Respuesta :
Answer:
a) 99.24% chance HLI will find a sample mean between 5.5 and 7.1 hours.
b) 81.64% probability that the sample mean will be between 5.9 and 6.7 hours.
Step-by-step explanation:
To solve this question, it is important to know the Normal probability distribution and the Central Limit Theorem
Normal probability distribution
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.
Central Limit Theorem
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].
In this problem, we have that:
[tex]\mu = 6.3, \sigma = 2.1, n = 49, s = \frac{2.1}{\sqrt{49}} = 0.3[/tex]
A) What is the chance HLI will find a sample mean between 5.5 and 7.1 hours?
This is the pvalue of Z when X = 7.1 subtracted by the pvalue of Z when X = 5.5.
By the Central Limit Theorem, the formula for Z is:
[tex]Z = \frac{X - \mu}{s}[/tex]
X = 7.1
[tex]Z = \frac{7.1 - 6.3}{0.3}[/tex]
[tex]Z = 2.67[/tex]
[tex]Z = 2.67[/tex] has a pvalue of 0.9962
X = 5.5
[tex]Z = \frac{5.5 - 6.3}{0.3}[/tex]
[tex]Z = -2.67[/tex]
[tex]Z = -2.67[/tex] has a pvalue of 0.0038
So there is a 0.9962 - 0.0038 = 0.9924 = 99.24% chance HLI will find a sample mean between 5.5 and 7.1 hours.
B) Calculate the probability that the sample mean will be between 5.9 and 6.7 hours.
This is the pvalue of Z when X = 6.7 subtracted by the pvalue of Z when X = 5.9
X = 6.7
[tex]Z = \frac{6.7 - 6.3}{0.3}[/tex]
[tex]Z = 1.33[/tex]
[tex]Z = 1.33[/tex] has a pvalue of 0.9082
X = 5.9
[tex]Z = \frac{5.9 - 6.3}{0.3}[/tex]
[tex]Z = -1.33[/tex]
[tex]Z = -1.33[/tex] has a pvalue of 0.0918.
So there is a 0.9082 - 0.0918 = 0.8164 = 81.64% probability that the sample mean will be between 5.9 and 6.7 hours.
