Question

The volume, L liters, of paint in a plastic tub may be assumed to be normally distributed with mean 10.25 and variance σ^2.
(a) assuming that variance = 0.04, determine P(L<10).
(b) Find the value of standard deviation so that 98% of tubs contain more than 10 liters of paint.

Answer 1

Suppose $L$ is a variable representing the volume of paint in a plastic tub.

$L$ is normally distributed with mean $\mu=10.25$ and variance $\sigma^2:L\sim N(10.25, \sigma^2).$

If $L\sim(N, \sigma^2),$ then $Z={L-\mu \over \sigma}\sim N(0, 1)$

(a) Assume that $\sigma^2=0.04$

$$Z={L-\mu \over \sigma}={10-10.25 \over \sqrt{0.04}}=-1.25$$

Then

$$P(L<10)=P(Z<-1.25)=0.105650$$ $$P(L<10)=0.105650$$

(b) Find the value of standard deviation so that 98% of tubs contain more than 10 liters of paint.

Given that

$$P(L>10)=0.98$$

Then

$$P(L>10)=P(Z>Z^*)=0.98=>$$ $$=>Z^*=-2.053749$$

$${L-\mu \over \sigma}=Z^*$$ $$\sigma={L-\mu \over Z^*}$$

$$\sigma={10-10.25 \over -2.053749}\approx0.1217\approx \sqrt{0.0148}$$

If the value of standard deviation is $0.1217,$ then 98% of tubs contain more than 10 liters of paint.