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

Answer :

Answer:

2630 g

Step-by-step explanation:

From the given information:

Given that:

mean (μ) = 3750 g

Standard deviation (σ) = 500

Suppose the hospital officials demand special treatment with a percentage of  lightest 3% (0.03) for newborn babies;

Then, the weight of birth that differentiates the babies that needed special treatment from those that do not can be computed as follows;

P(Z < z₁) = 0.03

Using the Excel Formula; =NORMSINV(0.03) = -1.88

z₁ = - 1.88

Using the test statistics z₁ formula:

[tex]z_1 = \dfrac{X-\mu}{\sigma}[/tex]

[tex]-1.88 = \dfrac{X-3570}{500}[/tex]

By cross multiply, we have:

-1.88 × 500 = X - 3570

-940 = X - 3570

-X = -3570 + 940

-X = -2630

X = 2630 g

Hence, 2630 g is the required weight of birth   that differentiates the babies that needed special treatment from those that do not

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