Chapter 4.3, Problem 49E

Consider babies born in the “normal” range of 37–43 weeks gestational age. Extensive data supports the assumption that for such babies born in the United States, birth weight is normally distributed with mean 3432 g and standard deviation 482 g. [The article “Are Babies Normal?” (The American Statistician, 1999: 298–302) analyzed data from a particular year; for a sensible choice of class intervals, a histogram did not look at all normal, but after further investigations it was determined that this was due to some hospitals measuring weight in grams and others measuring to the nearest ounce and then converting to grams. A modified choice of class intervals that allowed for this gave a histogram that was well described by a normal distribution.]

a. What is the probability that the birth weight of a randomly selected baby of this type exceeds 4000 g? Is between 3000 and 4000 g?

b. What is the probability that the birth weight of a randomly selected baby of this type is either less than 2000 g or greater than 5000 g?

c. What is the probability that the birth weight of a randomly selected baby of this type exceeds 7 lb?

d. How would you characterize the most extreme .1% of all birth weights?

e. If X is a random variable with a normal distribution and a is a numerical constant (a ≠ 0), then Y = aX also has a normal distribution. Use this to determine the distribution of birth weight expressed in pounds (shape, mean, and standard deviation), and then recalculate the probability from part (c). How does this compare to your previous answer?

To determine

Find the probabilities that the birth weight of such a baby exceeds 4,000 g and birth weight is between 3,000 g and 4,000 g.

Answer to Problem 49E

The probability that the birth weight of such a baby exceeds 4,000 g is 0.1190.

The probability that the birth weight of a baby is between 3,000 and 4,000 g, is 0.6969.

Explanation of Solution

Given info:

The probability distribution of the birth weight of babies who are born in the normal range of 37 to 43 weeks gestational period in the United States is normal with mean 3,432 g and standard deviation 482 g.

Calculation:

Let X denote the birth weight of a randomly selected baby from such babies as described above. The random variable X is normally distributed with parameters $\mu =3,432$ g and $\sigma =482$ g.

The probability that the birth weight of such a baby exceeds 4,000 g is:

$\begin{array}{c}P\left(X>4,000\right)=P\left(\frac{X-\mu }{\sigma }>\frac{4,000-\mu }{\sigma }\right),\left(\text{by standardizing}\right)\\ =P\left(Z>\frac{4,000-3,432}{482}\right)\\ =1-P\left(Z\le \frac{4,000-3,432}{482}\right)\\ =1-\Phi \left(\frac{4,000-3,432}{482}\right)\end{array}$

$=1-\Phi \left(1.18\right)$.

Use Table A-3 “Standard Normal Curve Areas” to find the required probability.

Procedure:

• Locate the value of z as 1.1 in the first column of the table.
• Move along the corresponding row to obtain the value in that row, which is exactly in the column of .08.

Thus, $\Phi \left(1.18\right)=0.8810$.

As a result,

$\begin{array}{c}P\left(X>4,000\right)=1-\Phi \left(1.18\right)\\ =1-0.8810\\ =\mathrm{0.1190.}\end{array}$

Hence, the probability that the birth weight of such a baby exceeds 4,000 g is 0.1190.

Theprobability that the birth weight of a baby is between 3,000 and 4,000 g is:

$\begin{array}{c}P\left(3,000\le X\le 4,000\right)=P\left(\frac{3,000-\mu }{\sigma }\le \frac{X-\mu }{\sigma }\le \frac{4,000-\mu }{\sigma }\right),\left(\text{by standardizing}\right)\\ =P\left(Z\le \frac{4,000-3,432}{482}\right)-P\left(Z<\frac{3,000-3,432}{482}\right)\\ =\Phi \left(1.18\right)-\Phi \left(-0.90\right).\end{array}$

Use Table A-3 “Standard Normal Curve Areas” to find the required probability.

Procedure:

• Locate the value of z as –0.9 in the first column of the table.
• Move along the corresponding row to obtain the value in that row, which is exactly in the column of .00.

Thus, $\Phi \left(-0.90\right)=0.1841$.

