Question

(1) According to a report by the U.S. Fish and Wildlife Service, the mean length of six-year-old rainbow trout in the Arolik River in Alaska is 481 millimeters with a standard deviation of 41 millimeters. Assume these lengths are normally distributed. (a) What percent of six-year-old rainbow trout are less than 450 millimeters long? (b) What percent of six-year-old rainbow trout are between 400 and 500 millimeters long? (c) What percent of six-year-old rainbow trout are greater than 492 millimeters long? (d) Find the 58th percentile of the lengths. (e) Find the first quartile of the lengths. 1

(2) A process manufactures ball bearings with diameters that are normally distributed with mean 25.1 millimeters and standard deviation 0.08 millimeter. (a) What percentage of the diameters are greater than 25.3 millimeters? (b) What percentage of the diameters are less than 24.7 millimeters? (c) To meet a certain specification, a ball bearing must have a diameter between 25.0 and 25.2 millimeters. What percentage of the ball bearings meet the specification? (d) Find the third quartile of the diameters. (e) Find the 34th percentile of the diameters.

Answer 1

1)

a)

µ =    481          
σ =    41          
              
P( X ≤    450   ) = P( (X-µ)/σ ≤ (450-481) /41)      
=P(Z ≤   -0.76   ) =   0.2248   (answer)

b)

we need to calculate probability for ,                                      
P (   400   < X <   500   )                      
=P( (400-481)/41 < (X-µ)/σ < (500-481)/41 )                                      
                                      
P (    -1.976   < Z <    0.463   )                       
= P ( Z <    0.463   ) - P ( Z <   -1.98   ) =    0.6785   -    0.0241   =    0.6544   (answer)

c)

P ( X ≥   492   ) = P( (X-µ)/σ ≥ (492-481) / 41)              
= P(Z ≥   0.27   ) = P( Z <   -0.268   ) =    0.3942   (answer)

d)

Z value at    0.58   =   0.2019   (excel formula =NORMSINV(   0.58   ) )
z=(x-µ)/σ                      
so, X=zσ+µ=   0.202   *   41   +   481  
X   =   489.28   (answer)          

e)

P(X≤x) =   0.25                  
                      
Z value at    0.25   =   -0.6745   (excel formula =NORMSINV(   0.25   ) )
z=(x-µ)/σ                      
so, X=zσ+µ=   -0.674   *   41   +   481  
X   =   453.35   (answer)          

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2)

a)

µ =    25.1                  
σ =    0.08                  
                      
P ( X ≥   25.3   ) = P( (X-µ)/σ ≥ (25.3-25.1) / 0.08)              
= P(Z ≥   2.50   ) = P( Z <   -2.500   ) =    0.0062   (answer)

b)

P( X ≤    24.7   ) = P( (X-µ)/σ ≤ (24.7-25.1) /0.08)      
=P(Z ≤   -5.00   ) =   0.0000   (answer)

c)

we need to calculate probability for ,                                      
P (   25   < X <   25.2   )                      
=P( (25-25.1)/0.08 < (X-µ)/σ < (25.2-25.1)/0.08 )                                      
                                      
P (    -1.250   < Z <    1.250   )                       
= P ( Z <    1.250   ) - P ( Z <   -1.25   ) =    0.8944   -    0.1056   =    0.7887   (answer)

d)

P(X≤x) =   0.75                  
                      
Z value at    0.75   =   0.6745   (excel formula =NORMSINV(   0.75   ) )
z=(x-µ)/σ                      
so, X=zσ+µ=   0.674   *   0.08   +   25.1  
X   =   25.15   (answer)          

e)

P(X≤x) =   0.34                  
                      
Z value at    0.34   =   -0.4125   (excel formula =NORMSINV(   0.34   ) )
z=(x-µ)/σ                      
so, X=zσ+µ=   -0.412   *   0.08   +   25.1  
X   =   25.07   (answer)