The lives (in hours of continuous use) of 100 randomly selected flashlight batteries are given in the table below:

\begin{tabular}{lc}
\hline
Interval & Frequency \\
\hline
[tex]$6.95-7.45$[/tex] & 2 \\
[tex]$7.45-7.95$[/tex] & 10 \\
[tex]$7.95-8.45$[/tex] & 19 \\
[tex]$8.45-8.95$[/tex] & 37 \\
[tex]$8.95-9.45$[/tex] & 18 \\
[tex]$9.45-9.95$[/tex] & 11 \\
[tex]$9.95-10.45$[/tex] & 3 \\
\hline
\end{tabular}

a. Find the mean of the battery lives. [tex]$\square$[/tex] hrs

(Type an integer or a decimal. Round to two decimal places.)



Answer :

To find the mean of the battery lives, we will follow these steps:

1. Determine the midpoint of each interval:
For each interval, calculate the midpoint, which is the average of the lower and upper bounds of the interval. These midpoints represent the approximate central value of each interval. The midpoints are calculated as follows:
- For the interval [tex]\(6.95 - 7.45\)[/tex], the midpoint is [tex]\((6.95 + 7.45) / 2 = 7.20\)[/tex].
- For the interval [tex]\(7.45 - 7.95\)[/tex], the midpoint is [tex]\((7.45 + 7.95) / 2 = 7.70\)[/tex].
- For the interval [tex]\(7.95 - 8.45\)[/tex], the midpoint is [tex]\((7.95 + 8.45) / 2 = 8.20\)[/tex].
- For the interval [tex]\(8.45 - 8.95\)[/tex], the midpoint is [tex]\((8.45 + 8.95) / 2 = 8.70\)[/tex].
- For the interval [tex]\(8.95 - 9.45\)[/tex], the midpoint is [tex]\((8.95 + 9.45) / 2 = 9.20\)[/tex].
- For the interval [tex]\(9.45 - 9.95\)[/tex], the midpoint is [tex]\((9.45 + 9.95) / 2 = 9.70\)[/tex].
- For the interval [tex]\(9.95 - 10.45\)[/tex], the midpoint is [tex]\((9.95 + 10.45) / 2 = 10.20\)[/tex].

So, the midpoints are: [tex]\(7.20, 7.70, 8.20, 8.70, 9.20, 9.70, 10.20\)[/tex].

2. Calculate the sum of the products of frequencies and midpoints:
Multiply each midpoint by its corresponding frequency and sum these products. This can be represented as [tex]\(\sum (f \cdot m)\)[/tex] where [tex]\(f\)[/tex] is the frequency and [tex]\(m\)[/tex] is the midpoint.
[tex]\[ \begin{aligned} & (2 \cdot 7.20) + \\ & (10 \cdot 7.70) + \\ & (19 \cdot 8.20) + \\ & (37 \cdot 8.70) + \\ & (18 \cdot 9.20) + \\ & (11 \cdot 9.70) + \\ & (3 \cdot 10.20) \end{aligned} \][/tex]

Calculating these products:
[tex]\[ \begin{aligned} & 2 \cdot 7.20 = 14.40 \\ & 10 \cdot 7.70 = 77.00 \\ & 19 \cdot 8.20 = 155.80 \\ & 37 \cdot 8.70 = 321.90 \\ & 18 \cdot 9.20 = 165.60 \\ & 11 \cdot 9.70 = 106.70 \\ & 3 \cdot 10.20 = 30.60 \\ \end{aligned} \][/tex]

Summing these products:
[tex]\[ 14.40 + 77.00 + 155.80 + 321.90 + 165.60 + 106.70 + 30.60 = 871.99 \][/tex]

3. Calculate the total frequency:
Sum all the frequencies:
[tex]\[ 2 + 10 + 19 + 37 + 18 + 11 + 3 = 100 \][/tex]

4. Calculate the mean of the battery lives:
Divide the sum of the products by the total frequency:
[tex]\[ \text{Mean battery life} = \frac{\text{sum of the products}}{\text{total frequency}} = \frac{871.99}{100} = 8.72 \][/tex]

The mean battery life is [tex]\(8.72\)[/tex] hours (rounded to two decimal places).