Answer to Question #130232 in Statistics and Probability for Azie

Question

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:

More than 28.8 hours
Between 18.6 and 27.5 hours
Between 25.0 and 34.0 hours

Answer 1

Given the following values under normal distribution,

mean m = 25.4

standard deviation $\sigma$ = 4.1 and

consider variable x = time taken to assemble a car

We can convert this into a standard normal distribution by using the below formula

$(score) Z = \frac{x - m}{\sigma} \sim N(0,1)$

CASE 1 :

We have to find the sum of all the probabilities 'Z' when the car takes more than 28.8 hours to get assembled which can be represented as

$p(x>28.8) = p(\frac{x-m}{\sigma} > \frac{28.8 - m}{\sigma})$

$p(x>28.8) =p(Z>\frac{28.8-25.4}{4.1}) = p(Z>0.829268)$

Using the table for area under the curve for a standard normal distribution of 'Z' we get that

$p(x>28.8) = 0.5 - 0.2939 = 0.2061 = 0.20$

Hence the probability p(x>28.8) = 0.2

CASE 2 :

We have to find the sum of all the probabilities 'Z' when the car takes more than 18.6 hours and less than 27.5 hours to get assembled which can be represented as

$p(18.6<x<27.5) = p(\frac{18.6-m}{\sigma}< \frac{x-m}{\sigma} < \frac{27.5-m}{\sigma})$

$p(18.6<x<27.5) = p(\frac{18.6-25.4}{4.1}< Z < \frac{27.5-25.4}{4.1})$

$p(18.6<x<27.5) = p(-1.658< Z < 0.512)$

Using the table for area under the curve for a standard normal distribution of 'Z' we get that

$p(18.6<x<27.5) = 0.4505 + 0.1950 = 0.6455 = 0.64$

Hence the probability p(18.6<x<27.5) = 0.64

CASE 3 :

We have to find the sum of all the probabilities 'Z' when the car takes more than 25 hours and less than 34 hours to get assembled which can be represented as

$p(25<x<34) = p(\frac{25-m}{\sigma}< \frac{x-m}{\sigma} < \frac{34-m}{\sigma})$

$p(25<x<34) = p(\frac{25-25.4}{4.1}< Z < \frac{34-25.4}{4.1})$

$p(25<x<34) = p(-0.097< Z < 2.097)$

Using the table for area under the curve for a standard normal distribution of 'Z' we get that

$p(25<x<34) = 0.0398 + 0.4817 = 0.519 = 0.52$

Hence the probability p(25<x<34) = 0.52