As a result,

$\begin{array}{c}P\left(3,000\le X\le 4,000\right)=\Phi \left(1.18\right)-\Phi \left(-0.90\right)\\ =0.8810-0.1841\\ =\mathrm{0.6969.}\end{array}$

Hence, the probability that the birth weight of a baby is between 3,000 and 4,000 g, is 0.6969.

To determine

Find the probability that the birth weight of a randomly selected baby is either less than 2,000 g or more than 5,000 g.

Answer to Problem 49E

The probability that the birth weight of a randomly selected baby is either less than 2,000 g or more than 5,000 g is 0.0021.

Explanation of Solution

Calculation:

Theprobability that the birth weight of a randomly selected baby is either less than 2,000 g or more than 5,000 g is:

$\begin{array}{c}P\left(X<2,000\text{ or }X>5,000\right)=1-P\left(2,000\le X\le 5,000\right)\\ =1-P\left(\frac{2,000-\mu }{\sigma }\le \frac{X-\mu }{\sigma }\le \frac{5,000-\mu }{\sigma }\right),\left(\text{by standardizing}\right)\\ =1-\left[P\left(Z\le \frac{5,000-3,432}{482}\right)-P\left(Z<\frac{2,000-3,432}{482}\right)\right]\\ =1-\left[\Phi \left(3.25\right)-\Phi \left(-2.97\right)\right].\end{array}$

Use Table A-3 “Standard Normal Curve Areas” to find $\Phi \left(3.25\right)$.

Procedure:

• Locate the value of z as 3.2 in the first column of the table.
• Move along the corresponding row to obtain the value in that row, which is exactly in the column of .05.

Thus, $\Phi \left(3.25\right)=0.9994$.

Use Table A-3 “Standard Normal Curve Areas” to find $\Phi \left(-2.97\right)$.

Procedure:

• Locate the value of z as –2.9 in the first column of the table.
• Move along the corresponding row to obtain the value in that row, which is exactly in the column of .07.

Thus, $\Phi \left(-2.97\right)=0.0015$.

As a result,

$\begin{array}{c}P\left(X<2,000\text{ or }X>5,000\right)=1-\left[\Phi \left(3.25\right)-\Phi \left(-2.97\right)\right]\\ =1-\left[0.9994-0.0015\right]\\ =1-0.9979\\ =\mathrm{0.0021.}\end{array}$

Hence, the probability that the birth weight of a randomly selected baby is either less than 2,000 g or more than 5,000 g is 0.0021.

To determine

Find the probability that the birth of a randomly selected baby exceeds 7 lb.

Answer to Problem 49E

The probability that the birth of a randomly selected baby exceeds 7 lb is 0.7019.

Explanation of Solution

Calculation:

Use the conversion: $1\text{ lb}\approx 454\text{ g}$.

Thus,

$\begin{array}{c}7\text{ lb}\approx \left(7\times 454\right)\text{ g}\\ \text{=3,178 g}.\end{array}$

The probability that the birth of a randomly selected baby exceeds 7 lb is the same as the probability that the birth of a randomly selected baby exceeds 3,178 g:

$\begin{array}{c}P\left(X>3,178\right)=P\left(\frac{X-\mu }{\sigma }>\frac{3,178-\mu }{\sigma }\right),\left(\text{by standardizing}\right)\\ =P\left(Z>\frac{3,178-3,432}{482}\right)\\ =1-P\left(Z\le \frac{3,178-3,432}{482}\right)\\ =1-\Phi \left(\frac{3,178-3,432}{482}\right)\end{array}$

$=1-\Phi \left(-0.53\right)$.

Use Table A-3 “Standard Normal Curve Areas” to find the required probability.

Procedure:

• Locate the value of z as –0.5 in the first column of the table.
• Move along the corresponding row to obtain the value in that row, which is exactly in the column of .03.

Thus, $\Phi \left(-0.53\right)=0.2981$.

As a result,

$\begin{array}{c}P\left(X>3,178\right)=1-\Phi \left(-0.53\right)\\ =1-0.2981\\ =\mathrm{0.7019.}\end{array}$

Hence, the probability that the birth of a randomly selected baby exceeds 7 lb is 0.7019.

