IQ scores on the wais test approximate a normal curve with a means of 100 and a

Question

IQ scores on the wais test approximate a normal curve with a means of 100 and a standard deviation of 15 what IQ score is identified with. The upper 2% that is 2% to the right and 98% to the left? the lower 10% ? the upper 60%? the middle 95% remember the middle 95% straddles the line perpendicular to the means or the 50 percentile with of 95% or 4 7.5% above this line in the remaining 47.5% below this line the middle 99%?

G between 1300 and 1400 hours Finding scores 5.13 scores on the WAS lest approximate a normal curve with a mean 5 What IQ score is identtied wth 00 and a standard deviation at al the upper 2 percent, tatis.2 percent to he right and Eperoattottelety? (b) the lower 10 percent? ch the upper 60 paroent? (d) the middle g5 pertent? IRemember, the midle gs peroent the perpendicalar to the man lot 50th percem MIN nar or 475 percem abort this line and the remaning 475 percent below this line a) the middle percanr? 5.14 For the normal of taming times of igtt bulbs a man eaual 1200 hours and deviaton tqual 120 hours What burning with the (ak upper 50 percent?

Explanation / Answer

Mean ( u ) =100
Standard Deviation ( sd )=15
Normal Distribution = Z= X- u / sd ~ N(0,1)                  
a.
P ( Z > x ) = 0.02
Value of z to the cumulative probability of 0.02 from normal table is 2.0537
P( x-u/ (s.d) > x - 100/15) = 0.02
That is, ( x - 100/15) = 2.0537
--> x = 2.0537 * 15+100 = 130.8062                  
b.
P ( Z < x ) = 0.1
Value of z to the cumulative probability of 0.1 from normal table is -1.282
P( x-u/s.d < x - 100/15 ) = 0.1
That is, ( x - 100/15 ) = -1.28
--> x = -1.28 * 15 + 100 = 80.7767      
c.
P ( Z > x ) = 0.6
Value of z to the cumulative probability of 0.6 from normal table is -0.2533
P( x-u/ (s.d) > x - 100/15) = 0.6
That is, ( x - 100/15) = -0.2533
--> x = -0.2533 * 15+100 = 96.1998                  
d.
P ( Z < x ) = 0.25
Value of z to the cumulative probability of 0.25 from normal table is -0.674
P( x-u/s.d < x - 100/15 ) = 0.25
That is, ( x - 100/15 ) = -0.67
--> x = -0.67 * 15 + 100 = 89.8827                  
P ( Z > x ) = 0.25
Value of z to the cumulative probability of 0.25 from normal table is 0.6745
P( x-u/ (s.d) > x - 100/15) = 0.25
That is, ( x - 100/15) = 0.6745
--> x = 0.6745 * 15+100 = 110.1173                  

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