Answer:
The proportion of new typists that take at most two hours to learn the computer system is 0.9525 = 95.25%.
Step-by-step explanation:
Normal Probability Distribution:
Problems of normal distributions can be solved using 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.
Normal distribution with a mean of 90 minutes and a standard deviation of 18 minutes.
This means that [tex]\mu = 90, \sigma = 18[/tex]
The proportion of new typists that take at most two hours to learn the computer system is
This is the pvalue of Z when X = 120 minutes = 2 hours. So
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{120 - 90}{18}[/tex]
[tex]Z = 1.67[/tex]
[tex]Z = 1.67[/tex] has a pvalue of 0.9525.
The proportion of new typists that take at most two hours to learn the computer system is 0.9525 = 95.25%.
