Question

4. Suppose the birth weights of babies in the USA are normally distributed, with mean 7.47...

4. Suppose the birth weights of babies in the USA are normally distributed, with mean 7.47 lb and standard deviation 1.21 lb. a. Find the probability that a randomly chosen baby weighed between 6.4 and 8.1 pounds. (Show work.) b. Suppose a hospital wants to try a new intervention for the smallest 4% of babies (those with the lowest birth weights). What birth weight in pounds is the largest that would qualify for this group? (Show your work.)

Homework Answers

Answer #1

Solution :

Given that ,

mean = = 7.47

standard deviation = = 1.21

P(6.4< x <8.1 ) = P[(6.4 -7.47) /1.21 < (x - ) / < (8.1 -7.47) / 1.21)]

= P( -0.88< Z < 0.52)

= P(Z < 0.52) - P(Z <-0.88 )

Using z table   

= 0.6985-0.1894

probability= 0.5091

(b)

Using standard normal table,

P(Z < z) = 4%

= P(Z < z) = 0.04  

= P(Z <-1.75 ) = 0.04

z = -1.75   

Using z-score formula  

x= z * +

x=-1.75 *1.21+7.47

x= 5.3525

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