Chapter 6.3, Problem 1E

To determine

a.

To find:

The probability, from the given information, that a randomly chosen Norwegian baby’s birth weight was less than 4000 g.

Answer to Problem 1E

Solution:

The required probability is $P\left(\text{birth}\text{weight}\text{less}\text{than}4000g\right)=0.7422$.

Explanation of Solution

In statistics, the normal Gaussian distribution with the following characteristics:

$\mu =0$, and

$\sigma =1$

is called standard normal distribution with probability density function:

$f\left(x|0,1\right)=\frac{1}{\sqrt{2\pi }}{e}^{-\frac{x^2}{2}}$

When this function is plotted on the $x$-axis, the area under between the curve and $x$-axis is equal to one, because the sum of probabilities is unity.

Area on the left of a given $z$-value (say $z=a$) is the probability $P\left(X\le a\right)$ given by the integration:

${\displaystyle \underset{-\infty }{\overset{a}{\int }}f\left(x\right)dx}$,

which can be obtained from the normal distribution table.

Area on the right of a given $z$-value (say $z=a$) is the probability $P\left(X\ge a\right)=1-P\left(X\le a\right)$ given by the integration:

${\displaystyle \underset{a}{\overset{\infty }{\int }}f\left(x\right)dx}=1-{\displaystyle \underset{-\infty }{\overset{a}{\int }}f\left(x\right)dx}$,

which can be obtained by subtracting a certain value from the normal distribution table from one.

Area in between of given $z$-values (say $z=a$ and $z=b$) is the probability $P\left(a\le X\le b\right)$ given by the integration:

${\displaystyle \underset{a}{\overset{b}{\int }}f\left(x\right)dx}={\displaystyle \underset{-\infty }{\overset{b}{\int }}f\left(x\right)dx}-{\displaystyle \underset{-\infty }{\overset{a}{\int }}f\left(x\right)dx}$,

both of which can both be obtained from the normal distribution table.

Area below a smaller given $z$-value (say $z=a$) above a larger given $z$-value (say $z=b$) is the probability $P\left(X\le a\text{or}X\ge b\right)$ given by the integration:

${\displaystyle \underset{-\infty }{\overset{a}{\int }}f\left(x\right)dx}+{\displaystyle \underset{b}{\overset{\infty }{\int }}f\left(x\right)dx}=1-{\displaystyle \underset{-\infty }{\overset{b}{\int }}f\left(x\right)dx}+{\displaystyle \underset{-\infty }{\overset{a}{\int }}f\left(x\right)dx}$,

both of which can both be obtained from the normal distribution table.

In order to get the above values from the table, the standard scores are to be obtained using the relation:

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

and then the tables are to be accessed to find the values of the integrations and thus the probabilities.

It is to note that the probabilities when multiplied by 100 give the percentage chance.

Given:

Birth weights of Norwegians are normally distributed with the following characteristics:

mean $\mu =3668$

standard deviation $\sigma =511$

Calculation:

The standard value corresponding to the $x$-value $x=4000$ is:

$\begin{array}{rll}a& =& \frac{4000-3668}{511}\\ & =& \frac{332}{511}\\ & =& \mathrm{0.64970...}\end{array}$

The obtained value is the $z$-value, on whose left the area represents the required probability.

Since the digit at $4^{\text{th}}$ place is 7 and $7\ge 5$, therefore the $z$-value is accordingly adjusted by first increasing the $3^{\text{rd}}$ place by 1 which subsequently gives the $z$-value as:

$a=0.650$.

From the table, the area on the left of the above $z$-value is:

$P\left(z\le 0.650\right)=0.74215$.

Since the digit at the $5^{\text{th}}$ place is 5, thus the $4^{\text{th}}$ decimal place is increased by 1, which gives the required probability as:

$P\left(\text{birth}\text{weight}\text{less}\text{than}4000g\right)=0.7422$.

To determine

b.

To find:

The probability, from the given information, that a randomly chosen Norwegian baby’s birth weight was greater than 3750 g.

Answer to Problem 1E

Solution:

The required probability is $P\left(\text{birth}\text{weight}\text{greater}\text{than}3750g\right)=0.4364$.

Explanation of Solution

Given:

Birth weights of Norwegians are normally distributed with the following characteristics:

mean $\mu =3668$

standard deviation $\sigma =511$

Calculation:

The standard value corresponding to the $x$-value $x=3750$ is:

$\begin{array}{rll}a& =& \frac{3750-3668}{511}\\ & =& \frac{82}{511}\\ & =& \mathrm{0.16047...}\end{array}$

The obtained value is the $z$-value, on whose right the area represents the required probability.

Since the digit at $4^{\text{th}}$ place is 4 and $4\le 5$, therefore the $z$-value is approximated as:

$a=0.160$.

From the table, the area on the right of the above $z$-value is:

$P\left(z\ge 0.160\right)=1-0.56356$,

which simplifies to:

$P\left(z\ge 0.160\right)=0.43644$.

Since the digit at the $5^{\text{th}}$ place is 4, thus the $4^{\text{th}}$ decimal place is kept unchanged, which gives the required probability as:

$P\left(\text{birth}\text{weight}\text{greater}\text{than}3750g\right)=0.4364$

To determine

c.

