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Sedbober [7]
4 years ago
11

Can someone help please ❤️❤️

Mathematics
1 answer:
loris [4]4 years ago
8 0

(A) The mean is 82

(B) The mean absolute deviation is approximately 9.71

(D) The IQR is 11

<u>Explanation:</u>

<u />

(A) 89, 87, 54, 78, 87, 99, 80

Mean of the data = \frac{sum of the data}{total number of data}

Mean = \frac{89 + 87+54+78+87+99+80}{7} \\\\Mean = \frac{574}{7} \\\\Mean = 82

Therefore, mean of the data is 82

(B) The mean absolute deviation, MAD is

data - 54, 78, 80, 87, 87, 89, 99

Mean of the data, x' is 82

|x - x'| : 54 - 82 = 28

           78 - 82 = 4

           80 - 82 = 2

           87 - 82 = 5

           87 - 82 = 5

           89 - 82 = 7

           99 - 82 = 17

Total of |x - x'| = 68

MAD = |x - x'| / n

MAD = 68/7

MAD = 9.71

(D) IQR is the interquartile range

Data = 54, 78, 80, 87, 87, 89, 99

IQR = median of lower quartile range - median of upper quartile range

IQR = 89 - 78

IQR = 11

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