾\frac{85-75}{10}\right)\\ & =& P\left(Z⩾1\right)\\ & =& 1-P\left(Z⩽1\right)\\ & =& 1-0.8413\\ & =& 0.1587\\ & & \end{array}$

Hence,$P\left(x⩾85\right)=$$0.1587$

(c) Calculate P70⩽x⩽75

$\begin{array}{rcl}P\left(70⩽x⩽75\right)& =& P\left(\frac{70-75}{10}⩽\frac{x-\mu }{\sigma }⩽\frac{75-75}{10}\right)\\ & =& P\left(\text{- 0.5}⩽Z⩽0\right)\\ & =& P\left(Z⩽0\right)-P\left(Z⩽-0.5\right)\\ & =& P\left(Z⩽0\right)-1+P\left(Z⩽0.5\right)\\ & =& 0.5-1+0.6915\\ & =& 0.1915\\ & & \end{array}$

Hence,$P\left(70⩽x⩽75\right)=$$0.1915$

(d) Calculate Px>80

$\begin{array}{rcl}P\left(x>80\right)& =& P\left(\frac{x-\mu }{\sigma }>\frac{80-75}{10}\right)\\ & =& P\left(Z>0.5\right)\\ & =& 1-P\left(Z<0.5\right)\\ & =& 1-0.6915\\ & =& 0.3085\\ & & \end{array}$

Hence,$P\left(x>80\right)=0.3085$

(e) Calculate Px=78

$P\left(x=78\right)=0$[Since, Probability of any single point is 0]

Hence,$P\left(x=78\right)=0$

(f) Calculate Px⩽110

$\begin{array}{rcl}P\left(x⩽110\right)& =& P\left(\frac{x-\mu }{\sigma }⩽\frac{110-75}{10}\right)\\ & =& P\left(Z⩽3.5\right)\\ & =& 0.9998\\ & & \end{array}$

Hence,$P\left(x⩽110\right)=$$0.9998$