NORMAL DISTRIBUTION<br />Given the standard normal deviation, find the area under
 To the left of z=1.39
Answer:P( x<-1.39) = 0.0823
To the right of z=1.96Answer:z=1.960.9750z=1-0.9750 = 0.0250 thereforeP(x<1.96)<br /> Between z=-0.48 and z=1.74Answer: P(-0.48<x<1.74)z1=-0.480.3156z2 =0.9591z2-z1=0.9591-0.3156=0.6435<br /> Given the normal distribution with u=200 and σ=10, Find the area of a curve<br />below 214
Z=1.4  0.9192Therefore P(x<214)= 0.9192z=x-μσ
z=214-20010
z=1.4
Z=-2.1 0.0179Z=1-0.0179= 0.9821Therefore P(x>179)= 0.9821area above 179 and z=x-μσ<br />z=179-20010<br />z=-2.1<br />(c) area between188 and 206Z1=--1.2 0.1151Z2=0.60.7257Therefore P(188<x<206)= 0.7257-0.1151=0.6106z1=x-μσ<br />z1=188-20010<br />z1=-1.2<br />z2=x-μσ<br />z2=206-20010<br />z2=0.6<br />the x value that has 80% of the area below it Z1=--1.2 0.1151Z2=0.60.7257Therefore P(188<x<206)= 0.7257-0.1151=0.6106z=x-μσ; area 0.80.84(x)<br />z=x-μσ

Inferential statistics

  • 1.
    NORMAL DISTRIBUTION<br />Giventhe standard normal deviation, find the area under
  • 2.
    To theleft of z=1.39
  • 3.
  • 4.
    To the rightof z=1.96Answer:z=1.960.9750z=1-0.9750 = 0.0250 thereforeP(x<1.96)<br /> Between z=-0.48 and z=1.74Answer: P(-0.48<x<1.74)z1=-0.480.3156z2 =0.9591z2-z1=0.9591-0.3156=0.6435<br /> Given the normal distribution with u=200 and σ=10, Find the area of a curve<br />below 214
  • 5.
    Z=1.4 0.9192ThereforeP(x<214)= 0.9192z=x-μσ
  • 6.
  • 7.
  • 8.
    Z=-2.1 0.0179Z=1-0.0179= 0.9821ThereforeP(x>179)= 0.9821area above 179 and z=x-μσ<br />z=179-20010<br />z=-2.1<br />(c) area between188 and 206Z1=--1.2 0.1151Z2=0.60.7257Therefore P(188<x<206)= 0.7257-0.1151=0.6106z1=x-μσ<br />z1=188-20010<br />z1=-1.2<br />z2=x-μσ<br />z2=206-20010<br />z2=0.6<br />the x value that has 80% of the area below it Z1=--1.2 0.1151Z2=0.60.7257Therefore P(188<x<206)= 0.7257-0.1151=0.6106z=x-μσ; area 0.80.84(x)<br />z=x-μσ