Get clear, concise, and accurate answers to your questions on IDNLearn.com. Discover in-depth and trustworthy answers from our extensive network of knowledgeable professionals.
Sagot :
Answer:
0.8413 = 84.13% probability of a bulb lasting for at most 605 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 590 hours.
This means that [tex]\sigma = 15, \mu = 590[/tex]
Find the probability of a bulb lasting for at most 605 hours.
This is the pvalue of Z when X = 605. So
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{605 - 590}{15}[/tex]
[tex]Z = 1[/tex]
[tex]Z = 1[/tex] has a pvalue of 0.8413
0.8413 = 84.13% probability of a bulb lasting for at most 605 hours.
We appreciate every question and answer you provide. Keep engaging and finding the best solutions. This community is the perfect place to learn and grow together. For trustworthy answers, visit IDNLearn.com. Thank you for your visit, and see you next time for more reliable solutions.
