Lengths (in mm) of 45 leaves collected for a Shimla herbarium project: 100-107: 5, 107-114: 9, 114-121: 14, 121-128: 11, 128-135: 6. The class width is the awkward value 7. Using the step-deviation method with a = 117.5, find the mean length.

Question

Accepted Answer

118.12. Class marks 103.5, 110.5, 117.5, 124.5, 131.5 give u_i = -2, -1, 0, 1, 2 . f_i u_i = -10 - 9 + 0 + 11 + 12 = 4 , f_i = 45 . Mean = 117.5 + 7 (4/45 ) = 117.5 + 0.62 = 118.12 mm. An awkward h costs nothing here — the u_i are still small integers. "117.59" leaves out the factor h = 7; "116.88" reverses the sign; "123.1" divides by 5 classes instead of by 45.

Lengths (in mm) of 45 leaves collected for a Shimla herbarium project:
100-107: 5, 107-114: 9, 114-121: 14, 121-128: 11, 128-135: 6.
The class width is the awkward value 7. Using the step-deviation method with a = 117.5, find the mean length.

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More Class 10 Mean of Grouped Data practice

Lengths (in mm) of 45 leaves collected for a Shimla herbarium project:100-107: 5, 107-114: 9, 114-121: 14, 121-128: 11, 128-135: 6.The class width is the awkward value 7. Using the step-deviation method with a = 117.5, find the mean length.

Step-by-Step Solution

Master this chapter in 30 minutes a day.

More Class 10 Mean of Grouped Data practice

Lengths (in mm) of 45 leaves collected for a Shimla herbarium project:
100-107: 5, 107-114: 9, 114-121: 14, 121-128: 11, 128-135: 6.
The class width is the awkward value 7. Using the step-deviation method with a = 117.5, find the mean length.