Answer to Question #305656 in Statistics and Probability for sunoovrse

Question

Situation: You volunteered to be part of the medical mission in your barangay and in-charge

to get weight of the children. The weights of children have an average of kilograms

and a standard deviation of kilograms.

a. How many children weight 35 kg to 55 kg?

b. How many are in the upper 10% of the population?

c. Estimate the range of the number of children weighing

c.1. 40kg to 50kg

c.2. below 55kg

c.3. between 25kg to 58kg

Answer 1

Let $X=$ weight of children: $X\sim N(\mu, \sigma^2)$

Given $\mu=45kg, \sigma=5kg$

a.

$$P(35<X<55)=P(X<55)-P(X\le35)$$

$$=P(Z<\dfrac{55-45}{5})-P(Z\le\dfrac{35-45}{5})$$

$$=P(Z<2)-P(Z\le-2)$$

$$\approx0.97725-0.02275=0.9545$$

b.

$$1-P(Z\le\dfrac{x-45}{5})=0.1$$

$$\dfrac{x-45}{5}\approx1.28155$$

$$x\approx51.408$$

c.

c.1.

$$P(40<X<50)=P(X<50)-P(X\le40)$$

$$=P(Z<\dfrac{50-45}{5})-P(Z\le\dfrac{40-45}{5})$$

$$=P(Z<1)-P(Z\le-1)$$

$$\approx0.84134-0.15866=0.6827$$

c.2.

$$P(X<55)=P(Z<\dfrac{55-45}{5})$$

$$=P(Z<2)\approx0.97725$$

c.3.

$$P(25<X<58)=P(X<58)-P(X\le25)$$

$$=P(Z<\dfrac{58-45}{5})-P(Z\le\dfrac{25-45}{5})$$

$$=P(Z<2.6)-P(Z\le-4)$$

$$\approx0.99534-0.00003=0.9953$$