Question

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.)

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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