Respuesta :
Answer:
The maximum lifetime a bulb can have and still be among the lowest 1% of all lifetimes is 572 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 lifetime of the bulbs is a normally distributed random variable with mean equal to 600 hours and standard deviation of 12 hours.
This means that [tex]\mu = 600, \sigma = 12[/tex]
What is the maximum lifetime a bulb can have (hrs) and still be among the lowest 1% of all lifetimes?
This is the 1st percentile of lifetimes, which is X when Z has a pvalue of 0.01, that is, X when Z = -2.327.
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]-2.327 = \frac{X - 600}{12}[/tex]
[tex]X - 600 = -2.327*12[/tex]
[tex]X = 572[/tex]
The maximum lifetime a bulb can have and still be among the lowest 1% of all lifetimes is 572 hours.
