Can you find the standard deviation with without knowing the mean?
Sure you can! If this list is x[1], x[2], …, x[n], let Q = sum(i < j) ( x[i] - x[j] )^2. Then sqrt[ Q / (n(n-1) ] is the standard deviation.This is not the easiest way to compute the standard deviation.
What is the difference between variance and standard deviation?
The central tendency mean gives you the idea of average of the data points( i.e centre location of the distribution)And now you want to know how far are your data points from meanSo, here comes the concept of variance to calculate how far are your data points from mean ( in simple terms, it is to calculate the variation of your data points from mean)Population variance : [math]\sum\limits_{i=1}^n \frac{(x_i-\mu)^2}{N}[/math]Sample variance : [math]\sum\limits_{i=1}^n \frac{(x_i-\overline{x})^2}{n-1}[/math]Standard deviation is simply the square root of variance . And standard deviation is also used to calculate the variation of your data points(And you may be asking, why do we use standard deviation , when we have variance. Because, in order to maintain the calculations in same units i.e suppose mean is in [math]cm/m,[/math] then variance is in [math]cm^2/m^2[/math] , whereas standard deviation is in [math]cm/m[/math] , so we use standard deviation most)population standard deviation:[math]\sqrt{\sum\limits_{i=1}^n \frac{(x_i-\mu)^2}{N}}[/math]sample standard deviation :[math] \sqrt{\sum\limits_{i=1}^n \frac{(x_i-\overline{x})^2}{n-1}}[/math]
Statistics question, confidence intervals- my answers are included?
1.) D. Is wrong. Consider that a confidence interal has the form x̄ ± cs/√n, where c is a constant (often 1.96). Clearly if either s or c increases, the confidence interval gets wider. Notice that as n gets larger, dividing by √n makes the ± amount smaller, hence decreasing the confidence interval size. The formula holds the key to seeing that the correct answer is C. 2.) x̄ ± cs/√n = 452 ± 1.96(12)/√4 = 452 ± 11.76 = (440.24, 463.76) which is C 3.)They want n such that the CI is 452 ± 2. In other words, then want 1.96(12)/√n = 2. So solve for n and get n = (1.96(12)/2)² = 11.76² = 138.30. So since it is greater than 138, make it1 39 because 138 won't be enough. Note that if we used 138 we'd get this: 1.96(12)/√138 = 2.002 which is not "within 2" so we must go up to n = 139. 4.) is B, because a wider interval (more conmfidence) means less risk. Think that you are casting a bigger net so you are more likely to catch the fish (the mean). 5.) Is done like #3. Here σ² = 2.1, so σ = √2.1 and they want ± 1.96(√2.1)/√n = 0.3. Again, solve for n and get √n = 1.96(√2.1)/0.3 = 9.4677 so n = 9.4677² = 89.64 approx. So we go up to 90 as before. I hope you have the idea by now. The cs/√n is the half-width of the confidence interval. For 90%, use c=1.645 instead of c=1.96. P.S. I like bad corny jokes. You will understand this if you just keep looking at what the formula for a CI tells you. I like your attitude. Good luck.
Probability with mean and standard deviation.?
A grocery supplier believes that in a dozen eggs, the mean number of broken eggs is .8 with a standard deviation of .5 eggs. You buy 3 dozen eggs with out checking them. A) How many broken eggs do you expects to get? B) What's the standard deviation? C) What assumption did you have to make about the eggs in order to answer this question? That the egg cartoons are ... Independent, continuously random variable or discrete random variable? It would be appreciated if you could explain this to me and not just give me an answers and let me know if any of this is possible on a TI-84 Plus Calculator?
What does standard deviation tell you about a data set?
The standard deviation of a dataset gives you a measure of how spread out it is. On an average, it helps you ascertain how close each point is from the mean.To understand this in a practical scenario, let’s look at test scores across two exams. Let’s say the mean for each is 70 and 80 points, out of 100, respectively. If the standard deviation is say, 5 for the first test and 10 for the other test, what can we say about the tests? We can say that the results for the second exam were more spread out, and that means that the test takers got scores vastly different to each other, in comparison to the first exam, in which the scores were packed more tightly.
Why is there a degree of freedom of n-1 for sample standard deviation?
Degree of FreedomDegree of freedom is nothing but number of observations (number of independent piece of information) in the data that are free to vary when estimating the parameters.For example:1. Sample meanYou try to calculate a sample mean for a particular random variable. You have collected 5 data points to calculate the sample mean using below formula.[math]\bar{x} =\sqrt{\frac{\sum x}{n}} \tag{1}[/math]If you sample one more time with the same sample size, you are not at all surprised to get different estimate due to random variability. All the data points are free to vary in calculating the sample mean.Data points: 5, 4, 3, 6, 10.You have a plan to sample 5 data points and sampled a first data point. Still, you need to sample four more data points and all the four data points can be anything. It means that all the data points are free to vary. Therefore, the degree of freedom for a sample mean is n.2. Standard deviationIf we know the population mean, the standard deviation is given by[math]\sigma=\sqrt{\frac{\sum(x-\mu)}{n}^2} \tag{2}[/math]If we don’t know the population mean, we can use sample mean to calculate the standard deviation.[math]s=\sqrt{\frac{\sum(x-\bar x)}{n-1}^2} \tag{3}[/math]Why different degree of freedom for the sample standard deviation and the population standard deviation?Assume that we don’t know the population mean for the above sample. So, we need to calculate the sample standard deviation for those data points.Data points: 5, 4, 3, 6, 10.Sample mean using the equation 1 [math](\bar{x}): 5.6[/math]In this scenario, four data points are free to vary. But, the fifth data point is fixed automatically due to a constraint that [math]\bar{x} = 5.6.[/math] This constraint arises only if we use the sample mean to calculate the standard deviation. If we know the population mean for the above data points, there is no constraint that sample mean of data point is equal to population mean. So, all the five data points are free to vary. This is the reason, degree of freedom for the equation 2 is n and degree of freedom for the equation 3 is n-1.