The probability that a random sample of 34 smoke detectors will have a mean lifetime between 58 and 63 months, using the normal distribution, is of:
0.9135 = 91.35%.
How to obtain probabilities using the normal distribution?
The z-score of a measure X of a variable that has mean symbolized by [tex]\mu[/tex] and standard deviation symbolized by [tex]\sigma[/tex] is obtained by the rule presented as follows:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
- The z-score represents how many standard deviations the measure X is above or below the mean of the distribution, depending if the obtained z-score is positive or negative.
- Using the z-score table, the p-value associated with the calculated z-score is found, and it represents the percentile of the measure X in the distribution.
- By the Central Limit Theorem, the sampling distribution of sample means of size n has standard deviation [tex]s = \frac{\sigma}{\sqrt{n}}[/tex].
The parameters in the context of this problem are given as follows:
[tex]\mu = 60, \sigma = 8, n = 34, s = \frac{8}{\sqrt{34}} = 1.37[/tex]
The probability that a random sample of 34 smoke detectors will have a mean lifetime between 58 and 63 months is the p-value of Z when X = 63 subtracted by the p-value of Z when X = 58, hence:
X = 63:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
By the Central Limit Theorem
[tex]Z = \frac{X - \mu}{s}[/tex]
Z = (63 - 60)/1.37
Z = 2.19
Z = 2.19 has a p-value of 0.9857.
X = 58:
[tex]Z = \frac{X - \mu}{s}[/tex]
Z = (58 - 60)/1.37
Z = -1.46
Z = -1.46 has a p-value of 0.0722.
0.9857 - 0.0722 = 0.9135 = 91.35%.
More can be learned about the normal distribution at https://brainly.com/question/25800303
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