Joel is mailing a large envelope to his cousin. The envelope has pictures inside, so he doesn't want to bend it.
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Answer:
Depth:
μ =20.2025 km
M = 15.625 km
Range = 47.15 km
σ ≈ 15.92 km
Q₁ = 5.7375 km
Q₃ = 34.6675 km
Magnitude:
μ = 2.08375
M = 1.465
Range, R = 5.17
σ = 1.801485 ≈ 1.8
Q₁ = 0.5625
Q₃ = 3.925
Step-by-step explanation:
The given data are;
Depth Magnitude
0.76 0.84
4.93 0.47
8.16 0.35
33.58 1.32
21.2 1.61
35.03 4.57
10.05 5.52
47.91 1.99
For the Depth, we have;
The mean, μ = (0.76+4.93+8.16+33.58+21.2+35.03+10.05+47.91)/8 =20.2025 km
The median, M = The (n + 1)/2th term after arranging the term in increasing order as follows;
0.76, 4.93, 8.16, 10.05, 21.2, 33.58, 35.03, 47.91 , the median is therefore;
(8 + 1)/2th term or the 4.5th term which is 10.05 + (21.2 - 10.05)/2 = 15.625 km
The Range = The highest - The lowest value = 47.91 - 0.76 = 47.15 km
The Standard deviation of, σ, is given as follows;
Where;
= The individual data point = (0.76, 4.93, 8.16, 10.05, 21.2, 33.58, 35.03, 47.91
)
N = The total number of data point = 8
Substituting, (using Microsoft Excel) we get;
Q₁ = The first quartile = The (n + 1)/4th = term arranged in increasing order
Q₁ = The (8 + 1)/4th term = The 2.25th term = 4.93 + (8.16 - 4.93)×0.25) = 5.7375 km
Q₃ = The first quartile = The 3×(n + 1)/4th = term arranged in increasing order
Q₃ = The 3×(8 + 1)/4th term = The 6.75th term = 33.58 + 3×(35.03 - 33.58)×0.25) = 34.6675 km
For the magnitude, we have, using the same formulas and procedures as above;
μ = 2.08375
M = 1.465
Range, R = 5.17
σ = 1.801485 ≈ 1.8
Q₁ = 0.5625
Q₃ = 3.925