Newborn babies: A study conducted by the Center for Population Economics at the University of Chicago studied the birth weights of 732 babies born in New York. The mean weight was 3311 grams with a standard deviation of 860 grams. Assume that birth weight data are approximately bell-shaped. Part 1 of 3 (a) Estimate the number of newborns whose weight was less than 4171 grams. Approximately of the 732 newborns weighed less than 4171 grams. Part 2 of 3 (b) Estimate the number of newborns whose weight was greater than 1591 grams. Approximately of the 732 newborns weighed more than 1591 grams. Part 3 of 3 (c) Estimate the number of newborns whose weight was between 3311 and 5031 grams. Approximately of the 732 newborns weighed between 3311 and 5031 grams.

a) 615

b) 715

c) 344

Step-by-step explanation:

According to the Question,

Given that,  A study conducted by the Center for Population Economics at the University of Chicago studied the birth weights of 732 babies born in New York. The mean weight was 3311 grams with a standard deviation of 860 gramsSince the distribution is approximately bell-shaped, we can use the normal distribution and calculate the Z scores for each scenario.

Z = (x - mean)/standard deviation

Now,

For x = 4171,  Z = (4171 - 3311)/860 = 1  

P(Z < 1) using Z table for areas for the standard normal distribution, you will get 0.8413.

Next, multiply that by the sample size of 732.

Therefore 732(0.8413) = 615.8316, so approximately 615 will weigh less than 4171  

 

For part b, use the same method except x is now 1591.    

Z = (1581 - 3311)/860 = -2    

P(Z > -2) , using the Z table is 1 - 0.0228 = 0.9772 . Now 732(0.9772) = 715.3104, so approximately 715 will weigh more than 1591.

 

For part c, we now need to get two Z scores, one for 3311 and another for 5031.

Z1 = (3311 - 3311)/860 = 0

Z2 = (5031 - 3311)/860= 2  

P(0 ≤ Z ≤ 2) = 0.9772 - 0.5000 = 0.4772

  approximately 47% fall between 0 and 1 standard deviation, so take 0.47 times 732 ⇒ 732×0.47 = 344.