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Thread: Normal distribution excercise

  1. #1
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    Normal distribution excercise

    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 $2X \sim N(6,8)$.

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

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

    $$P(Y>1.6)=0.70884$$
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  2. #2
    MHF Contributor
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    Re: Normal distribution excercise

    Quote Originally Posted by feli3105 View Post
    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:

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

    $$P(-0.5<Z<0.025)=P(Z<0.0025)-P(Z<-0.5)=0.50798-0.30854=0.19944$$
    I get $P(Z<0.0025)=0.509973$

    giving an answer of $P(-0.5<Z<0.025)=0.201435\approx 20.1\%$


    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 $a=6.36 \;mm^{2}$.
    rounding properly it's $a=6.37mm$ but your method is correct.

    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 $2X \sim N(6,8)$.

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

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

    $$P(Y>1.6)=0.70884$$
    correct
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  3. #3
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    Re: Normal distribution excercise

    Excellent job by the OP showing work. This is what asking for help should look like. Notice how easy it was and how quickly it took to respond to this question. Bravo!
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