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castortr0y [4]
3 years ago
5

The time taken to assemble a laptop computer in a certain plant is a random variable having a normal distribution of 20 hours an

d 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?

Mathematics
1 answer:
ludmilkaskok [199]3 years ago
4 0

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 \mu and standard deviation \sigma, the zscore of a measure X is given by:

Z = \frac{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 = \frac{X - \mu}{\sigma}

Z = \frac{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 = \frac{X - \mu}{\sigma}

Z = \frac{22 - 20}{2}

Z = 1

Z = 1 has a pvalue of 0.8413

X = 20

Z = \frac{X - \mu}{\sigma}

Z = \frac{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.

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