91Ó°ÊÓ

Consider the given question,

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

The table is given below,

(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 $62,63$,

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

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

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.

(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 $65,70$,

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

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

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.