Q. 6.98 - Elementary Statistics

Question

According to the National Health and Nutrition Examination Survey, published by the National Center for Health Statistics, the serum total cholesterol level of U.S. females $20$ years old or older is normally distributed with a mean of $206mg/dL$ and a standard deviation of $44.7mg/dL$.

Part (a): Determine the percentage of U.S. females $20$ years old or older who have a serum total cholesterol level between $150mg/dL$ and $250mg/dL$.

Part (b): Determine the percentage of U.S. females $20$ years or older who have a serum total cholesterol level below $220mg/dL$.

Part (c): Obtain and interpret the quartiles for serum total cholesterol level of U.S. females $20$ years old or older.

Part (d): Find the interpret the fourth decile for serum total cholesterol level of U.S. females $20$years old or older.

Step-by-Step Solution

Answer

$20$Part (a): The percentage of U.S. females $20$ years old or older who have a serum total cholesterol level between $150 mg/dL$ and $250 mg/dL$ is $73.09\%$.

Part (b): The percentage of U.S. females $20$ years old years or older who have a serum total cholesterol level below $220 mg/dL$ is $62.17\%$.

Part (c): The first quartile is $176.05mg/dL$. $25\%$of the females have serum cholesterol level below $176.05mg/dL$.

The second quartile is $206mg/dL$. $50\%$of the females have serum cholesterol level below $206mg/dL$.

The third quartile is $235.95mg/dL$. $75\%$ of the females have serum cholesterol level below $235.95mg/dL$.

Part (d): $40\%$of the female have serum cholesterol level below $194.83mg/dL$.

Part (a) Step 1. Given information.

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

Part (a) Step 2. Draw the figure showing the required shaded region.

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 $250$.

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

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 $150 mg/dL$ and $250 mg/dL$.

Part (b) Step 1. Draw the figure showing the required shaded region.

Draw the figure showing the required shaded region,

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

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

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 $62.17\%$.

Part (c) Step 1. Determine the first and second quartiles for serum total cholesterol level of U.S. females.

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 $176.05 mg/dL$. On interpreting, we can say, $25\%$ of the females have serum cholesterol level below $176.05 mg/dL$.

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 $206mg/dL$.

Part (c) Step 2. Determine the third quartiles for serum total cholesterol level of U.S. females.

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 $235.95mg/dL$.

Part (d) Step 1. Determine the fourth decile for serum total cholesterol level of U.S. females.

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 $194.83mg/dL$.

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