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Normal Distributions Homework: Problem 20 Previous Problem List Next (1 point) Scores on a certain intelligence test for chil
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Answer #1

Solution:

Given: x = scores on certain intelligence test for children ages 13 and 15 years are approximately Normally distributed.

Mean =u= 101

Standard Deviation = 0 = 21

Part a) Find:
P( X > 83 ) =..........?

Find z score for x = 83

エール 2 %3D

3-101 -18 -= -= -0.86 21 91

Thus we get:

P( X > 83 ) = P( Z> -0.86)

P( X > 83 ) = 1 - P( Z < -0.86)

Look in z table for z = -0.8 and 0.06 and find area.

.09 | 3.4 -3.3 -3.2 -3.1 -3.0 -2.9 -2.8 -2.7 -2.6 -2.5 -2.4 -2.3 -2.2 2.1 -2.0 -19 -1.8 -1.7 -1.6 -1.5 -1.4 -1.3 -1.2 -1.1 -1

P( Z < -0.86) = 0.1949

Thus

P( X > 83 ) = 1 - P( Z < -0.86)

P( X > 83 ) = 1 - 0.1949

P( X > 83 ) = 0.8051

Part b)

Find x value such that:
P( X < x ) = 20%

P( X < x ) = 0.2000

thus find z value such that:

P( Z < z) = 0.2000

Look in z table for Area = 0.2000 or its closest area and find z value.

2 .00 1-3.4 .0003 -3.3 .0005 -3.2 .0007 -3.1 .0010 1-3.0 .0013 -2.9 .0019 -2.8 .0026 -2.7 .0035 -2.6 .0047 -2.5 .0062 1-2.4 .

Area 0.2005 is closest to 0.2000 and it corresponds to -0.8 and 0.04

thus z= -0.84

Now use following formula to find x value:
= μ +Χσ

r = 101 + (-0.84) X 21

r = 101 – 17.64

x = 83.36

Part c)

Find x value such that:
P( X > x ) =10%

P( X > x ) = 0.1000

thus find z value such that:

P( Z > z) = 0.1000

that is find z value such that:

P( Z < z ) = 1 - P( Z > z )

P( Z < z ) = 1 - 0.1000

P( Z < z ) = 0.9000

Look in z table for area = 0.9000 or its closest area and find z value:

.00 .01 07 | .08 .09 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0 1.1 1.2 1.3 15000 .5040 .5398 .5438 .5793 .5832 .6179 .6217 .655

Area 0.8997 is closest to 0.9000 and it corresponds to 1.2 and 0.08

thus z = 1.28

Thus

= μ +Χσ

r = 101 + 1.28 x 21

x= 101 + 26.88

x= 127.88

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