Question

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.

Answer 1

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 $\mu$ and standard deviation $\sigma$ is given by:

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

• 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:

$\mu = 540, \sigma = \sqrt{625} = 25$

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

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

$Z = \frac{569 - 540}{25}$

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 brainly.com/question/24663213

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