Answer to Question #292895 in Statistics and Probability for normaldistribution

Question

The time taken to assemble a car in certain plant is a random variable having a normal distribution of 20 hours and a standard deviation of 2 hours. What is the probability the car can be assembled at this plant in a period of time

a) less than 19.5 hours?

b) between 20 and 22 hours?

Answer 1

Let X denote the time taken to assemble a car in a certain plant. Given $\mu =20, \sigma=2$.

Let $Z = \dfrac{X - 20}{2}$.

a)

$\begin{aligned} P(X<19.5) &= P\left(\dfrac{X - 20}{2} < \dfrac{19.5 -20}{2}\right)\\ &= P(Z < -0.25)\\ &= 1 - P(Z<0.25)\\ &= 1 - 0.5987 ~~(\text{Using Standard Normal Table})\\ &=0.4013 \end{aligned}$

b)

$\begin{aligned} P(20<X<22) &= P\left(\dfrac{20-20}{2}<\dfrac{X - 20}{2} < \dfrac{22 -20}{2}\right)\\ &= P(0< Z < 1)\\ &= P(Z<1)-P(Z<0)\\ &=0.8413-0.5\\ &= 0.3413 ~~(\text{Using Standard Normal Table})\\ \end{aligned}$