Chapter 14.2, Problem 8CFU

(a)

To determine

To find: Thenumber of members that used the centres fitness facilities during the month.

Answer to Problem 8CFU

The number of members is 120.

Explanation of Solution

Given:

The table for the given data set is shown in the Table 1

Table 1

Visits	1-5	5-9	9-13	13-17	17-21	21-25	25-29	29-33
Members	2	8	15	6	38	31	13	7

Calculation:

Consider the sum of the number of members is,

  $2+8+15+6+38+31+13+7=120$

As, the sum of the number of member is $120$ then the number of members that used the centre’s fitness facilities during the month is 120.

(b)

To determine

To find: The mean of the data.

Answer to Problem 8CFU

The mean of the data is $19.33$ .

Explanation of Solution

Consider the table shown in Table 2

Table 2

Visits	1-5	5-9	9-13	13-17	17-21	21-25	25-29	29-33
Members	2	8	15	6	38	31	13	7
Mid values $x$	3	7	11	15	19	23	27	31
$f\cdot x$	6	56	165	90	722	713	351	217

From the above table consider the value of the mean is obtained by the formula,

  $\begin{array}{c}\overline{X}=\frac{{\displaystyle \sum _{i=1}^8{f}_i{x}_i}}{{\displaystyle \sum _{i=1}^8{f}_i}}\\ =\frac{2320}{120}\\ =19.33\end{array}$

(c)

To determine

To find: The median class of the data.

Answer to Problem 8CFU

Thus, the median is $17-21$ .

Explanation of Solution

Consider the table shown in Table 1, as the scores 120 in the frequency distribution and the 60 scores are below the median and 60 are above the median.

Then, from the Table 1, the least cumulative frequency that is more than or equal to the 60 to 69 is $17-21$ .

(d)

To determine

To find: The median of the data.

Answer to Problem 8CFU

The value of the mode is $25.89$ .

Explanation of Solution

Consider the proportion to find the value of the medina by finding the ratio of the differences the cumulative frequencies and the upper limits of the classes.

Consider the figure for the proportion is shown in Figure 1

  

Figure 1

Then, by taking the proportion the mode is,

  $\begin{array}{l}\frac{29}{x}=\frac{9}{4}\\ x=12.89\end{array}$

Then, the mode is,

  $\begin{array}{l}{M}_d-13=12.89\\ M_d=25.89\end{array}$