Answer:
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.
Step-by-step explanation:
When the distribution is normal, we use the z-score formula.
In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the zscore of a measure X is given by:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
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:
[tex]\mu = 20, \sigma = 2[/tex]
a)Less than 19.5 hours?
This is the pvalue of Z when X = 19.5. So
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{19.5 - 20}{2}[/tex]
[tex]Z = -0.25[/tex]
[tex]Z = -0.25[/tex] 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
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{22 - 20}{2}[/tex]
[tex]Z = 1[/tex]
[tex]Z = 1[/tex] has a pvalue of 0.8413
X = 20
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{20 - 20}{2}[/tex]
[tex]Z = 0[/tex]
[tex]Z = 0[/tex] 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.
