Question

Increasing joint extension is one goal of post-surgical rehabilitation therapy. In a study to evaluate the effect of therapy which uses ultrasound and stretching, passive knee extension was measured after each treatment. Passive knee extension data is given (in degrees) for a sample of 10 therapy participants.

63 41 62 55 66 57 39 55 68 52

Compute the following:

a) Mean

b) Median

c) Range

d) Standard Deviation

e) What would be the z-score for a patient that experienced a knee extension of 62 degrees?

f) Should a therapist be concerned about a patient who experienced a knee extension of less than 35 degrees? Why?

Answer 1

39 41 52 55 55 57 62 63 66 68

a) Mean = (39 + 41 + 52 + 55 + 55 + 57 + 62 + 63 + 66 + 68)/10 = 55.8

b) Median = (55 + 57)/2 = 56

c) Range = 68 - 39 = 29

d) Standard deviation (s) = sqrt(((39 - 55.8)^2 + (41 - 55.8)^2 + (52 - 55.8)^2 + (55 - 55.8)^2 + (55 - 55.8)^2 + (57 - 55.8)^2 + (62 - 55.8)^2 + (63 - 55.8)^2 + (66 - 55.8)^2 + (68 - 55.8)^2)/9) = 9.784

e) Z - score = (X - )/s

= (62 - 55.8)/9.784

= 0.63

f) Z - score = (X - )/s

= (35 - 55.8)/9.784

= -2.13

Since the z-score is less than -2, so it will be an unusual value.

So the therapist should be concerned about a patient who experienced a knee extension of less than 35 degrees.