Question

Please provide all the calculation steps and answers should either be exact or given to 4 decimal places.

The ED mointor SO2 emission of XYZ powerplant every hour. Assume the concentration of SO2 from XYZ powerplant is normal distributed. The powerplant claims that the average SO2 emission is 75 ppb (parts per billion) and the standard deviation is 0.75 ppb. The ED set the standard of SO2 concentration for all powerplants is 77 ppb.

(a) Find the probability that the SO2 concentration of a randomly selected hour between 74.34 ppb to 74.82 ppb.

(b) Calculate the SO2 concentration interval for the middle 95% of the emission and explain the result.

(c) Find the probability that the powerplant will exhaust more than the standard.

(d) There are randomly selected 11 powerplants in this country with mean of 75.2 ppb and standard deviation of 0.4 ppb.

(di) Construct a 90% confidence interval for the population mean and interpret the result.

(dii) Determine the assumption in d(i).

Answer 1

a)

µ =    75                                  
σ =    0.75                                  
we need to calculate probability for ,                                      
P (   74.34   < X <   74.82   )                      
=P( (74.34-75)/0.75 < (X-µ)/σ < (74.82-75)/0.75 )                                      
                                      
P (    -0.880   < Z <    -0.240   )                       
= P ( Z <    -0.240   ) - P ( Z <   -0.880   ) =    0.4052   -    0.1894   =    0.2157   (answer)

b)

µ =    75                          
σ =    0.75                          
proportion=   0.9500                          
proportion left    0.0500   is equally distributed both left and right side of normal curve                       
z value at   0.025   = ±   1.960   (excel formula =NORMSINV(   0.05   / 2 ) )      
                              
z = ( x - µ ) / σ                              
so, X = z σ + µ =                              
X1 =   -1.960   *   0.75   +   75   =   73.530  
X2 =   1.960   *   0.75   +   75   =   76.470  
SO2 concentration interval for the middle 95% of the emission is (73.53 , 76.47)

c)

P ( X > 77   ) = P( (X-µ)/σ ≥ (77-75) / 0.75)              
= P(Z > 2.667   ) = P( Z <   -2.667   ) =    0.0038   (answer)

d)

di)

population std dev ,    σ =    0.7500
Sample Size ,   n =    11
Sample Mean,    x̅ =   75.2000

Level of Significance ,    α =    0.1          
'   '   '          
z value=   z α/2=   1.645   [Excel formula =NORMSINV(α/2) ]      
                  
Standard Error , SE = σ/√n =   0.7500   / √   11   =   0.2261
margin of error, E=Z*SE =   1.6449   *   0.2261   =   0.3720
                  
confidence interval is                   
Interval Lower Limit = x̅ - E =    75.20   -   0.371957   =   74.8280
Interval Upper Limit = x̅ + E =    75.20   -   0.371957   =   75.5720
90%   confidence interval is (   74.83   < µ <   75.57   )

dii)assumption: population from which sample is taken is normally distributed

please revert for doubts...