Respuesta :
Answer:
There is a 91.15% probability that their mean life will be longer than 12 years.
Step-by-step explanation:
The Central Limit Theorem estabilishes that, for a random variable X, with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], a large sample size can be approximated to a normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]\frac{\sigma}{\sqrt{n}}[/tex]
Problems of normally distributed samples can be solved using the z-score formula.
In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the zscore of a measure X is given by:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
The Z-score 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. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.
In this problem, we have that:
A manufacturer knows that their items have a normally distributed lifespan, with a mean of 13.2 years, and standard deviation of 3.1 years. This means that [tex]\mu = 13.2, \sigma = 3.1[/tex].
If you randomly purchase 12 items, what is the probability that their mean life will be longer than 12 years?
There are 12 items, so [tex]n = 12[/tex].
This is 1 subtracted by the pvalue of Z when [tex]X = 12[/tex].
By the Central Limit Theorem, we use the standard deviation of the sample in the Z score formula. That is:
[tex]s = \frac{\sigma}{\sqrt{n}} = \frac{3.1}{\sqrt{12}} = 0.89[/tex]
[tex]Z = \frac{X - \mu}{s}[/tex]
[tex]Z = \frac{12-13.2}{0.89}[/tex]
[tex]Z = -1.35[/tex]
[tex]Z = -1.35[/tex] has a pvalue of 0.0885
This means that there is a 1-0.0885 = 0.9115 = 91.15% probability that their mean life will be longer than 12 years.
