Question

Assume the change in serum vitamin E levels for a study group of 16 subjects follows a normal distribution with mean 0.80 mg/dl and standard deviation 0.48. What percentage of the study group would be expected to show a change of not more than 1.741 mg/dl? Identify the threshold of change that will assure 20% of the study group will have a change less than this threshold.

Answer 1

Let X be the random variable denoting the change in serum Vitamin E levels (in mg/dl).

X ~ N(0.80, 0.48) i.e. Z = (X - 0.80)/0.48 ~ N(0,1).

The percentage of study group to show a change of not more than 1.741 mg/dl = P(X <= 1.741) = P[(X - 0.80)/0.48 <= (1.741 - 0.80)/0.48] = P[(X - 0.80)/0.48 <= 1.9604] = (1.9604) = 0.975. (Ans).

Let, P(Z < a) = 0.2 i.e. (a) = 0.2 i.e. a = (0.2) = - 0.842.

i.e. P[(X - 0.80)/0.48 < -0.842] = 0.2 i.e. P(X < 0.39584) = 0.2.

Hence, the required threshold is 0.39584 mg/dl. (Ans).