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$$\begin{gathered} 渭=3668 \\ 蟽=511 \end{gathered}$$

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:

$$\begin{gathered} 渭鈭�3蟽=3668鈭�3\left(511\right)=2135 \\ 渭+3蟽=3668+3\left(511\right)=5201 \end{gathered}$$

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.

The formula used:$z=\frac{x鈭�渭}{蟽}$

The $z-$score is the

$$\begin{gathered} z=\frac{x鈭�渭}{蟽} \\ 蟽=\frac{2500鈭�3668}{511} \\ =鈭�2.29 \end{gathered}$$

Using the normal probability table, get the associated probability.

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

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

The formula used:$z=\frac{x鈭�渭}{蟽}$

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$-0.6+.07=-0.67.$

$z=鈭�0.67$

The $z-$score is

$z=\frac{x鈭�渭}{蟽}=\frac{x鈭�3668}{511}$

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

$\frac{x鈭�3668}{511}=鈭�0.67$

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

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