Midtermexam Fall2017 pdf Word AAndrews, Renjamin File Home Insert Desiqn Layout

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Midtermexam Fall2017 pdf Word AAndrews, Renjamin File Home Insert Desiqn Layout References Mailings Review wew Tell me what you want to do Share Title TITLE Tite Title Effects Themes Colis Fonts Watermark Pg Pae Set as Default Culur Durders Psqr Backgro nd (8) Assume that adults have IQ scores that are normally distributed with a mean of 100 and a standard deviation of 15 (as on the Wechsler test). Let Xbe the 1Q scores for adults. So XN 100,15* (a) Find the probability that a randomly selected adult has an 1Q greater than 131.5 (the requirement for membership in the Mensa organization). Show the area under the normal curve (b) rind the probability that a randomly selected adult has an IQ betweern 90 and 110 (referred to as the normal range). Show the area under the normal curve (c) what is the IQ score separating the bottom 20% from the top 80%. the normal curve to explain your answer. Type here to search 1019/2017 2

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 ) = 100
standard Deviation ( sd )= 15
a.
P(X > 131.5) = (131.5-100)/15
= 31.5/15 = 2.1
= P ( Z >2.1) From Standard Normal Table
= 0.0179
b.
To find P(a < = Z < = b) = F(b) - F(a)
P(X < 90) = (90-100)/15
= -10/15 = -0.6667
= P ( Z <-0.6667) From Standard Normal Table
= 0.2525
P(X < 110) = (110-100)/15
= 10/15 = 0.6667
= P ( Z <0.6667) From Standard Normal Table
= 0.7475
P(90 < X < 110) = 0.7475-0.2525 = 0.495
c.
P ( Z < x ) = 0.2
Value of z to the cumulative probability of 0.2 from normal table is -0.8416
P( x-u/s.d < x - 100/15 ) = 0.2
That is, ( x - 100/15 ) = -0.8416
--> x = -0.8416 * 15 + 100 = 87.3757

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