The Chapin Social Insight Test evaluates how accurately the subject appraises ot

Question

The Chapin Social Insight Test evaluates how accurately the subject appraises other people. In the reference population used to develop the test, scores are approximately normally distributed with mean 25 and population standard deviation five. The range of possible scores is between 0 to 41. What is the probability of a score of 33 or more (Please write your answer as a decimal. The testing syste m is not able to recognize a % sign. So if you have 1.23% then this should be written as .0123 which is the decimal equivalent)

Explanation / Answer

NORMAL DISTRIBUTION
the PDF of normal distribution is = 1/ * 2 * e ^ -(x-u)^2/ 2^2
standard normal distribution is a normal distribution with a,
mean of 0,
standard deviation of 1
equation of the normal curve is ( Z )= x - u / sd ~ N(0,1)
mean ( u ) = 25
standard Deviation ( sd )= 5

probability of score of 33 or more
P(X < 33) = (33-25)/5
= 8/5= 1.6
= P ( Z <1.6) From Standard Normal Table
= 0.9452
P(X > = 33) = (1 - P(X < 33)
= 1 - 0.9452 = 0.0548

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