Answer this question fast

Given a population with a mean of muμequals=100100 and a variance of sigma squaredσ2equals=3636, the central limit theorem appl

lakkis [162]

Answer:

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

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

b) $P(\bar X >101)=1-P(\bar X$

c) $P(\bar X$

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:

$P(\bar X >101)=1-P(\bar X$

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:

$P(\bar X$

5 0

4 years ago

HELP MUST BE DONE BY 4:00

Leya [2.2K]

Answer:

the antelope is 5280 feet the cougar is 4224 feet the hare is 4136 feet the kangaroo is 3520 feet and the coyote is 3773 feet the ostrich is 3773 feet hope this helps the answer is the antelope

8 0

3 years ago

Solve the system of linear equations by substitution

Bond [772]

Answer:

(-4, 5)

Step-by-step explanation (work shown in attached picture):

1) Since x is already isolated in the first equation, substitute that value for x into the other equation to find y. So, substitute 16-4y for the x in 3x + 4y = 8, then solve for y. This gives us y = 5.

2) Now, substitute that given value for y back into any one of the equations to find x. I chose to do it in the first equation. Substitute 5 for the y in x = 16-4y, then solve for x this time. This gives us x = -4.

Since x = -4 and y = 5, the solution is (-4, 5).