To determine

Characterize the most extreme 0.1% of all birth weights.

Answer to Problem 49E

The values that characterize the most extreme 0.1% of the birth weights are 1,846.22 g, which separates the observations from the smallest 0.05% of weights and 5,017.78 g, which separates the observations from the largest 0.05% of weights.

Explanation of Solution

Calculation:

The normal distribution is symmetric. Thus, for any constant k, $P\left(Y<-k\right)=P\left(Y>k\right)$.

The most extreme 0.1% of all birth weights can be characterized by obtaining a value c, such that $99.9\%\left(=100\%-0.1\%\right)$ of all the observations fall within the interval (-c, c), that is, only 0.1% of the observations fall outside said interval.

Let $p=P\left(Y<-k\right)$.

As a result, $\left(100p\right)\%$ of all observations of Y fall below the value –k. It automatically ensures that $\left(100p\right)\%$ of all observations of Y lie above the value k. Thus, $\left(200p\right)\%$ of the observations fall outside the interval (–k, k).

Here,

$\begin{array}{c}\left(200p\right)\%=0.1\%\\ p=\frac{0.1}{200}\\ =0.0005\\ \left(100p\right)\%=0.05\%.\end{array}$

Here, let –c and c be the values that separate the middle portion of the distribution, with the smallest and largest 0.1% of the distribution. As a result, 0.05% of all values of the distribution must lie below –c. Moreover, 0.05% of all values must lie above c or $99.95\%\left(=100\%-0.05\%\right)$ of all values lie below c.

Thus,

$\begin{array}{c}0.999=P\left(-c\le X\le c\right)\\ =P\left(\frac{-c-\mu }{\sigma }\le \frac{X-\mu }{\sigma }\le \frac{c-\mu }{\sigma }\right)\\ =P\left(\frac{-c-3,432}{482}\le Z\le \frac{c-3,432}{482}\right)\\ =P\left(Z\le \frac{-c-3,432}{482}\right)-P\left(Z\le \frac{c-3,432}{482}\right)\end{array}$

$=\Phi \left(\frac{c-3,432}{482}\right)-\Phi \left(\frac{-c-3,432}{482}\right)$.

Here, c separates the distribution from the largest 0.05% of the values, implying that c is the 99.95^th percentile of the distribution. As a result, $\Phi \left(\frac{c-3,432}{482}\right)=0.9995$.

Thus, the corresponding percentile is $\Phi \left(\eta \left(0.9995\right)\right)=0.9995$.

In Table A-3 “Standard Normal Curve Areas” find the value of the variable which corresponds to the probability 0.9995.

Procedure:

• Locate the probability value 0.9995 in the body of the table.

The exact value 0.9995 is not unique, as itappears in the table for 6 times. Take the lowest z value for which 0.9995 occurs for the first time as follows:

• Move horizontally from the very first occurrence in increasing order of the probability value 0.9995 till the first column is reached and note the value of z as 3.2.
• Move vertically from the probability value 0.9995 till the first rowis reached and note the value of z as .09.

Thus,

$\Phi \left(3.29\right)=0.9995$.

Now, $\Phi \left(\frac{c-3,432}{482}\right)=\Phi \left(\eta \left(0.9995\right)\right)=0.9995$. Thus,

$\begin{array}{c}\frac{c-3,432}{482}=3.29\\ c-3,432=482\times 3.29\\ =\text{1,585}\text{.78}\\ x=3,432+\text{1,585}\text{.78}\\ =\text{5,017}\text{.78}\end{array}$

Hence, the 99.95^th percentile of the distribution is 5,017.78.

Similarly, –c separates the distribution from the smallest 0.05% of the values, implying that–c is the 0.05^th percentile of the distribution. As a result, $\Phi \left(\frac{-c-3,432}{482}\right)=0.0005$.

Thus, the corresponding percentile is $\Phi \left(\eta \left(0.0005\right)\right)=0.0005$.

