Respuesta :
Using the normal distribution and the central limit theorem, it is found that there is a 0.9878 = 98.78% probability that the mean lifetime of components in this sample will be longer than 99 months.
In a normal distribution 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]
- It 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, which is the percentile of X.
- By the Central Limit Theorem, the sampling distribution of sample means of size n has standard deviation [tex]s = \frac{\sigma}{\sqrt{n}}[/tex].
In this problem:
- The mean is of 9 years, hence, in months, [tex]\mu = 9(12) = 108[/tex]
- The standard deviation is of 1 year, hence, in months, [tex]\sigma = 12[/tex].
- Sample of 9 pacemakers, hence [tex]n = 9, s = \frac{12}{\sqrt{9}} = 4[/tex].
The probability is 1 subtracted by the p-value of Z when X = 99, hence:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
By the Central Limit Theorem
[tex]Z = \frac{X - \mu}{s}[/tex]
[tex]Z = \frac{99 - 108}{4}[/tex]
[tex]Z = -2.25[/tex]
[tex]Z = -2.25[/tex] has a p-value of 0.0122.
1 - 0.0122 = 0.9878.
0.9878 = 98.78% probability that the mean lifetime of components in this sample will be longer than 99 months.
A similar problem is given at https://brainly.com/question/24663213
