Respuesta :
To solve this question, the normal distribution and the central limit theorem are used.
Doing this, there is an 0.0129 = 1.29% probability that the mean installation time for 31 computers is less than 43 minutes.
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First, these concepts are presented, then we identify mean, standard deviation and sample size, and then, we find the desired probability.
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Normal Probability Distribution
Problems of normal distributions can be solved using the z-score formula.
In a set 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]
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 p-value, we get the probability that the value of the measure is greater than X.
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Central Limit Theorem
The Central Limit Theorem establishes 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
For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.
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Mean of 45, standard deviation of 5:
This means that [tex]\mu = 45, \sigma = 5[/tex]
Sample size of 31:
This means that [tex]n = 31, s = \frac{5}{\sqrt{31}}[/tex]
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Probability the sample mean is less than 43:
This is the p-value of Z when X = 43, so:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
By the Central Limit Theorem
[tex]Z = \frac{X - \mu}{s}[/tex]
[tex]Z = \frac{43 - 45}{\frac{5}{\sqrt{31}}}[/tex]
[tex]Z = -2.23[/tex]
[tex]Z = -2.23[/tex] has a p-value of 0.0129.
Thus, 0.0129 = 1.29% probability that the mean installation time for 31 computers is less than 43 minutes.
A similar question is given at: https://brainly.com/question/15020228
