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A normal population has mean = 9 and standard deviation -5. (a) What proportion of the population is less than 19? (b) What i
A normal population has mean = 60 and standard deviation o=18. (a) What proportion of the population is greater than 987 (b)
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Answer #1

1)

a)

µ =    9          
σ =    5          
              
P( X ≤    19   ) = P( (X-µ)/σ ≤ (19-9) /5)      
=P(Z ≤   2.000   ) =   0.9772   (answer)

b)

µ =    9                  
σ =    5                  
                      
P ( X > 4   ) = P( (X-µ)/σ ≥ (4-9) / 5)              
= P(Z ≥   -1.000   ) = P( Z <   1.000   ) =    0.8413   (answer)

========================

2)

a)

µ =    60                  
σ =    18                  
                      
P ( X > 98   ) = P( (X-µ)/σ ≥ (98-60) / 18)              
= P(Z ≥   2.111   ) = P( Z <   -2.111   ) =    0.0174   (answer)

b)

P( X < 85   ) = P( (X-µ)/σ ≤ (85-60) /18)      
=P(Z ≤   1.389   ) =   0.9176   (answer)
=====================

3.1)

µ =    11.3                  
σ =    2.6                  
                      
P ( X ≥   13   ) = P( (X-µ)/σ ≥ (13-11.3) / 2.6)              
= P(Z ≥   0.654   ) = P( Z <   -0.654   ) =    0.2566   (answer)

3.2)

P( X ≤    13   ) = P( (X-µ)/σ ≤ (13-11.3) /2.6)      
=P(Z ≤   0.654   ) =   0.7434   (answer)

3.3)

we need to calculate probability for ,                                      
P (   11   < X <   12   )                      
=P( (11-11.3)/2.6 < (X-µ)/σ < (12-11.3)/2.6 )                                      
                                      
P (    -0.115   < Z <    0.269   )                       
= P ( Z <    0.269   ) - P ( Z <   -0.115   ) =    0.6061   -    0.4541   =    0.1521   (answer)

3.4)

P ( X ≥   16   ) = P( (X-µ)/σ ≥ (16-11.3) / 2.6)              
= P(Z ≥   1.808   ) = P( Z <   -1.808   ) =    0.0353   (answer)

answer: Yes

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