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Q.84

Expert-verified
Found in: Page 393

### Introductory Statistics

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 $0.024$.

See the step by step solution

## Step 1: Given Information

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

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

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

## Step 2: Explanation:

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

$=-3.077,$

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

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

$=2$

## Step 3: Explanation

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 $X:N\left(280,13\right)$. We need to find the likelihood that standardizing variable $X*:N\left(0,1\right)$ 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\left(-3.077\right)$$=0.977-0.001$

$=0.976$.

## Step 4: Explanation

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)$

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

$=0.001+1-0.976$

$=0.024$