To find:

The probability, from the given information, that a randomly chosen Norwegian baby’s birth weight was greater than 3750 g.

Answer to Problem 1E

Solution:

The required probability is $P\left(\text{birth}\text{weight}\text{greater}\text{than}3750g\right)=0.4364$.

Explanation of Solution

Given:

Birth weights of Norwegians are normally distributed with the following characteristics:

mean $\mu =3668$

standard deviation $\sigma =511$

Calculation:

The standard value corresponding to the $x$-value $x=3000$ is:

$\begin{array}{rll}{a}_1& =& \frac{3000-3668}{511}\\ & =& \frac{\left(-668\right)}{511}\\ & =& -\frac{668}{511}\\ & =& -1.30724\mathrm{...}\end{array}$

and that corresponding to the $x$-value $x=4000$ is:

$\begin{array}{rll}{a}_2& =& \frac{4000-3668}{511}\\ & =& \frac{332}{511}\\ & =& \mathrm{0.64970...}\end{array}$

The obtained values are the $z$-values, between which the area represents the required probability.

Since the digit at $4^{\text{th}}$ place of $a_1$ is 2 and $2\le 5$, therefore the $z$-value is unchanged and approximate $z$-value is:

$a_1=-1.307$.

Since the digit at $4^{\text{th}}$ place of $a_2$ is 7 and $7\ge 5$, therefore the $z$-value is accordingly adjusted by first increasing the $3^{\text{rd}}$ place by 1 which subsequently gives the $z$-value as:

$a_2=0.650$.

From the table, the area on the left of the $z$-value $a_1$ is the average of table values corresponding to $-1.30$ and $-1.31$ which is given as:

$\begin{array}{rll}P\left(z\le -1.307\right)& =& \frac{0.09680+0.09510}{2}\\ & =& \frac{0.1919}{2}\\ & =& 0.09595\end{array}$

Similarly, the area on the left of the $z$-value $a_2$ is given as:

$P\left(z\le 0.650\right)=0.74215$.

Thus, the probability that $z$-value lies between $-1.307$ and $0.650$ is:

$\begin{array}{rll}P\left(-1.307\le z\le 0.650\right)& =& 0.74215-0.09595\\ & =& 0.6462\end{array}$

This gives the required probability as:

$P\left(\text{birth}\text{weight}\text{between}3000g\text{and}4000g\right)=0.6462$.

To determine

d.

To find:

The probability, from the given information, that a randomly chosen Norwegian baby’s birth weight was greater than 3750 g.

Answer to Problem 1E

Solution:

The required probability is $P\left(\text{less}\text{than}2650g\text{or}\text{greater}\text{than}4650g\right)=0.05014$.

Explanation of Solution

Given:

Birth weights of Norwegians are normally distributed with the following characteristics:

mean $\mu =3668$

standard deviation $\sigma =511$

Calculation:

The standard value corresponding to the $x$-value $x=2650$ is:

$\begin{array}{rll}{a}_1& =& \frac{2650-3668}{511}\\ & =& \frac{\left(-1018\right)}{511}\\ & =& -\frac{1018}{511}\\ & =& -1.99217\mathrm{...}\end{array}$

and that corresponding to the $x$-value $x=4650$ is:

$\begin{array}{rll}{a}_2& =& \frac{4650-3668}{511}\\ & =& \frac{982}{511}\\ & =& \mathrm{1.92172...}\end{array}$

The obtained values are the $z$-values, between which the area represents the required probability.

Since the digit at $4^{\text{th}}$ place of $a_1$ is 1 and $1\le 5$, therefore the $z$-value is unchanged and approximate $z$-value is:

$a_1=-1.992$.

Since the digit at $4^{\text{th}}$ place of $a_2$ is 7 and $7\ge 5$, therefore the $z$-value is accordingly adjusted by first increasing the $3^{\text{rd}}$ place by 1 which subsequently gives the $z$-value as:

$a_2=1.922$.

From the table, the area on the left of the $z$-value $a_1$ is the average of table values corresponding to $-1.99$ and $-2.00$ which is given as:

$\begin{array}{rll}P\left(z\le -1.992\right)& =& \frac{0.02330+0.02275}{2}\\ & =& \frac{0.04605}{2}\\ & =& 0.023025\end{array}$

Similarly, the area on the right of the $z$-value $a_2$ is given as:

$\begin{array}{rll}P\left(z\ge 1.922\right)& =& 1-P\left(z\le 1.922\right)\\ & =& 1-\frac{0.97257+0.97320}{2}\\ & =& 1-\frac{1.94577}{2}\\ & =& 1-0.972885\\ & =& 0.027115\end{array}$.

Thus, the probability that $z$-value lies between $-1.992$ and $1.922$ is:

$\begin{array}{rll}P\left(z\le -1.992\text{or}z\ge 1.922\right)& =& 0.023025+0.027115\\ & =& 0.05014\end{array}$

Here, since the $5^{\text{th}}$ digit is 6, thus the $4^{\text{th}}$ digit is raised by one and the subsequent approximation gives the required probability as:

$P\left(\text{less}\text{than}2650g\text{or}\text{greater}\text{than}4650g\right)=0.05014$.