
Mauricio M. answered 09/22/22
Credentialed Secondary Math Teacher
Hello Grace,
Since the infants born from 37 to 43 weeks whose mothers did not have a history of diabetes follow a normal distribution with mean 3432 g and standard deviation 482, we can find the proportion of infants with weights above 4430 g using a standard normal distribution and corresponding table of areas and z-scores (or an excel formula such as =1- NORM.DIST(4430,3432,482,TRUE) = 0.019201 ≈ 0.019)
First find the z-score using:
z = (x - μ) / σ = 2.07
Hence, P( X > 4430) = 1 - P(X < 4430) OR equivalently,
P(Z > 2.07) = 1 - P(Z < 2.07) = 1 - 0.9808 = 0.0192 ≈ 0.019
Notice how the excel result is more accurate than the table result, however results are the same when rounded to three decimal places.
I expect that you can substitute the necessary information to confirm that z = 2.07 and use the standard normal table to verify that the area to the left of 2.07 is indeed 0.9808. Furthermore, since the question is asking for fraction (or probability) of being greater than 4430 g the complement rule applies.
Hope this helps. Good Luck.