Now,

$\begin{array}{c}\Phi \left(\eta \left(0.0005\right)\right)=0.0005\\ P\left(Z\le \eta \left(0.0005\right)\right)=0.0005\\ 1-P\left(Z>\eta \left(0.0005\right)\right)=0.0005\\ P\left(Z>\eta \left(0.0005\right)\right)=1-0.0005\\ =\mathrm{0.9995.}\end{array}$

It is already shown that $P\left(Y<-k\right)=P\left(Y>k\right)$. Earlier in part d, it is found that $\Phi \left(3.27\right)\approx 0.9995$, that is, $P\left(Z\le 3.29\right)=0.9995$. This implies that:

$P\left(Z>-3.29\right)\approx 0.9995$.

Now, $P\left(Z>\eta \left(0.9995\right)\right)\approx 0.9995$. Moreover, $\Phi \left(\frac{-c-3,432}{482}\right)=\Phi \left(\eta \left(0.0005\right)\right)=0.0005$ Thus,

$\begin{array}{c}\frac{-c-3,432}{482}=-3.2\\ -c-3,432=482\times \left(-3.27\right)\\ =-\text{1,585}\text{.78}\\ x=3,432-\text{1,585}\text{.78}\\ =\text{1,846}\text{.22}\end{array}$

Hence, the 0.05^th percentile of the distribution is 1,846.22.

Thus, the values that characterize the most extreme 0.1% of the birth weights are 1,846.22 g, which separates the observations from the smallest 0.05% of weights and 5,017.78 g, which separates the observations from the largest 0.05% of weights.

To determine

Find the birth weight distribution in pound.

Find the probability of part c and compare the two cases- when weight is measured in pounds and when weight is measured in g.

Answer to Problem 49E

Thebirth weight distribution in pound has a normal distribution, with mean7.56 lband standard deviation 1.06 lb.

The probability that the birth of a randomly selected baby exceeds 7 lb is 0.7019.

The probability that the birth weight exceeds 7 lb is the same for when weight is measured in g and when weight is measured in lb.

Explanation of Solution

Calculation:

Consider $Y=aX$, where Y is the weight measured in pounds, X is the weight measured in g and a is a constant, which transforms the measurement unit from g to lb (pound).

As observed in part c, $1\text{ lb}\approx 454\text{ g}$.

Thus, the weight in lb, Y, can be expressed as:

$Y=\frac{1}{454}X$, that is, $\underset{\_}{Y=0.002203X}$, that is, $a=0.002203$.

Now, the transformation $Y=aX$ is a linear transformation of the random variable X, which is normally distributed. It is known that, in case of a linear transformation of the form $Y=aX$ of a normal random variable, the variable Y also has a normal distribution with parameters ${\mu }_Y=a\mu$ and ${\sigma }_Y=a\sigma$.

Here, $\mu =3,432$ and $\sigma =482$. Thus,

$\begin{array}{c}{\mu }_Y=0.002203\times 3,432\\ =\underset{\_}{7.56\text{ lb}}.\end{array}$

$\begin{array}{c}{\sigma }_Y=0.002203\times 482\\ =\underset{\_}{1.06\text{ lb}}.\end{array}$

Hence, $\underset{\_}{Y=0.002203X}$ has a normal distribution.

The probability to be found is theprobability that the birth of a randomly selected baby exceeds 7 lb, which is:

$\begin{array}{c}P\left(Y>7\right)=P\left(\frac{Y-{\mu }_Y}{{\sigma }_Y}>\frac{7-{\mu }_Y}{{\sigma }_Y}\right),\left(\text{by standardizing}\right)\\ =P\left(Z>\frac{7-7.56}{1.06}\right)\\ =1-P\left(Z\le \frac{7-7.56}{1.06}\right)\\ =1-\Phi \left(\frac{7-7.56}{1.06}\right)\end{array}$

$=1-\Phi \left(-0.53\right)$.

From part c, $\Phi \left(-0.53\right)=0.2981$.

As a result,

$\begin{array}{c}P\left(X>3,178\right)=1-\Phi \left(-0.53\right)\\ =1-0.2981\\ =\mathrm{0.7019.}\end{array}$

Hence, the probability that the birth of a randomly selected baby exceeds 7 lb is 0.7019. The probability that the birth weight exceeds 7 lb is the same for when weight is measured in g and when weight is measured in lb.