Consider babies born in the “normal” range of 37–43 weeks gestational age. Extensive data supports the assumption that for such babies born in the United States, birth weight is normally distributed with mean 3432 g and standard deviation 482 g. [The article “Are Babies Normal?” (?The American Statistician, 1999: 298–302) analyzed data from a particular year; for a sensible choice of class intervals, a histogram did not look at all normal, but after further investigations it was determined that this was due to some hospitals measuring weight in grams and others measuring to the nearest ounce and then converting to grams. A modified choice of class intervals that allowed for this gave a histogram that was well described by a normal distribution.] a.? ?What is the probability that the birth weight of a randomly selected baby of this type exceeds 4000 g? Is between 3000 and 4000 g? b.? ?What is the probability that the birth weight of a randomly selected baby of this type is either less than 2000 g or greater than 5000 g? c.? ?What is the probability that the birth weight of a randomly selected baby of this type exceeds 7 lb? d.? ow would you characterize the most extreme .1% of all birth weights? e.? ?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 to determine the distribution of birth weight expressed in pounds (shape, mean, and standard deviation), and then recalculate the probability from part (c). How does this compare to your previous answer?

Answer : Step 1 of 5 : Given, Consider babies born in the “normal” range of 37–43 weeks gestational age. Extensive data supports the assumption that for such babies born in the United States, birth weight is normally distributed with mean 3432 g and standard deviation 482 g. a) The claim is to find the probability that the birth weight of a randomly selected baby of this type exceeds 4000 g and Is between 3000 and 4000 g P(x > 4000) = 1 - P(x 4000) = 1 - (z x ) Where, x = 4000, = 3432 and = 482 Therefore, P(x > 4000) = 1 - (z 4000 3432) 482 = 1 - (z 1.178) = 1 - 0.8810 ( from the area under normal curve) = 0.119 Therefore, P(x > 4000) = 0.119 Then, between 3000 and 4000 P(3000 < x < 4000) = P(x 4000) - P(x3000) 4000 3432 3000 3432 = (z 482 ) - (z 482 ) = (z 1.178) - (z -0.896) = 0.8810 - 0.1867 = 0.6943 Therefore, P(3000 < x < 4000) = 0.6943 Step 2 of 5 : b) The claim is to find the probability that the birth weight of a randomly selected baby of this type is either less than 2000 g or greater than 5000 g P(2000 < x >5000) = 1 - P(2000 x 5000) = 1 - [ P(x 5000) - P(x 2000) ] = 1 - [ (z 5000 343) - (z 2000 343) 482 482 = 1 - [(z 3.25) - (z -2.97) ] = 1 - [ 0.9994 - 0.0015 ] = 1 - 0.9979 = 0.0021 Therefore, P(2000 < x >5000) = 0.0021. Step 3 of 5 : c) The claim is to find the probability that the birth weight of a randomly selected baby of this type exceeds 7 lb Where, 7lb = 3175 g 3175 3432 P( x > 3175) = (z > 482 ) = (z > 0.53) = 0.2981 ( from the area under normal curve) Therefore , P( x < 3175) = 1 - P( x > 3175) = 1 - 0.2981 = 0.7019