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Sagot :
Using the normal distribution and the central limit theorem, it is found that there is a 0.1896 = 18.96% probability that the average waiting time for a random sample of ten customers is between 4.0 and 4.2 minutes.
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 4.1 minutes, hence [tex]\mu = 4.1[/tex].
- The standard deviation is of 1.3 minutes, hence [tex]\sigma = 1.3[/tex].
- Sample of 10 customers, hence [tex]n = 10, s = \frac{1.3}{\sqrt{10}} = 0.411[/tex]
The probability is the p-value of Z when X = 4.2 subtracted by the p-value of Z when X = 4, hence:
X = 4.2
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
By the Central Limit Theorem
[tex]Z = \frac{X - \mu}{s}[/tex]
[tex]Z = \frac{4.2 - 4.1}{0.411}[/tex]
[tex]Z = 0.24[/tex]
[tex]Z = 0.24[/tex] has a p-value of 0.5948.
X = 4
[tex]Z = \frac{X - \mu}{s}[/tex]
[tex]Z = \frac{4 - 4.1}{0.411}[/tex]
[tex]Z = -0.24[/tex]
[tex]Z = -0.24[/tex] has a p-value of 0.4052.
0.5948 - 0.4052 = 0.1896
0.1896 = 18.96% probability that the average waiting time for a random sample of ten customers is between 4.0 and 4.2 minutes.
A similar problem is given at https://brainly.com/question/25779119
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