Statistics Help I Have The Mean Standard Distribution Sample Size Of 10. I Need To Know What

Need help with statistics question about distribution for sample mean.?

This is an application of the Central Limit Theorem (also known as the Fundamental Theorem of Statistics). It states that when taking a larg enough sample from a population of any distribution, the means of those samples will form a normal model with a mean the same as the parent distribution and standard deviation = (population standard deviation) / (square root of sample size).

Emphasis on "large enough." If the population is normal, any sample size will do. Typically, once you get a sample size around 40, that will work for a population of any shape. So n=50 will do just fine.

The standard deviation of a sampling distribution of means for this situation, then, equals 2.3/sqrt(50) which, to 4 decimal places, is 0.3253.

The z-score of a mean of 15 = (15-12)/0.3253 = 9.222

The corresponding probability is really small, on the order of 1.5x10^-20. In other words pretty much impossible.



if you take a step back to think, that should make sense. The probability that exactly one corn husk will be 15 inches is P(z>(15-12)/2.3)=0.096. Less than 10%. In order to get FIFTY corn husks to average more than 15 inches long, you need to get a less than 1 in 10 chance outcome to happen 50 times in a row. That's some small odds.

Statistics need help finding standard deviation?

A random sample is to be selected from a population that has a proportion of successes p = 0.69. Determine the mean and standard deviation of the sampling distribution of p̂ for each of the following sample sizes. (Round your standard deviations to four decimal places.)

n = 10

Mean= .69 I know this
Standard deviation= ????

But how do i find standard dev

Statistics help! I have the mean, standard distribution, sample size of 10. I need to know what 68.3% of the means will be in what interval?

A normally distributed population has a mean of 25 and a standard deviation of 6. For a set samples of size 10 taken from this population, about 68.3% of the means will be in what interval?

- 22.10,25.90
- 23.10,26.90
- 20.00,22, 00
- 23.23,27.27

Any help will be helpful. I've been reading over my notes, school book, but I can't seem to figure this one out. Any help will be highly appreciated!

Statistics x-bar sampling distribution normal?

hi need help with this problem please do not say the answer i just want no know how i find it i know the larger n is the grater the chances for it to be normal but is there a formula to do this???


A random sample is selected from a population with mean μ = 99 and standard deviation σ = 10. For which of the sample sizes would it be reasonable to think that the x sampling distribution is approximately normal in shape? (Select all that apply.)


n = 12
n = 17
n = 41
n = 65
n = 130
n = 520

thanks

Statistics, last questions, Please help with whichever one you know, Thanks!?

1) Which of the following statements about the sampling distribution of the sample mean, x-bar, is correct? < e >

2) In June 2005, a survey was conducted... < b >

3) Which of the following is true regarding the sample proportion, p-hat, of "yes" responses? < a >

4) If the study were conducted over and over (selecting different samples of people each time), which one of the following would be true regarding the resulting sample proportions of "yes" responses? < c >

5) Which one of the following is true regarding the standard deviation of the sampling distribution of the sample proportion, p-hat, of "yes" responses? < d >

6) The sampling distribution of a statistic is (select the best answer): < d >

7) Suppose that 20% of the residents in a certain state support an increase in the property tax. An opinion poll will randomly sample 400 state residents and will then compute the proportion in the sample that support a property tax increase. How likely is the resulting sample proportion to be within .04 of the true proportion (i.e., between .16 and .24)?

This can be done using a normal distribution probabilities:

p_hat = 20% = 0.20 so q_hat = 1 - 0.20 = 0.80 and the standard error (aka s.e.) is = sqrt (0.20 x 0.80 / 400) = 0.4 / 20 = 0.02

Now, the z-scores = +/- (0.04) / s.e. = +/- (0.04) / 0.02 = +/- 2.

The desired probability is the area under the standard normal curve between the z-scores -2 and +2. Use a table or technology (like I did) to find this area is about 0.95 or 95%.

Note that np = 400 (0.20) = 80 and nq = 400 (0.80) = 320, both are larger than 5 so the normal approximation to the binomial distribution which is needed in the above is valid.

Need help with a Statistics problem....please!!!?

the answer is b. here is how to find it this solution.



the middle 80% of the data is between the 10th and 90th percentiles

For any normal random variable X with mean μ and standard deviation σ , X ~ Normal( μ , σ ), (note that in most textbooks and literature the notation is with the variance, i.e., X ~ Normal( μ , σ² ). Most software denotes the normal with just the standard deviation.)

You can translate into standard normal units by:
Z = ( X - μ ) / σ

Moving from the standard normal back to the original distribuiton using:
X = μ + Z * σ

Where Z ~ Normal( μ = 0, σ = 1). You can then use the standard normal cdf tables to get probabilities.

If you are looking at the mean of a sample, then remember that for any sample with a large enough sample size the mean will be normally distributed. This is called the Central Limit Theorem.

If a sample of size is is drawn from a population with mean μ and standard deviation σ then the sample average xBar is normally distributed

with mean μ and standard deviation σ /√(n)

An applet for finding the values
http://www-stat.stanford.edu/~naras/jsm/...

calculator
http://stattrek.com/Tables/normal.aspx

how to read the tables
http://rlbroderson.tripod.com/statistics...


the 10th percentile is found by:

We know from the standard normal that
P(Z < z) = 0.1
= P(Z > z) = 0.9
when z = -1.281552

x = μ + z * σ
x = 80 + -1.281552 * 8
x = 69.74759



the 90th percentile is found by:
We know from the standard normal that
P(Z < z) = 0.9
= P(Z > z) = 0.1
when z = 1.281552

x = μ + z * σ
x = 80 + 1.281552 * 8
x = 90.25241


the solution is (69.74759 - 90.25241) which rounds to 70 - 90 degrees, answer b.

Statistics problem, need help quickly!! Mean, shape and standard error of distribution?

According to the Alzheimer's association, Alzheimer's disease affects 1 to 10 people over the age of 65. A study was conducted on health issues faced by people over the age of 65. What would be the shape, mean, and standard error of the sampling distribution of the proportion who suffer from Alzheimer's disease if the sample size is 200 compared to if the sample size is 800?

Statistics Problem Need help?

Answer was the 3rd choice I don't get why :
5. For air travelers, one of the biggest complaints involves the waiting time between when the airplane taxis away from the terminal until the flight takes off. This waiting time is known to have a skewed-right distribution with a mean of 10 minutes and a standard deviation of 8 minutes. Suppose 100 flights have been randomly sampled. Describe the sampling distribution of the mean waiting time between when the airplane taxis away from the terminal until the flight takes off for these 100 flights.
a) Distribution is skewed-right with mean = 10 minutes and standard error = 0.8 minutes.
b) Distribution is skewed-right with mean = 10 minutes and standard error = 8 minutes.
c) Distribution is approximately normal with mean = 10 minutes and standard error = 0.8 minutes.
d) Distribution is approximately normal with mean = 10 minutes and standard error = 8 minutes.