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Q: What is the median of 73 75 78 82 93?

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The median of these fourteen numbers is 84.

The median is (89 + 91)/2 = 90

The standard deviation of the scores 100 92 94 95 81 82 87 89 71 71 73 61 62 68 51 55 is: 15.1921

The first step in constructing a box-and-whisker plot is to first find the median, thelower quartile and the upper quartile of a given set of data. Example: The following set of numbers are the amount of marbles fifteen different boys own (they are arranged from least to greatest).18 27 34 52 54 59 61 68 78 82 85 87 91 93 100First find the median. The median is the value exactly in the middle of an ordered set of numbers. *68 is the medianNext, we consider only the values to the left of the median: 18 27 34 52 54 59 61. We now find the median of this set of numbers. Remember, the median is the value exactly in the middle oa an ordered set of numbers. Thus 52 is the median of the scores less than the median of all scores, and therefore is the lower quartile.52 is the lower quartileNow consider only the values to the right of the median: 78 82 85 87 91 93 100. We now find the median of this set of numbers. The median 87 is therefore called the upper quartile.87 is the upper quartile(*If you're finding the median in an ordered set with an even number of values, you must take the average of the two middle numbers. For example: 3, 5, 7, and 10. Add the two middle numbers. 5 + 7 = 12. Divided 12 by 2 to get the average. 12 / 2 = 6. Therefore 6 is the median for the ordered set of 3, 5, 7, and 10.)You are now ready to find the interquartile range (IQR).The interquartile range is the difference between the upper quartile and the lower quartile. In our case the IQR = 87 - 52 = 35. The IQR is a very useful measurement. It is useful because it is less influenced by extreme values, it limits the range to the middle 50% of the values.35 is the interquartile rangehere is a sucessful site called Jen's Box and Whisker plot, and here is a example of a whisker plot from that very own site. because itso hard to describe! i hope you find this helpful good luck!!

A 75 B 2 C 44 D 27 E 8 F 91 G 18 H 46 I 38 J 62? First put the numbers/terms in rank order. That is lowest to highest, or vice versa. 2,8,18,27,38,44,46,62,75,91. The MEDIAN is the absolute middle middle number. Since there are an even number of terms, there is no absolute middle term. So we take the two nearest to middle terms, they are 38 & 44. (NB This leaves four terms to the left of 38, and four terms to the right of 44. We then add the two terms and divide by '2'. Hence ( 38 + 44) / 2 = 82/2 = 41 '41' is the median.

Related questions

Since the set of data is arranged in numerical order, first we need to find the median (also called the second quartile), which separates the data into two equal groups, in our case there are 6 numbers in each group.54 65 66 68 73 75 | 75 78 82 82 87 97The first quartile (also called the lower quartile) is the middle value of numbers that are below the median, in our case is 67.54 65 66 | 68 73 75 | 75 78 82 82 87 97The third quartile (also called the upper quartile) is the middle value of numbers that are above the median, in our case is 82.54 65 66 | 68 73 75 | 75 78 82 | 82 87 97The interquartile range is the difference between the first and third quartiles, which is 15, (82 - 67).

83

Median is average so 80.

The median is 86.

If final is 25% of grade and your average was 82 before final that part is 75%. So we have (.75(82) + .25 (73)) = 80

4+78=82-7=75

there is no mode in this set or data of numbersmode means the number that occurs or shows most often and none of these numbers repeat

66

It is 82

Life expectancy for a male born in 1941 was 65 and for females 70.

Listing the 47 numbers in increasing value, the 23rd, 24th, 25th, and 26th numbers are 73. There are 22 scores higher and 22 scores lower, making 73 the median.

One possible set is 62, 62, 75, 76 and 82