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Question

A distribution of values is normal with a mean of 36.8 and a standard deviation of 61.6.

Find the probability that a randomly selected value is greater than 203.1.

P(X > 203.1) = 

Enter your answer as a number accurate to 4 decimal places. Answers obtained using exact z-scores or z-scores rounded to 3 decimal places are accepted.

Answer 1

First, we standardize the distribution by determining the z-score:

z = (x - μ)/σ = (203.1-36.8)/61.6 ≈ 2.7

Now, we can determine P(Z>2.7) with a z-table. According to the table, P(Z>2.7) corresponds to 1-0.9965 = 0.0035 = 0.35% of the distribution.

Thus, P(X>203.1) ≈ 0.0035

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