Respuesta :
Answer:
For the exponential distribution:
[tex] \mu = \frac{1}{\lambda}[/tex]
[tex] \sigma^2 = \frac{1}{\lambda^2}[/tex]
We know that the exponential distribution is skewed but the sample mean for this case using a sample size of 60 would be approximately normal, so then we can conclude that if we have a sample size like this one and an exponential distribution we can approximate the sample mean to the noemal distribution and indeed use the Central Limit theorem.
[tex] \bar X \sim N(\mu_{\bar X} , \frac{\sigma}{\sqrt{n}})[/tex]
[tex] \mu_{\bar X} = \bar X[/tex]
[tex]\sigma_{\bar X}= \frac{\sigma}{\sqrt{n}}[/tex]
Step-by-step explanation:
The central limit theorem states that "if we have a population with mean μ and standard deviation σ and take sufficiently large random samples from the population with replacement, then the distribution of the sample means will be approximately normally distributed. This will hold true regardless of whether the source population is normal or skewed, provided the sample size is sufficiently large".
For this case we have a large sample size n =60 >30
The exponential distribution is the probability distribution that describes the time between events in a Poisson process.
For the exponential distribution:
[tex] \mu = \frac{1}{\lambda}[/tex]
[tex] \sigma^2 = \frac{1}{\lambda^2}[/tex]
We know that the exponential distribution is skewed but the sample mean for this case using a sample size of 60 would be approximately normal, so then we can conclude that if we have a sample size like this one and an exponential distribution we can approximate the sample mean to the noemal distribution and indeed use the Central Limit theorem.
[tex] \bar X \sim N(\mu_{\bar X} , \frac{\sigma}{\sqrt{n}})[/tex]
[tex] \mu_{\bar X} = \bar X[/tex]
[tex]\sigma_{\bar X}= \frac{\sigma}{\sqrt{n}}[/tex]
