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).
78
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
83
One possible set is 62, 62, 75, 76 and 82
78
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).
78 appears more than any other number, so it is the mode.
78
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
83
One possible set is 62, 62, 75, 76 and 82
Life expectancy for a male born in 1941 was 65 and for females 70.
1, 73 1, 2, 41, 82
Mean = Sum/Count = 461/6 = 76.833... recurring.
69 + 78 + 82 = 229