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Sagot :
Answer:
0.7031 = 70.31% probability of a bulb lasting for at most 528 hours.
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 z-score 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 p-value, we get the probability that the value of the measure is greater than X.
The standard deviation of the lifetime is 15 hours and the mean lifetime of a bulb is 520 hours.
This means that [tex]\sigma = 15, \mu = 520[/tex]
Find the probability of a bulb lasting for at most 528 hours.
This is the p-value of Z when X = 528. So
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{528 - 520}{15}[/tex]
[tex]Z = 0.533[/tex]
[tex]Z = 0.533[/tex] has a p-value of 0.7031
0.7031 = 70.31% probability of a bulb lasting for at most 528 hours.
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