Answer:
(a) Probability that Y falls into the dangerous region is 0.0013.
(b) Probability that the mean Y-bar falls into the dangerous region is 0.00001. 
Step-by-step explanation:
We are given that a certain capsule is manufactured so that the dosage of the active ingredient follows the distribution Y ~ N(μ = 10 mg, σ = 1 mg). 
A dosage of 13 mg is considered dangerous.
Let Y = <u><em>dosage of the active ingredient </em></u>
The z-score probability distribution for normal distribution is given by;
                                  Z  =   ~ N(0,1)
  ~ N(0,1) 
where,  = population mean = 10 mg
 = population mean = 10 mg
              = standard deviation = 1 mg
 = standard deviation = 1 mg
(a) Probability that Y falls into the dangerous region is given by = P(Y  13 mg)
 13 mg)
        P(Y  13 mg) = P(
 13 mg) = P(  
  
  ) = P(Z
 ) = P(Z  3) = 1 - P(Z < 3)
 3) = 1 - P(Z < 3)   
                                                           = 1 - 0.9987 = <u>0.0013</u>
The above probability is calculated by looking at the value of x = 3 in the z table which has an area of 0.9987.
(b) We are given that a dosage of 13 mg is considered dangerous. And we sample 49 capsules at random.
Let  = sample mean dosage
 = sample mean dosage
The z-score probability distribution for sample mean is given by;
                                  Z  =   ~ N(0,1)
  ~ N(0,1) 
where,  = population mean = 10 mg
 = population mean = 10 mg
              = standard deviation = 1 mg
 = standard deviation = 1 mg 
             n = sample of capsules = 49 
So, Probability that the mean Y-bar falls into the dangerous region is given by = P( 
  13 mg)
 13 mg) 
           P(Y  13 mg) = P(
 13 mg) = P(  
  
  ) = P(Z
 ) = P(Z  21) = 1 - P(Z < 21)
 21) = 1 - P(Z < 21)   
                                                              = <u>0.00001</u>