91Ó°ÊÓ

The given mean is $206mg/dL$and standard deviation is$44.7mg/dL$.

Draw the figure showing the required shaded region and its delimiting x-values, which are $150$and $250$.

Now, we need to compute the z-scores for the x-values $150$and $250$.

$$\begin{gathered} x=150→ \\ z=\frac{150-206}{44.7} \\ =-1.25 \\ x=250→ \\ z=\frac{250-206}{44.7} \\ =0.98 \end{gathered}$$

We need to find the area under the standard normal curve that lies between $-1.25$and $0.98$. The area to the left of $-1.25$is $0.1056$, and the area to the left of $0.98$is $0.8365$. The required area, shaded in the figure is $0.8365-0.1056=0.7309$.

On interpreting, we can say,$73.09\%$of females$20years$old or older have a serum total cholesterol level between$150mg/dL$and$250mg/dL$.

Draw the figure showing the required shaded region,

We need to compute the z-score for the x-value $19$,

$$\begin{gathered} x=220→ \\ z=\frac{220-206}{44.7} \\ z=0.31 \end{gathered}$$

We need to find the area under the standard normal curve that lies below $0.31$. The area to the left of $0.31$is $0.6217$. The required area shaded in the figure is $0.6217$.

The required percentage of U.S. females 20 years is $62.17\%$.

The z-score corresponding to $P_{25}$is the one having an area $0.25$ to its left under the standard normal curve. From standard normal table, that z-score is $-0.67$approximately.

We must find the x-value having the z-score is $0.67$, the length that is $0.67$standard deviations below the mean. It is $176.05$.

The first quartile or $25$th percentile for serum cholesterol level is $176.05mg/dL$. On interpreting, we can say, $25\%$of the females have serum cholesterol level below $176.05mg/dL$.

The z-score corresponding to $P_{50}$is the one having an area $0.5$ to its left under the standard normal curve. From standard normal table, that z-score is $0$.

We must find the x-value having the z-score is $0$, the length that is $0$standard deviations below the mean. It is $206$mg/dL. On interpreting, we can say, $50\%$of the females have serum cholesterol level below $206mg/dL$.

The z-score corresponding to $P_{75}$is the one having an area $0.75$ to its left under the standard normal curve. From standard normal table, that z-score is $0.67$.

We must find the x-value having the z-score is $0.67$, the length that is $0.67$standard deviations below the mean. It is $235.95$.

The third quartile or $75th$percentile for serum cholesterol level is $235.95mg/dL$.

On interpreting, we can say,$75\%$of the females have serum cholesterol level below $235.95mg/dL$.

The z-score corresponding to $P_{40}$, fourth decile is the one having an area $0.4$ to its left under the standard normal curve. From standard normal table, that z-score is $-0.25$.

We must find the x-value having the z-score is $-0.25$, the length that is $-0.25$ standard deviations below the mean. It is $194.83$.

The $40th$percentile for serum cholesterol level is localid="1652472851312" $194.83mg/dL$.

On interpreting, we can say, $40\%$of the female have serum cholesterol level below$194.83mg/dL$.