Question

The heights of the $3264$female students attending a midwestern college are approximately normally distributed with mean $64.4inches$and standard deviation $2.4inches$. Thus we can use the normal distribution of with role="math" localid="1652511343344" $\mu =64.4,\sigma =2.4$to approximate the percentage of these students having heights within any specified range. In each part, (i): Obtain the exact percentage from Table $6.1$.

(ii): Use the normal distribution to approximate the percentage.

(iii): Compare your answers.

Part (a): The percentage of female students with height between$62$and $63$inches.

Part (b): The percentage of female students with height between $65$and $70$inches.

Short answer

Part (a): (i) The percentage of female students who are between $62$and $63$inches is $11.7\%$.

(ii) $12.23\%$of female students have heights between $62$and $63$inches.

(iii) Both the results are approximately the same.

Part (b): (i) The percentage of female students who are between $65-70$inches is $39.83\%$.

(ii) $39.14\%$of female students have the heights between $65-70$inches.

(iii) Both the results are approximately the same.

Step-by-step solution

Part (a) Step 1. Given information.

Consider the given question,

The mean is $64.4$and standard deviation is $2.4$.

The table is given below,

Part (a) Step 2. Use the table.

(i) Consider the given data table,

The percentage of female students who are between $62$and $63$inches is $11.7\%$.

(ii) The normal curve associated with the variable is shown in the below figure. Note that tick marks are unit apart, that is the distance between successive tick marks is equal to the standard deviation. The figure below shows the required shaded region and its delimiting x-values which are $62,63$,

Part (a) Step 3. Compute the z-score for the given x-values.

We need to compute the z-score for the x-values $62,63$,

$$\begin{gathered} x=62\to \\ z=\frac{62-64.4}{2.4} \\ =-1 \\ x=63\to \\ z=\frac{63-64.4}{2.4} \\ =-0.58 \end{gathered}$$

We need to find the area under the standard normal curve that lies between $-1,-0.58$. The area to the left of $-1$is $0.1587$and the area to the left of $-0.58$is $0.281$. The required area shaded in the figure is $0.281-0.1587=0.1223$.

Therefore, we can say $12.23\%$of female students have heights between $62,63$inches.

(iii) Both the results are approximately the same.

Part (b) Step 1. Use the table.

(i) Consider the given data table,

The percentage of female students who are between $65,70$inches can be obtained by adding the corresponding relative frequencies of $65-66,66-67,...,69-70$.

$0.1575+0.11+0.0735+0.0374+0.0199=0.3983$

Therefore, the percentage of students who are between $65,70$inches is $39.83\%$.

(ii) The normal curve associated with the variable is shown in the below figure. Note that tick marks are unit apart, that is the distance between successive tick marks is equal to the standard deviation. The figure below shows the required shaded region and its delimiting x-values which are $65,70$,

Part (b) Step 2. Compute the z-score for the given x-values.

We need to compute the z-score for the x-values $65,70$,

$$\begin{gathered} x=65\to \\ z=\frac{65-64.4}{2.4} \\ =0.25 \\ x=70\to \\ z=\frac{70-64.4}{2.4} \\ =2.33 \end{gathered}$$

We need to find the area under the standard normal curve that lies between $0.25,2.33$. The area to the left of $0.25$is $0.5987$and the area to the left of $2.33$is $0.9901$. The required area shaded in the figure is role="math" localid="1652513679740" $0.9901-0.5987=0.3914$.

Therefore, $39.14\%$of female students have the heights between $65,70$inches.

(iii) Both the results are approximately the same.