Statistics - mid interval value of grouped data?
If the data is discrete then the mid interval value of the interval 1-5 is (1+5)/2 = 3 BUT the mid interval value is defined as 1/2*(lower class boundary + upper class boundary) AND 1-5 means 1 ≤ x < 6 SO the mid interval value should be 1/2*(1+6) = 3.5 why is the mid interval value 3 and not 3.5? why does the definition yields incorrect values ? please help me :)
In statistics, for different class intervals, how do we estimate the mid-points of extreme open classes like 500+?
In statistics, for different class intervals, how do we estimate the mid-points of extreme open classes like 500+?Good question. You can’t. The best bet is to choose some arbitrary value based on the shape of the frequency polygon. You can usually make a guess at where the number of observations would be small enough to ignore.
If xi's are the mid points of the class intervals of grouped data, fi's are the corresponding frequency and x̅ is mean, then ∑(fixi-x̅)?
OK let me explain you with layman terms. In frequency table, generally frequency are calculated within a specific boundary, that is called class boundary.Now you are said to find out average of some numerical numbers. What will you do? You would add all the numbers and then divide them with number counts.For the case of frequency table, there are no direct numbers available to add. What can we do? First we are taking mid value of a certain upper and lower boundary by adding upper and lower class boundary values and then dividing it by 2. Then we would multiply it with the corresponding frequency. Here actually, we are assuming all the corresponding counted frequency has the same value as the mid value of that boundaries. After progressing of calculation, at last we can get summation of all fx values. After that we can divide it with total number of counted frequency. Finally we got average.But if we again subtract the fx value from x bar(x with line drawn above of it) it shows deviation or difference between x bar(mean) and fx. And at last after progressing of this equation, we got summation of fx minus x b bar. Most astonishingly, whatever the data is, the result is always zero.Although it is another case, to get rid of zero, the square term of summation of fx minus x bar is taken. And after that if result is divided by N, then it's called variance. If we add extra square root along with the equation, then it called SD.
How do I take assumed mean in statistics?
TRICK(mostly works)if no. of observations is even then take assumed mean =(n/2 +1)th observationif no.of observation is odd then take assumed mean as=(n/2 + 0.5)th observatione.g if no. of obs=16 assumed mean=16/2+1=9th obsif no. of obs =15 assumed mean= 15/2+0.5=7.5+0.5=8th obs
Do ungrouped data or grouped data give an accurate mean and standard deviation? Why?
Assuming that the data are accurate, then ungrouped data will give an accurate mean and standard deviation and grouped data can only be approximately right and may be quite far off, if any of the groups are unbounded.Raw data give accurate values by definition. That’s how the mean and sd are calculated - on raw (ungrouped) data.Grouped data with bounded groups can give pretty good approximations of the mean and sd, but you have to make some additional assumptions to do the calculation. In particular, you have to assume some distribution of the values within each group - the usual thing is to assume a uniform distribution, but this may not be realistic; a more realistic (but more difficult to implement) assumption is that the distribution in each group is part of the larger distribution of the variable).However, if any group is unbounded, then anything can happen. If you have such data you should use the median or a trimmed mean instead of the mean, and either median absolute deviation or interquartile range instead of sd.
How do I calculate the mode for unequal class intervals?
Yes we can.. We need to do more computations and use an appropriate formula for that. we calculate height of the class interval by dividing the frequency by that class width. That class which has the maximum height will be the modal class, containing the mode.Class frequency height10 - 15 5 5/5= 115 - 18 9 9/3 = 3 call this H118 - 25 35 35/7= 525 - 29 16 16/4 = 4 Call this H230 - 32 4 4/2 = 2Now, class 18 - 25 contains mode.L = lower limit of the class 18.W = width of the class.height differences between classes on either side of modal class:h1 = 5 - 3 = 2h2 = 5 - 4 = 1Mode = L + h1 * w / (h1 + h2) = 18 + 2 * 7 /(2+1) = 68/3= 22.6667Also mode can be expressed as = L + H2 * w / (H1 + H2) = 18 + 4*7/(3+4)= 22.0Read more on http://Brainly.in - https://brainly.in/question/1910...
How do I find the midpoint of an interval?
The variable [math]x[/math] usually indicates a real variable. Midpoints of intervals are usually only considered for intervals of real numbers. The midpoint of the interval [math][30,40)[/math] is [math]35[/math]. The midpoint of the interval [math][30,39][/math] is [math]34.5[/math].On the other hand, if you have 10 data values 30, 31, 32, 33, 34, 35, 36, 37, 38, and 39, their median is 34.5 since there are an even number of data values, and the median in that case is taken to be the average of the middle two. Perhaps that’s what was meant rather than the midpoint of an interval.
How to find the probability for 5 intervals using the standard normal table?
I don't understand this either. It seems to be a bit of a nonsense question really. There is no way that these counts resemble a normal distribution (but perhaps that is the object of the exercise, to show this ?) Well, O K, the mean seems to be at 0, which corresponds to a normal distribution, but to use the tables, you need to know the standard deviation, which is not given. You could calculate it for the given data I suppose . . . . . [ for grouped data, you take the mid-point of each group (call it m), and multiply ] . . . . . [ it by the group frequency (call it f). Then calculate the variance (s²) from . . . . ] . . . . . [ {Σ f (m - μ)²} / n. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ] but there is another problem, because you do not have a lower limit for the under -0.8 group, nor an upper limit for the over +0.8 group, so you cannot find a mid-point for these groups. The frequencies do not even seem to be diminishing towards the upper and lower groups, as they would for a normal distribution, so you cannot even assume that the lower limit might be -1.4, say, and the upper limit +1.4 (making for equal group widths.) So there is really not much you can do with it. I would sketch a histogram with the given frequencies, and it should be obvious that it cannot be dealt with as a normal distribution.
What is the difference between class intervals and class boundaries?
In short, class interval is nothing but the difference between class boundariesClass boundaryClass boundary is the midpoint of the upper class limit of one class and the lower class limit of the subsequent class.Eg: class A 0-9 B 10-19 C 20-29For Class B, Lower class limit(LCL)=10 Upper class limit(UCL)=19 a) Lower class boundary = (LCL + UCL of the previous class)/2 [math]= ( 10 + 9 )/2 =9.5[/math]b) Upper class boundary = (UCL+ LCL of the Subsequent class)/2 [math]= ( 20 + 19 )/2 =19.5[/math]Class interval a) Inclusive Class Interval: When the lower and the upper class limit is includedEg: class A 0-9 B 10-19 C 20-29For Class B, Lower class boundary=9.5 Upper class boundary=19.5Inclusive class interval=Upper class boundary-lower class boundary [math]=19.5-9.5 =10[/math]b) Exclusive Class Interval: When the lower limit is included, but the upper limit is excludedEg: class A 0-10 B 10-20 C 20-30For Class B, Lower class limit(LCL)=10 Upper class limit(UCL)=20(not included) Exclusive class interval=UCL-LCL [math]=20-10 =10[/math]
How can you find the mean and median of a set of data using a histogram?
Use this free histogram calculator that calculates median, average, sum and other important statistical numbers like standard deviation.