Answer to Question #183047 in Statistics and Probability for Richard Sasunathan

Question

3. The time taken to assemble a car in a certain plant has a normal distribution with mean of 25.4 hours and a standard deviation of 4.1 hours. Calculate the probability that a car can be assembled at this plant in the following period of time:

a) More than 28.8 hours

b) Between 18.6 and 27.5 hours

c) Between 25.0 and 34.0 hours

Answer 1

$\mu = 25.4 \\ \sigma = 4.1$

a) P(X>28.8) = 1 -P(X<28.8)

$= 1 -P(Z< \frac{28.8 -25.4}{4.1}) \\ = 1 -P(Z<0.829) \\ = 1 -0.7964 \\ = 0.2036$

b) P(18.6<X<27.5) = P(X<27.5) -P(X<18.6)

$=P(Z< \frac{27.5-25.4}{4.1}) -P(Z< \frac{18.6-25.4}{4.1}) \\ = P(Z<0.512) -P(Z< -1.658) \\ = 0.6956 -0.0486 \\ = 0.647$

c) P(25.0<X<34.0) = P(X<34.0) -P(X<25.0)

$= P(Z< \frac{34.0-25.4}{4.1}) -P(Z< \frac{25.0 -25.4}{4.1}) \\ = P(Z<2.097) -P(Z<-0.097) \\ = 0.9820 -0.4613 \\ = 0.5207$