Observe the value of given mean, standard deviation and limits.
Find the z-value for the respective random variable by using the formula
z = $\frac{x - \mu}{\sigma}$
Where,
x = Normal random variable
$\mu$ = mean
$\sigma$ = standard deviation.
Now, find the area under the standard normal curve for the z-value obtained from step2 by using Z-table.
Subtract area of upper limit from area of lower limit gives the normal probabilty density function.
If mean($\mu$) = 40 and standard deviation($\sigma$) = 5, then find the P(40 $\leq$ x $\leq$ 45), where x is the normal random variable.
Given mean($\mu$) = 40 and standard deviation($\sigma$) = 5
P(40 $\leq$ x $\leq$ 45) = ?
For x = 40, z = $\frac{x - \mu}{\sigma}$
z = $\frac{40 - 40}{5}$ = 0
For x = 45, z = $\frac{x - \mu}{\sigma}$
z = $\frac{45 - 40}{5}$ = 1
P(40 $\leq$ x $\leq$ 45) = P(0 $\leq$ x $\leq$ 1) = [area to the left of z = 1] - [area to the left of z = 0]
P(0 $\leq$ x $\leq$ 1) = [area to the left of z = 1] - [area to the left of z = 0]
$\Rightarrow$ 0.8413 - 0.5
$\Rightarrow$ 0.3413
P(40 $\leq$ x $\leq$ 45) = 0.3413
If mean($\mu$) = 20 and standard deviation($\sigma$) = 4, then find the P(20 $\leq$ x $\leq$ 25), where x is the normal random variable.
Given mean($\mu$) = 20 and standard deviation($\sigma$) = 4
P(20 $\leq$ x $\leq$ 25) = ?
For x = 20, z = $\frac{x - \mu}{\sigma}$
z = $\frac{20 - 20}{4}$ = 0
For x = 25, z = $\frac{x - \mu}{\sigma}$
z = $\frac{25 - 20}{4}$ = 1.25
P(20 $\leq$ x $\leq$ 25) = P(0 $\leq$ x $\leq$ 1.25) = [area to the left of z = 1.25] - [area to the left of z = 0]
P(0 $\leq$ x $\leq$ 1.25) = [area to the left of z = 1.25] - [area to the left of z = 0]
$\Rightarrow$ 0.8944 - 0.5
$\Rightarrow$ 0.3944
P(20 $\leq$ x $\leq$ 25) = 0.3944