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he life of light bulbs is distributed normally. The variance of the lifetime is 625 and the mean lifetime of a bulb is 540 hours. Find the probability of a bulb lasting for at most 569 hours. Round your answer to four decimal places.

Sagot :

Using the normal distribution, it is found that there is a 0.877 = 87.7% probability of a bulb lasting for at most 569 hours.

Normal Probability Distribution

The z-score of a measure X of a normally distributed variable with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex] is given by:

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

  • The z-score measures how many standard deviations the measure is above or below the mean.
  • Looking at the z-score table, the p-value associated with this z-score is found, which is the percentile of X.

The mean and the standard deviation are given, respectively, by:

[tex]\mu = 540, \sigma = \sqrt{625} = 25[/tex]

The probability of a bulb lasting for at most 569 hours is the p-value of Z when X = 569, hence:

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

[tex]Z = \frac{569 - 540}{25}[/tex]

Z = 1.16

Z = 1.16 has a p-value of 0.877.

0.877 = 87.7% probability of a bulb lasting for at most 569 hours.

More can be learned about the normal distribution at https://brainly.com/question/24663213

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