Question

Low-birth-weight babies Researchers in Norway analyzed data on the birth

weights of $400,000$ newborns over a $6-$year period. The distribution of birth weights is approximately Normal with a mean of $3668$ grams and a standard deviation of $511$ grams.$17$ Babies that weigh less than $2500$ grams at birth are classified as “low birth weight.”

a. Fill in the blanks: About $99.7\%$ of the babies had birth weights between ____ and

_____ grams.

b. What percent of babies will be identified as having low birth weight?

c. Find the quartiles of the birth weight distribution.

Short answer

Part (a)$2135,5201$

Part (b) $1.10\%$

Part (c) $1st$ quartile: $3325.63$ grams

3rd quartile: $4010.37$ grams

Step-by-step solution

Part (a) Step 1: Given information

$$\begin{gathered} \mu =3668 \\ \sigma =511 \end{gathered}$$

Part (a) Step 2: Calculation

A normal distribution's mid $68$ percentile is one standard deviation away from the mean.

The mid-$95$ percentile of a normal distribution is two standard deviations away from the mean.

The middle $99.7\%$ of a normal distribution is $3$ standard deviations from the distribution's mean.

Finding the value that is three standard deviations away from the mean is as follows:

$$\begin{gathered} \mu -3\sigma =3668-3\left(511\right)=2135 \\ \mu +3\sigma =3668+3\left(511\right)=5201 \end{gathered}$$

This means that $99.7\%$ of the observations will fall between $2135$ and $5201$ grams, implying that $99.7\%$ of the newborns will have a birth weight of between $2135$ and $5201$ grams.

Part (b) Step 1: Concept

The formula used:$z=\frac{x-\mu }{\sigma }$

Part (b) Step 2: Calculation

The $z-$score is the

$$\begin{gathered} z=\frac{x-\mu }{\sigma } \\ \sigma =\frac{2500-3668}{511} \\ =-2.29 \end{gathered}$$

Using the normal probability table, get the associated probability.

$$\begin{gathered} P(X<2500)=P(Z<-2.29) \\ =0.0110 \\ =1.10\% \end{gathered}$$

As a result, $1.10$ percent of the babies were born with a birth weight of less than $2500$ grams, which is considered poor.

Part (c) Step 1: Concept

The formula used:$z=\frac{x-\mu }{\sigma }$

Part (c) Step 2: Calculation

The first quartile has the feature that $25\%$of the data values are lower than the first quartile.

To determine the $z-$score that corresponds to a probability of 25% or $0.25$in normal probability. The closest probability is $0.2514$which is found in the normal probability table's rows -0.6 and column$.07$As a result, the associating $z-$score is$-0.6+.07=-0.67.$

$z=-0.67$

The $z-$score is

$z=\frac{x-\mu }{\sigma }=\frac{x-3668}{511}$

The two found expressions of the $z-$score then have to be equal:

$\frac{x-3668}{511}=-0.67$

$$\begin{gathered} x-3668=0.67\left(511\right) \\ x=3668+0.67\left(511\right) \\ x=4010.37 \end{gathered}$$

Therefore the $3rd$ quartile is $4010.37$ grams.