Question

Consider babies born in the "normal" range of 37–43 weeks gestational age. A paper suggests that a normal distribution with mean μ = 3500 grams and standard deviation σ = 619 grams is a reasonable model for the probability distribution of the continuous numerical variable x = birth weight of a randomly selected full-term baby.

a. What is the probability that the birth weight of a randomly selected full-term baby is either less than 2000 g or greater than 5000 g? (Round your answer to four decimal places.)

b How would you characterize the most extreme 0.1% of all full-term baby birth weights? (Round your answers to the nearest whole number.) The most extreme 0.1% of birth weights consist of those greater than ____ grams and those less than ____ grams.

c. If x is a random variable with a normal distribution and a is a numerical constant (a ≠ 0), then y = ax also has a normal distribution. Use this formula to determine the distribution of full-term baby birth weight expressed in pounds (shape, mean, and standard deviation), and then recalculate the probability from 0.700. (Round your answer to four decimal places.)

Answer 1

a)

µ =    3500      
σ =    619      
          
P( X ≤    2000   ) = P( (X-µ)/σ ≤ (2000-3500) /619)  
=P(Z ≤   -2.423   ) =   0.0077

µ =    3500              
σ =    619              
                  
P ( X ≥   5000.00   ) = P( (X-µ)/σ ≥ (5000-3500) / 619)          
= P(Z ≥   2.423   ) = P( Z <   -2.423   ) =    0.0077

0.0077 + 0.0077 = 0.0154

b)

0.1% extreme cases means .05% on both side of the chart

µ=   3500                  
σ =    619                  
proportion=   0.0005                  
                      
Z value at    0.0005   =   -3.29   (excel formula =NORMSINV(   0.0005   ) )
z=(x-µ)/σ                      
so, X=zσ+µ=   -3.29   *   619   +   3500  
X   =   1463   

µ=   3500                  
σ =    619                  
proportion=   0.9995                  
                      
Z value at    0.9995   =   3.29   (excel formula =NORMSINV(   0.9995   ) )
z=(x-µ)/σ                      
so, X=zσ+µ=   3.29   *   619   +   3500  
X   =   5537   

lesser than 1463 and greater than 5537

c)

Shape = Normal distribution

Multiply by constant a will multiply mean and S.D by a constant too.

Mean = 3500 * 0.2205 = 771.75 Pound

S.D = 619 * 0.2205 = 136.4895 Pound

Please revert back in case of any doubt.

Please upvote. Thanks in advance.