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No. Not if 25% of the data set are all at the maximum value.

Q: Must the third quartile of a data set be less than the maximum value?

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No, interquartile range cannot be for any data. The lower quartile for data must be used below the lower quartile.

The median must be at least as great as the first quartile in any data set. Normally it would be greater.The median must be at least as great as the first quartile in any data set. Normally it would be greater.The median must be at least as great as the first quartile in any data set. Normally it would be greater.The median must be at least as great as the first quartile in any data set. Normally it would be greater.

A box and whisker plot is used to show range, you must first find out the quartiles. The first quartile is the left edge of the box, the third quartile is the right edge of the box and the median is the line in the middle. The whiskers are the highest and lowest values in your data set. Sometimes if you have a value in your data that is a long way out then you may not use it as a whisker, this is an outlier.A boxplot is a way of depicting groups of numerical data. They have many lines extending vertically from the whiskers (boxes).

To find the limits of outliers in box and whisker plots, you first must determine the Interquartile Range. The Interquartile Range is the difference between the Upper Quartile and the Lower Quartile. For instance, if my Upper Quartile = 87 and my Lower Quartile is 52, then 87 - 52= 35. 35 is the Interquartile Range (IQR).Next, you use the formula 1.5 x IQR to determine if you have any outliers.Example:1.5 x 35 = 52.5Now determine the limit for the Upper Quartile by adding 52.5 to the Upper Quartile.Example:52.5 + 87 = 139.5139.5 is the limit for the Upper Quartile.Next, determine the limit for the Lower Quartile by subtracting the Lower Quartile from 52.5Example52 - 52.5 = -0.5-0.5 is the limit for the Lower QuartileThus, the LIMITS are -0.5 and 139.5. In order for a number to be considered an outlier, it must either be less than -0.5 or greater than 139.5

If you want to get the median of numbers within the first quartile, you have to use the QUARTILE, MEDIAN and IF functions together and then enter it as an array formula. Say your values are in the cells from E2 to E27. =MEDIAN( IF( E2:E27<=QUARTILE( E2:E27,1 ), E2:E27 ) ) To enter it as an array formula, you must press Ctrl- Shift - Enter together. The formula will then appear with curly braces around it, indicating that it is an array function.

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No. The upper quartile, by definition, must be at least as large as the lower quartile.

No, interquartile range cannot be for any data. The lower quartile for data must be used below the lower quartile.

The median must be at least as great as the first quartile in any data set. Normally it would be greater.The median must be at least as great as the first quartile in any data set. Normally it would be greater.The median must be at least as great as the first quartile in any data set. Normally it would be greater.The median must be at least as great as the first quartile in any data set. Normally it would be greater.

A quartile is a given section in a range of data. To find the quartile, you must first find the median. Then find the "median of the median", using these to separate your data into sections, giving you a total of four sections of data.

A box and whisker plot is used to show range, you must first find out the quartiles. The first quartile is the left edge of the box, the third quartile is the right edge of the box and the median is the line in the middle. The whiskers are the highest and lowest values in your data set. Sometimes if you have a value in your data that is a long way out then you may not use it as a whisker, this is an outlier.A boxplot is a way of depicting groups of numerical data. They have many lines extending vertically from the whiskers (boxes).

You must raise the value in mm to the third power.

No. The range is the maximum value less the minimum value. If all the values are different, the maximum must be different from the minimum. Consequently, the max minus min must be greater than zero.

I personally hated these. So say you have a set of numbers. 78, 80, 81, 83 ,and 85. You need to find the lower quartile, upper quartile, and of course median. the highest and lowest points will be on the end. draw a box around the upper, and lower quartile and lines coming out from it connecting to 78 and 85.

To find the limits of outliers in box and whisker plots, you first must determine the Interquartile Range. The Interquartile Range is the difference between the Upper Quartile and the Lower Quartile. For instance, if my Upper Quartile = 87 and my Lower Quartile is 52, then 87 - 52= 35. 35 is the Interquartile Range (IQR).Next, you use the formula 1.5 x IQR to determine if you have any outliers.Example:1.5 x 35 = 52.5Now determine the limit for the Upper Quartile by adding 52.5 to the Upper Quartile.Example:52.5 + 87 = 139.5139.5 is the limit for the Upper Quartile.Next, determine the limit for the Lower Quartile by subtracting the Lower Quartile from 52.5Example52 - 52.5 = -0.5-0.5 is the limit for the Lower QuartileThus, the LIMITS are -0.5 and 139.5. In order for a number to be considered an outlier, it must either be less than -0.5 or greater than 139.5

If you want to get the median of numbers within the first quartile, you have to use the QUARTILE, MEDIAN and IF functions together and then enter it as an array formula. Say your values are in the cells from E2 to E27. =MEDIAN( IF( E2:E27<=QUARTILE( E2:E27,1 ), E2:E27 ) ) To enter it as an array formula, you must press Ctrl- Shift - Enter together. The formula will then appear with curly braces around it, indicating that it is an array function.

Suppose the revenue equation is of the form R = ax2 + bx + c where a, b and c are constants and x is the variable. To have a maximum, either a must be negative or x must lie within fixed limits. If a is negative then the maximum revenue is attained when x = -b/(2a). That is, find the value of R when x = -b/(2a). If a is positive, then find the value of R when x is at each end point of its domain. One of them will be larger and that is the maximum value of the revenue.

By definition, a quarter of the observations are at most as large as the lower quartile. Therefore it is possible to have an observation, X, which is smaller than the lower quartile, L. That is X <= L Again, by definition, a quarter of the observations are at least as large as the lower quartile. Therefore it is possible to have an observation, Y, which is larger than the upper quartile, U. That is U <= Y So X <= L <= U <= Y Therefore U - L <= Y - X That is, the IQR must be less than or equal to the range.