1)
a)
µ = 400
σ = 50
n= 100
P(µ-9<x<µ+9) = P(391 ≤ X ≤
409)
Z1 = (X1 - µ )/(σ/√n) = ( -9 ) / (
50 / √ 100 ) =
-1.80
Z2 = (X2 - µ )/(σ/√n) = ( 9 ) / (
50 / √ 100 ) =
1.80
P ( -1.80 < Z <
1.80 ) = P ( Z < 1.80 ) - P
( Z < -1.80 ) =
0.964 - 0.036 =
0.9281 (answer)
b)
we need to calculate probability for ,
P(µ-13<x<µ+13) = P(387 ≤ X ≤ 413)
X1 = 387 , X2 =
413
Z1 = (X1 - µ )/(σ/√n) = ( 387
- 400 ) / ( 50 /
√ 100 ) =
-2.60
Z2 = (X2 - µ )/(σ/√n) = ( 413
- 400 ) / ( 50 /
√ 100 ) =
2.60
P ( 387 < X <
413 ) = P ( -2.60
< Z < 2.60 )
= P ( Z < 2.60 ) - P ( Z
< -2.60 ) =
0.995 - 0.005 =
0.9907 (answer)
==================
2)
a)
population
proportion ,p= 0.3
n= 300
std error
, SE = √( p(1-p)/n ) = 0.0265
we need to
compute probability for
0.27 < p̂ < 0.33
Z1 =( p̂1
- p )/SE= ( 0.27 -
0.3 ) / 0.0265 =
-1.134
Z2 =( p̂2
- p )/SE= ( 0.33 -
0.3 ) / 0.0265 =
1.134
P( 0.27 < p̂ <
0.33 ) = P[( p̂1-p )/SE< Z
<(p̂2-p)/SE ] =P( -1.134
< Z < 1.134 )
= P ( Z < 1.134 ) - P (
-1.134 ) = 0.8716
- 0.1284 =
0.7432
b)
population
proportion ,p= 0.3
n= 300
std error
, SE = √( p(1-p)/n ) = 0.0265
we need to
compute probability for
0.25 < p̂ < 0.35
Z1 =( p̂1
- p )/SE= ( 0.25 -
0.3 ) / 0.0265 =
-1.890
Z2 =( p̂2
- p )/SE= ( 0.35 -
0.3 ) / 0.0265 =
1.890
P( 0.25 < p̂ <
0.35 ) = P[( p̂1-p )/SE< Z
<(p̂2-p)/SE ] =P( -1.890
< Z < 1.890 )
= P ( Z < 1.890 ) - P (
-1.890 ) = 0.9706
- 0.0294 =
0.9412
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