A lower quartile for a set of four observations is not a meaningful concept.
To find the mean of the numbers 45, 50, 47, 52, 53, 45, and 51, first add them together: 45 + 50 + 47 + 52 + 53 + 45 + 51 = 393. Then, divide the total by the number of values (7): 393 ÷ 7 = 56.14. Therefore, the mean is approximately 56.14.
The median is 50, the middle number when they are listed in order.
It is 58. the lower quartile is 50 while the median is 52.Email me at ericmal15@yahoo.com if you need more info and title it Answers.com stuff so i will read it.
To solve for the quartile deviation, first calculate the first quartile (Q1) and the third quartile (Q3) of your data set. The quartile deviation is then found using the formula: ( \text{Quartile Deviation} = \frac{Q3 - Q1}{2} ). This value represents the spread of the middle 50% of your data, providing a measure of variability.
45, 46, 47, 48, 50, 50, 53, 54, 55
The answer is 47 / 47.5. I calculated it by adding 45 and 50, then divided it by 2.
The median is 50, the middle number when they are listed in order.
43 45 47 49
yes, if all the data is the same number; when the range is zero. * * * * * That is not true. You need 25% of the values to be small, then 50% identical values, followed by 25% large values. Then the lower (first) quartile will be the same as the upper (third) quartile. The inter-quartile range (IQR) will be zero but the overall range can be as large as you like.
50%
It is 58. the lower quartile is 50 while the median is 52.Email me at ericmal15@yahoo.com if you need more info and title it Answers.com stuff so i will read it.
50%. The second quartile is the median.
The answer is 47+46+45+...+1 To work that out quickly, we do (47*48)/2, which is 1128.
The numbers 41, 43 and 47 are prime.
45. Utah 46. Oklahoma 47. New Mexico 48. Arizona 49. Alaska 50. Hawaii
50
45, 46, 47, 48, 50, 50, 53, 54, 55