Question

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

a)Less than 19.5 hours?
b)Between 20 hours and 22 hours?

Answer 1

a) 40.13% probability that a laptop computer can be assembled at this plant in a period of time of less than 19.5 hours.

b) 34.13% probability that a laptop computer can be assembled at this plant in a period of time between 20 hours and 22 hours.

Explanation:

When the distribution is normal, we use the z-score formula.

In a set with mean
`\mu` and standard deviation
`\sigma`, the zscore of a measure X is given by:

`Z = (X - \mu)/(\sigma)`

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this question:

`\mu = 20, \sigma = 2`

a)Less than 19.5 hours?

This is the pvalue of Z when X = 19.5. So

`Z = (X - \mu)/(\sigma)`

`Z = (19.5 - 20)/(2)`

`Z = -0.25`

`Z = -0.25` has a pvalue of 0.4013.

40.13% probability that a laptop computer can be assembled at this plant in a period of time of less than 19.5 hours.

b)Between 20 hours and 22 hours?

This is the pvalue of Z when X = 22 subtracted by the pvalue of Z when X = 20. So

X = 22

`Z = (X - \mu)/(\sigma)`

`Z = (22 - 20)/(2)`

`Z = 1`

`Z = 1` has a pvalue of 0.8413

X = 20

`Z = (X - \mu)/(\sigma)`

`Z = (20 - 20)/(2)`

`Z = 0`

`Z = 0` has a pvalue of 0.5

0.8413 - 0.5 = 0.3413

34.13% probability that a laptop computer can be assembled at this plant in a period of time between 20 hours and 22 hours.