Respuesta :
Answer:
0.9830 = 98.30% probability that the mean clock life would be greater than 12.5 years.
Step-by-step explanation:
To solve this question, we need to understand the normal probability distribution and the central limit theorem.
Normal probability distribution
Problems of normally distributed samples are solved using the z-score formula.
In a set with mean [tex]\mu[/tex] and standard deviation(square root of the variance) [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.
Central Limit Theorem
The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]s = \frac{\sigma}{\sqrt{n}}[/tex].
For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.
In this problem, we have that:
[tex]\mu = 14, \sigma = \sqrt{25} = 5, n = 50, s = \frac{5}{\sqrt{50}} = 0.7071[/tex]
What is the probability that the mean clock life would be greater than 12.5 years?
This is 1 subtracted by the pvalue of Z when X = 12.5. So
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{12.5 - 14}{0.7071}[/tex]
[tex]Z = -2.12[/tex]
[tex]Z = -2.12[/tex] has a pvalue of 0.0170.
1 - 0.0170 = 0.9830
0.9830 = 98.30% probability that the mean clock life would be greater than 12.5 years.
