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Q.84

Expert-verifiedFound in: Page 393

Book edition
OER 2018

Author(s)
Barbara Illowsky, Susan Dean

Pages
902 pages

ISBN
9781938168208

An expert witness for a paternity lawsuit testifies that the length of a pregnancy is normally distributed with a mean of $280$ days and a standard deviation of $13$ days. An alleged father was out of the country from $240$ to $306$ days before the birth of the child, so the pregnancy would have been less than $240$ days or more than $306$ days long if he was the father. The birth was uncomplicated, and the child needed no medical intervention. What is the probability that he was NOT the father? What is the probability that he could be the father? Calculate the localid="1653472319552" $z$-scores first, and then use those to calculate the probability.

His probability of being the father is $0.024$.

Let variable $\mathrm{X}$ has normal distribution with mean localid="1653469397337" $\mu $ and variance ${\sigma}^{2}$. Then probability density function is defined by:

$f\left(x\right)=\frac{1}{\sqrt{2\pi}\sigma}{e}^{-\frac{1}{2}\frac{(x-\mu {)}^{2}}{{\sigma}^{2}}},x\in R$

The localid="1653469465424" $z$ statistic for value $x$ is calculated as follows:

$z=\frac{x-\mu}{\sigma}$

If $z$ statistic is positive, $x$ value is $\left|z\right|$ standard deviation to the right of the mean, whereas $z$ statistic is negative, the $x$ value is $\left|z\right|$ standard deviation to the left of the mean.

The length of a pregnancy in this task has a normal distribution with a mean of $\mu =280$ and a standard deviation of $\sigma =13$ days. Now we must determine the likelihood that this individual is not a father, followed by the likelihood that he is a father. We must first locate the $z$ statistics for the days $240$ and $306$ days. We have:

localid="1653471405557" ${z}_{240}=\frac{240-280}{13}$

$=-3.077,$

${z}_{306}=\frac{306-280}{13}\phantom{\rule{0ex}{0ex}}$

$=\frac{306-280}{13}$

$=2$

We'll utilise these numbers to figure out whether he's the father or not. We'll start by calculating the odds that he isn't the father. Let's define the variable $X$, which represents the length of pregnancy, and then write $X:N(280,13)$. We need to find the likelihood that standardizing variable $X*:N(0,1)$ is between ${z}_{240}$ and ${z}_{306}$ to calculate that he is not the father. Now we have:

$P\left({z}_{240}<{X}^{*}<{z}_{306}\right)=P\left(-3.077<{X}^{*}<2\right)$ &

$=P\left(-3.077<{X}^{*}<2\right)$

$={\int}_{-3.077}^{2}\frac{1}{\sqrt{2\pi}}{e}^{-\frac{{u}^{2}}{2}}du$

$=NORMSDIST\left(2\right)-NORMSDIST(-3.077)$$=0.977-0.001$$=0.976$.

The probability that he is not the father is $0.976$. We must now determine the likelihood that he is a parent. We must calculate the probability that the standardize variable $X*$ is smaller than ${z}_{240}$ and greater than ${z}_{306}$. Now we have:

$P\left({X}^{*}<{z}_{240}\cup {X}^{*}>{z}_{306}\right)=P\left({X}^{*}<-3.077\cup {X}^{*}>2\right)$$=P\left({X}^{*}<-3.077\cup {X}^{*}>2\right)$

$=NORMSDIST(-3.077)+(1-\text{NORMSDIST(2))}$

$=0.001+1-0.976$

$=0.024$

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