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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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