If you are given the mean and standard deviation of a raw data, can you conclude whether the data is normal distribution or not? If yes, how so?
The mean and variance are the first two moments of the data, and they describe the center and spread respectively. They indicate nothing about the shape. So the answer is a resounding NO. If you are given the third (skewness) and fourth (kurtosis) moments, then YES, you can make a good conclusion about whether the data approximately follow a normal distribution.
Help finding the standard deviation of SAT scores using a sum of independent normals with correlation?
Letting R denote Reading, M denote Math and W denote writing... The SAT has the following stats: Mean R score = 497, S.D. R = 114 Mean M score = 514, SD M = 117 Mean W score = 489, SD W = 113 Furthermore the following correlations between the exams are: Corr(R,M) = Corr (W,M) = 0.72, Corr(R,W) = 0.84 Find the standard deviation of the total composite SAT score. I know the answer is 315.18, but I don't know how to get there
Scores are normally distributed with a mean of 86 and a standard deviation of 14. What is the probability that a random student scored below 72?
Hi,Use a Z Score calculator such as this one:Z Score or Standard Value CalculatorRandom Value = 72Mean (u) = 86Standard Deviation = 14You should get this resultUsing a Z Score lookup table:You will see a Z value of 1 = .3413 or 34.13% of the area under the curve.But… to finish the calculation we need to put all the numbers in the right context.The value of .3413 is the area under the curve from “0” (zero, the normalized average of 86) to the value of 100 or 72 (since normal distribution is symmetrical).The number you want is what is the probability of a random student getting a score below 72 (which is 1 standard deviation away from the average, by the way).What we need to do is calculate the percentage (probability) below 72 by subtracting .3413 from .5000 (.5000 is the probability of what is contained in the lower half of the normal distribution and the .3413 is the probability of what is above 72). So .5000 - .3413 = .1587 or the probability that a student scored below 72 is 15.87%.To calculate the probability of a student scoring 72 or better just add the probabilities above the 72 level which would be .3413 + .5000 (upper half of normal distribution which was not affected by our calculations) and we get .8413 or 84.13% probability a student scored above 72.I hope this helps…
If students test scores are normally distributed (mean 75 and standard deviation 15), what is the probability that a randomly selected group of 10 students will have a mean score greater than 80?
[math]X_i =[/math] the mean of group [math]i[/math] of [math]10[/math] randomly selected students.[math]X[/math] is normally distributed with mean [math]= 75[/math], and standard deviation [math]= \frac{15}{\sqrt{10}}[/math][math]\begin{align*} P\left(X>80\right) & = \textstyle P\left(Z > \frac{80-75}{15/\sqrt{10}}\right) \\ & = 1 - \textstyle P\left(Z < \frac{\sqrt{10}}{3}\right) \\ & = 1 - 0.8541 \\ & = 0.1459 \\ & = \boxed{\boldsymbol{14.59\%}} \end{align*}[/math]
Help. Solve the problem. Mean,Confidence interval, Null hypothesis?
Can some one please explain how to solve these problems. What are the steps? Solve the problem. The systolic blood pressures of the patients at a hospital are normally distributed with a mean of 136 mm Hg and a standard deviation of 13.8 mm Hg . Find the two blood pressures having these properties: the mean is midway between them and 90% of all blood pressures are between them. A) 114.3 mm Hg, 159.7 mm Hg B) 123.6 mm Hg, 148.4 mm Hg C) 113.3 mm Hg, 158.7 mm Hg D) 118.3 mm Hg, 153.7 mm Hg Use the given degree of confidence and sample data to construct a confidence interval for the population mean μ. Assume that the population has a normal distribution. A sociologist develops a test to measure attitudes about public transportation, and 27 randomly selected subjects are given the test. Their mean score is 76.2 and their standard deviation is 21.4. Construct the 95% confidence interval for the mean score of all such subjects. A) 74.6 < μ < 77.8 B) 67.7 < μ < 84.7 C) 64.2 < μ < 88.2 D) 69.2 < μ < 83.2 Express the null hypothesis H0 and the alternative hypothesis H1 in symbolic form. Use the correct symbol (μ, p, σ )for the indicated parameter. A skeptical paranormal researcher claims that the proportion of Americans that have seen a UFO, p, is less than 2 in every one thousand. A) H0: p > 0.002 H1: p ≤ 0.002 B) H0: p < 0.002 H1: p ≥ 0.002 C) H0: p = 0.002 H1: p < 0.002 D) H0: p = 0.002 H1: p > 0.002
Statistics help! (intro statistics)?
I was wondering anyone might be able to help me on this problem. I'M NOT POSTING THIS TO TRY AND GET SOMEONE TO DO MY HOMEWORK, I JUST NEED AN EXPLANATION OF WHAT I NEED TO DO TO GET TO THE ANSWER. Q: The Scholastic Achievement Test (SAT) is carefully constructed to produce results with known population parameters (mean and standard deviation). Suppose that a researcher gives the SAT to a random sample of men and an independent random sample of women. Based on the size of these samples and the known population standard deviation, the researcher calculates the standard error of the difference between means to be 20. What is the probability that the mean score for the sample of women will be at least 30 points higher than the mean for the sample of men? A: .067 (the answer was provided for me in the back of my packet, however it doesn't show the steps to getting the answer) If anyone can point me in the right direction and help explain how to get .067 as my answer will be greatly appreciated!