Math experts help solve a statistical calculation problem Urgent

First of all, I don't use English in case you can't read it.

The meaning of the title is that the mean age from the sample of 9 students is years, and the standard deviation of their age is years). The age distribution of students is uneven, with a large number of students younger than 19 years old, and some older than 50 years old.

First of all, I want to make sure that you know s 2=[(x1-x pull) 2+(x2-x pull) 2+(x3-x pull) 2+....+xn-xpluck) 2] n

This formula, which is very basic, merges each square term to get the above formula.

Then, substituting the numbers in the question to get (x1 2+x2 2+...).x9^2-9*

x1^2+x2^2+..x9^2=

x1+x2+.x9=

You might as well set x1<=x2<=x3....=x9, where at least x1<19,x2<19,x1,x2....are whole numbers (because it's age).

First of all, it can be determined that x9=50, and only x9=50, because 50*50=2500, if x8=50, 5000, this is not possible.

x1^2+x2^2+..x8^2=

x1+x2+..x8=

Calculate the variance of these 8 numbers, s' 2 = <0

How is this possible, the variance should be non-negative, do this, my heart is cold, is it the lz number input wrong, if it is not negative, you can also introduce, as the 10th number, analyze the variance, gradually narrow the range of values, and then use integers to verify, this is the idea, your problem number has a problem, so there is no solution for the time being, and others should not waste time...

Please, you can translate it!

1.Average = 100 * 35% + 120 * 8% + 98 * 20% + 121 * 12% + 105 * 25% = yuan.

2.The variance of the same product for the same product at 5 points of sale in East China is (the standard deviation is the difference below the root number, which is equal to.

The standard deviation of 5 points of sale in East China is 10 in North China, so the average ** of this commodity in East China is more representative.

(1) 1992: 102%.

1993 vs. 1992: 95 per cent

1994 is double that of 1991 --m1994=1994=1994=

1995 compared to 1994 = 110%.

1996 compared to 1995 = 30 25 = 120% 1997 compared to 1996 = 37 30 =

2) = Month-on-month development speed product =

4) Average development speed = fixed base development speed to the 6th power = 2(1) A: Yield individual index = 5500 5000 100% = 110% factory ** individual index = 21 20 100% = 105% B:

Individual index of output: 3600 3000 = 120% Ex-factory ** Individual index: 28 25 * 100% = 112% (2) Total output index = (5500 + 3600) 8000 = increased output value = 500 * 20 + 600 * 25 = 25000 (3) Ex-factory ** total index = (21 + 28) (20 + 25) = increased output value = 1 * 5000 + 3 * 3000 = 14000

Average monthly gross industrial output = (200 + 230 + 300) 3 = average monthly output value per worker = average monthly total output value Average monthly number of workers in the quarter = total output value of the quarter The sum of the number of workers at the end of each month in the quarter;

That is, the average monthly output value of each worker = [(200 + 230 + 300) 3] [ 210 + 220 + 230) 3] = (200 + 230 + 300) (210 + 220 + 230).

1. The cost of a single book should have increased, 112% 108%=,, the result of my calculation of this question is like this, and the answer you provided is not the same number.

2. Price index 1 Price index 1

My understanding of this question is this, let's talk about the next question first.

The stem prompts that the collected data conform to a normal distribution, with a mean of 10% and a standard deviation of 8%, and the next step is to find the area between those percentages, and the first, second, and third quartiles of the following data.

Then there is the answer, in part a, it is the mean plus or minus one, two, and three standard deviations, and then according to the characteristics of the normal distribution, it is easy to know its area and the order is %,

Part B arranges the data from small to large, so that it is easy to calculate the quartile number, and there are formulas in the book, so it is good to understand it.

Or translate into Chinese first, the average 90 standard deviation 28 take out a sample of 49, what is the probability that the sample mean value is greater than 95?

Make use of the central limit theorem.

p(x>95)=1-p(x<95)=1-p[(x-90)/(28/√49)<(95-90)/(28/√49)]

1-z[(95-90) (28 49)]=1-z(5 4)=I think you can look at the central limit theorem again, which requires the probability that the mean value is greater than 95, so you can directly convert it to the standard normal according to the central limit theorem.