Respuesta :
Using the normal distribution and the central limit theorem, it is found that the probability that the mean of these 100 sleep times is farther than 0.4 hours away from the population mean is:
- a) 0.0455
Normal Probability Distribution
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, as long as n > 30, is approximately normal and has standard deviation [tex]s = \frac{\sigma}{\sqrt{n}}[/tex].
In this problem:
- Mean of 10 hours, hence [tex]\mu = 10[/tex].
- Standard deviation of 2 hours, hence [tex]\sigma = 2[/tex].
- Sample of 100, hence [tex]n = 100, s = \frac{2}{\sqrt{100}} = 0.2[/tex].
We want the probability that the mean is farther than 0.4 hours away from the population mean, hence:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
By the Central Limit Theorem:
[tex]Z = \frac{0.4}{s}[/tex]
[tex]Z = \frac{0.4}{0.2}[/tex]
[tex]Z = 2[/tex]
The probability is P(|Z| > 2), which is 2 multiplied by the p-value of Z = -2.
- Z = -2 has a p-value of 0.0275.
- 2 x 0.0275 = 0.0455
Hence, option A is correct.
To learn more about the normal distribution and the central limit theorem, you can take a look at https://brainly.com/question/24663213
