Chapter 2, Problem R2.5RE

(a)

To determine

To find: the blanks of about 99.7 percent of the babies had birth weight between __ and __ grams.

Answer to Problem R2.5RE

2135, 5201

Explanation of Solution

Given:

  $\begin{array}{c}\mu =3668\\ \sigma =\text{ 511}\end{array}$

Calculation:

The mid 68% 0f a normal distribution comes in 1 standard deviation from the mean of the distribution.

The mid 95% 0f a normal distribution comes in 2 standard deviation from the mean of the distribution

The mid 99.7% 0f a normal distribution comes in 3 standard deviation from the mean of the distribution

Finding the value that lays 3 standard deviations from the mean:

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

This then means that 99.7% of the observations are predicted to be between 2135 and 5201, which means that about 99.7% of the babies has a birth weight between 2135 and 5201 grams.

(b)

To determine

To find: the percent of babies would be identified as low birth weight.

Answer to Problem R2.5RE

1.10%

Explanation of Solution

Given:

  $\begin{array}{c}\mu =3668\\ \sigma =\text{511}\\ x\text{=2500}\end{array}$

Formula used:

  $z=\frac{x-\mu }{\sigma }$

Calculation:

The z-score is the

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

Finding the associating probability using the normal probability table $\begin{array}{c}P\left(X<2500\right)=P\left(Z<-2.29\right)\\ =0.0110\\ =1.10\%\end{array}$

Therefore 1.10% of the babies had a birth weight less than 2500 grams, which expected as having a low birth weight.

(c)

To determine

To find: the quartiles of the birth weight distribution.

Answer to Problem R2.5RE

1^st quartile: 3325.63 grams

3^rd quartile: 4010.37 grams

Explanation of Solution

Given:

  $\begin{array}{c}\mu =3668\\ \sigma =\text{511}\end{array}$

Formula used:

  $z=\frac{x-\mu }{\sigma }$

Calculation:

The 1^st quartile has the property that 25 percent of the data values are below the first quartile.

So find the z-score that associate with a probability of 25% or 0.25 in the normal probability. It is observed the closest probability is 0.2514 which comes in the row -0.6 and in the column .07 of the normal probability table and therefore the associating z-score is then -0.6 + .07 = -0.67.

  $z=-0.67$

The z-score is

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

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

  $\frac{x-3668}{511}=-0.67$

  $x-3668=-0.67\left(511\right)$

  $x=3325.63$

Therefore the first quartile is 3325.63 grams.

Third quartile

The 3^rd quartile has the property that 75 percent of the data values are above the first quartile.

Finding the z-score that associates with a probability of 75

  $z=0.67$

The z-score is

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

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

  $\frac{x-3668}{511}=0.67$

  $x-3668=0.67\left(511\right)$

  $x=3668+0.67\left(511\right)$

  $x=4010.37$

Therefore the 3^rd quartile is 4010.37 grams.