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

[tex]\[
\begin{tabular}{lc}
Interval & Frequency \\
\hline
$6.95-7.45$ & 2 \\
$7.45-7.95$ & 8 \\
$7.95-8.45$ & 19 \\
$8.45-8.95$ & 37 \\
$8.95-9.45$ & 20 \\
$9.45-9.95$ & 11 \\
$9.95-10.45$ & 3 \\
\end{tabular}
\][/tex]

a. Find the mean of the battery lives. [tex]$\square$[/tex] hrs
(Type an integer or a decimal. Round to two decimal places.)

b. Find the standard deviation of the battery lives. [tex]$\square$[/tex] hrs
(Type an integer or a decimal. Round to two decimal places.)


Sagot :

Alright, let's solve this problem step by step.

### a. Finding the Mean of Battery Lives

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

1. Identify the Intervals and Frequencies:
- The intervals and corresponding frequencies are given as:
[tex]\[ \begin{array}{cc} \text{Interval} & \text{Frequency} \\ \hline 6.95 - 7.45 & 2 \\ 7.45 - 7.95 & 8 \\ 7.95 - 8.45 & 19 \\ 8.45 - 8.95 & 37 \\ 8.95 - 9.45 & 20 \\ 9.45 - 9.95 & 11 \\ 9.95 - 10.45 & 3 \\ \end{array} \][/tex]

2. Calculate Midpoints of Each Interval:
- The midpoint of each interval is calculated by taking the average of the lower and upper boundaries:
[tex]\[ \begin{array}{cc} \text{Interval} & \text{Midpoint} \\ \hline 6.95 - 7.45 & \frac{6.95 + 7.45}{2} = 7.20 \\ 7.45 - 7.95 & \frac{7.45 + 7.95}{2} = 7.70 \\ 7.95 - 8.45 & \frac{7.95 + 8.45}{2} = 8.20 \\ 8.45 - 8.95 & \frac{8.45 + 8.95}{2} = 8.70 \\ 8.95 - 9.45 & \frac{8.95 + 9.45}{2} = 9.20 \\ 9.45 - 9.95 & \frac{9.45 + 9.95}{2} = 9.70 \\ 9.95 - 10.45 & \frac{9.95 + 10.45}{2} = 10.20 \\ \end{array} \][/tex]

3. Calculate the Total Frequency:
- The total frequency is the sum of all frequencies:
[tex]\[ \text{Total Frequency} = 2 + 8 + 19 + 37 + 20 + 11 + 3 = 100 \][/tex]

4. Calculate the Weighted Sum of Midpoints:
- Multiply each midpoint by the corresponding frequency and sum the results:
[tex]\[ \text{Weighted Sum} = (7.20 \times 2) + (7.70 \times 8) + (8.20 \times 19) + (8.70 \times 37) + (9.20 \times 20) + (9.70 \times 11) + (10.20 \times 3) \][/tex]

5. Calculate the Mean:
- The mean is the weighted sum divided by the total frequency:
[tex]\[ \text{Mean} = \frac{\text{Weighted Sum}}{\text{Total Frequency}} \approx \frac{875}{100} = 8.75 \][/tex]

Hence, the mean of the battery lives is [tex]\( 8.75 \)[/tex] hours.

### b. Finding the Standard Deviation of Battery Lives

To find the standard deviation, we follow these steps:

1. Calculate the Difference of Each Midpoint from the Mean:
- Subtract the mean (8.75) from each midpoint and square the result:
[tex]\[ (7.20 - 8.75)^2, (7.70 - 8.75)^2, (8.20 - 8.75)^2, (8.70 - 8.75)^2, (9.20 - 8.75)^2, (9.70 - 8.75)^2, (10.20 - 8.75)^2 \][/tex]

2. Multiply Each Squared Difference by the Corresponding Frequency:
- Multiply each squared difference by its corresponding frequency and sum the results:
[tex]\[ S = 2 \times (7.20 - 8.75)^2 + 8 \times (7.70 - 8.75)^2 + 19 \times (8.20 - 8.75)^2 + 37 \times (8.70 - 8.75)^2 + 20 \times (9.20 - 8.75)^2 + 11 \times (9.70 - 8.75)^2 + 3 \times (10.20 - 8.75)^2 \][/tex]

3. Calculate the Variance:
- Divide the sum [tex]\( S \)[/tex] by the total frequency minus 1 to get the variance:
[tex]\[ \text{Variance} = \frac{S}{100 - 1} = \frac{S}{99} \][/tex]

4. Calculate the Standard Deviation:
- The standard deviation is the square root of the variance:
[tex]\[ \text{Standard Deviation} \approx 0.63 \][/tex]

Hence, the standard deviation of the battery lives is [tex]\( 0.63 \)[/tex] hours.

### Final Answers:
a. The mean of the battery lives is [tex]\( 8.75 \)[/tex] hours. \\
b. The standard deviation of the battery lives is [tex]\( 0.63 \)[/tex] hours.