Let $X$ be a random variable that follows a normal distribution $N(3,4)$ which measures the surface area in $mm^{2}$ of a skin sample.

a) For the skin sample to be considered acceptable to conduct a test it must be between $1\;mm^{2}$ and $3.1\;mm^{2}$. Find the probability of getting a sample within these standards.

My answer:

$$P(1<X<3.1)=P\bigg(\frac{1-3}{4}<\frac{X-3}{4}<\frac{3.1-3}{4}\bigg)=P(-0.5<Z<0.025)$$

$$P(-0.5<Z<0.025)=P(Z<0.0025)-P(Z<-0.5)=0.50798-0.30854=0.19944$$

$$P(1<X<3.1)=0.19944$$

The probability of getting a sample between $1\;mm^{2}$ and $3.1\;mm^{2}$ is $19.9\%$.

b) Find the value $a$ (the maximum sample area) for which the probability of discarding a sample is $0.2$. The minimum area should remain $1\;mm^{2}$.

$$P(X>a)=0.20$$

$$P\bigg(\frac{X-3}{4}>\frac{a-3}{4}\bigg)=0.20$$

$$P\bigg(Z>\frac{a-3}{4}\bigg)=0.20$$

$$P\bigg(Z<\frac{a-3}{4}\bigg)=0.80$$

On the body of the normal table I found that $P(Z<0.84)=0.79955$ which is $\sim 0.80$.

$a=Z\sigma+\mu$, $a=0.84\cdot4+3$, therefore $a=6.36 \;mm^{2}$.

c) After treating the sample with a certain chemical a new sample is drawn whose volume is related to the skin area by a random variable $Y$. If $X\leqslant1.55$, $Y=2X$ and if $X>1.55$, $Y=0$. Find $P(Y>1.66)$.

I used the property: $aX \sim N(a\mu, a\sigma)$. So $2X \sim N(6,8)$.

$$P(Y>1.6)=P\bigg(\frac{Y-6}{8}>\frac{1.6-6}{8}\bigg)=P(Z>-0.55)$$

$$P(Z>-0.55)=1-P(Z<-0.55)=1-0.29116=0.70884$$

$$P(Y>1.6)=0.70884$$