Birth weights in norway are normally distributed with a mean of 3570 g and a standard deviation of 500 g. if the ulleval university hospital officials plan to require special treatment for the lightest 3% of newborn babies, what birth weight separates those requiring special treatment from those who do not?

We are given that:

μ = mean = 3570

σ = standard deviation = 500

From the normal probability table, find the value of z which corresponds to the largest 3%.

at P = 0.03, z = 1.88

The formula for z score is:

z = (x - μ) / σ

Finding for the sample value x:

1.88 =(x-3570)/500

x=(1.88)(500)+3570

x = 4510

The birth weight that separates those who do and do not require is 4510 g