Question

Given a population with a mean of muμequals=100100 and a variance of sigma squaredσ2equals=3636​, the central limit theorem applies when the sample size is ngreater than or equals≥25. A random sample of size nequals=2525 is obtained. a. What are the mean and variance of the sampling distribution for the sample​ means? b. What is the probability that x overbarxgreater than>101101​? c. What is the probability that 9898less than

Answer 1

Answer:

a) $\bar X \sim N(100,\frac{6}{\sqrt{25}}=1.2)$

$\mu_{\bar X}=100$ $\sigma^2_{\bar X}=1$

b) $P(\bar X >101)=1-P(\bar X <101) =1-P(Z<0.833)=1-0.798=0.202$

c) $P(\bar X <98)=P(Z<-1.67)=0.0478$

Step-by-step explanation:

Previous concepts

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 Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations 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".  

The complement rule is a theorem that provides a connection between the probability of an event and the probability of the complement of the event. Lat A the event of interest and A' the complement. The rule is defined by: $P(A)+P(A') =1$

Let X the random variable that represent the variable of interest on this case, and for this case we know the distribution for X is given by:  

$X \sim N(\mu=100,\sigma=6)$  

And let $\bar X$ represent the sample mean, by the central limit theorem, the distribution for the sample mean is given by:  

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

a. What are the mean and variance of the sampling distribution for the sample​ means?

$\bar X \sim N(100,\frac{6}{\sqrt{25}}=1.2)$

$\mu_{\bar X}=100$ $\sigma^2_{\bar X}=1.2^2=1.44$

b. What is the probability that x overbarxgreater than>101

First we can to find the z score for the value of 101. And in order to do this we need to apply the formula for the z score given by:  

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

If we apply this formula to our probability we got this:  

$z=\frac{101-100}{\frac{6}{\sqrt{25}}}=0.833$  

And we want to find this probability:

$P(\bar X >101)=1-P(\bar X <101) =1-P(Z<0.833)=1-0.798=0.202$

On this last step we use the complement rule.  

c. What is the probability that x bar 98less than

First we can to find the z score for the value of 98.

$z=\frac{98-100}{\frac{6}{\sqrt{25}}}=-1.67$  

And we want to find this probability:

$P(\bar X <98)=P(Z<-1.67)=0.0478$