Question

The average income in a certain region in 2013 was ​$59 comma 000 per person per year. Suppose the standard deviation is ​$26 comma 000 and the distribution is​ right-skewed. Suppose we take a random sample of 625 residents of the region. We want to find the probability that the sample mean will be more than ​$2 comma 080 away from the population mean. Use the technology output given below to complete parts​ (a) through​ (c).

Answer 1

Answer:

a) Yes. it is large enough because the sample size of 100 is greater than 25.

b) $\bar X \sim N(\mu=59000, \frac{26000}{\sqrt{625}}=1040)$

$\mu_{\bar X}=59000$ $\sigma_{\bar X}=1040$

c) $P(\bar X >61080)=1-P(\bar X<61080)=1-P(Z<\frac{61080-59000}{1040})=1-P(Z<2)=1-0.9773=0.0228=2.28\%$

Step-by-step explanation:

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".  

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".

Let X the random variable average income in a certain region in 2013 . We know from the problem that the distribution for the random variable X is given by:

$X\sim N(\mu =59000,\sigma =26000)$

We take a sample of n=625 . That represent the sample size.

Part a: Is the sample size large enough to use Central Limit Theorem ? Explain

Yes. it is large enough because the sample size of 100 is greater than 25.

Part b: The mean is___ and the standard error_____

From the central limit theorem we know that the distribution for the sample mean $\bar X$ is also normal and is given by:

$\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})$

$\bar X \sim N(\mu=59000, \frac{26000}{\sqrt{625}}=1040)$

$\mu_{\bar X}=59000$ $\sigma_{\bar X}=1040$

Part c: We want to find the probability that the sample mean will be more than ​$2080 away from the population mean

We want this probability:

$P(\bar X >59000+2080)=P(\bar X >61080)$

So we can find the z score for the two values with the following formula:

$z=\frac{\bar X- \mu}{\frac{\sigma}{\sqrt{n}}}$

And we want to find this probability:

$P(\bar X >61080)=1-P(\bar X<61080)=1-P(Z<\frac{61080-59000}{1040})=1-P(Z<2)=1-0.9773=0.0228=2.28\%$