Calculate the mean, median, mode, standard deviation, and coefficient of variation for the following light bulb lifespan data for Brand A and Brand B.

To find the mean, sum the values and divide by $n$.

$$\text{Mean (Brand A)} = \frac{1134 + 1157 + 1811 + 1858 + 1958}{5} = \frac{7918}{5} = 1583.6 \text{ hours}$$

To find the median, arrange the data in ascending order: $1134, 1157, 1811, 1858, 1958$. Since there are 5 values, the median is the middle ($3^{\text{rd}}$) value.

$$\text{Median (Brand A)} = 1811 \text{ hours}$$

To find the mode, identify the most frequent value. No value repeats in Brand A's data.

$$\text{Mode (Brand A)} = \text{None}$$

To find the mean, sum the values and divide by $n$.

$$\text{Mean (Brand B)} = \frac{1456 + 1787 + 1611 + 1872 + 1853}{5} = \frac{8579}{5} = 1715.8 \text{ hours}$$

To find the median, arrange the data in ascending order: $1456, 1611, 1787, 1853, 1872$. Since there are 5 values, the median is the middle ($3^{\text{rd}}$) value.

$$\text{Median (Brand B)} = 1787 \text{ hours}$$

To find the mode, identify the most frequent value. No value repeats in Brand B's data.

$$\text{Mode (Brand B)} = \text{None}$$

The formula for sample standard deviation is $s = \sqrt{\frac{\sum (x_i - \bar{x})^2}{n-1}}$.
The sum of squared differences from the mean for Brand A is:
$(1134 - 1583.6)^2 + (1157 - 1583.6)^2 + (1811 - 1583.6)^2 + (1858 - 1583.6)^2 + (1958 - 1583.6)^2$
$= (-449.6)^2 + (-426.6)^2 + (227.4)^2 + (274.4)^2 + (374.4)^2$
$= 202140.16 + 182087.56 + 51711.76 + 75295.36 + 140175.36 = 651410.2$

$$\text{Standard Deviation (Brand A)} = \sqrt{\frac{651410.2}{5-1}} = \sqrt{\frac{651410.2}{4}} = \sqrt{162852.55} \approx 403.55 \text{ hours}$$

The coefficient of variation (CV) is $CV = \frac{s}{\bar{x}} \times 100\%$.

$$\text{CV (Brand A)} = \frac{403.55}{1583.6} \times 100\% \approx 0.2548 \times 100\% \approx 25.48\%$$

The sum of squared differences from the mean for Brand B is:
$(1456 - 1715.8)^2 + (1787 - 1715.8)^2 + (1611 - 1715.8)^2 + (1872 - 1715.8)^2 + (1853 - 1715.8)^2$
$= (-259.8)^2 + (71.2)^2 + (-104.8)^2 + (156.2)^2 + (137.2)^2$
$= 67496.04 + 5069.44 + 10983.04 + 24398.44 + 18824.64 = 126771.6$

$$\text{Standard Deviation (Brand B)} = \sqrt{\frac{126771.6}{5-1}} = \sqrt{\frac{126771.6}{4}} = \sqrt{31692.9} \approx 177.99 \text{ hours}$$ $$\text{CV (Brand B)} = \frac{177.99}{1715.8} \times 100\% \approx 0.1037 \times 100\% \approx 10.37\%$$

A better brand of bulb would have a longer average lifespan and more consistent performance (lower variability).

• Mean Lifespan: Brand A (1583.6 hours) vs. Brand B (1715.8 hours). Brand B has a higher average lifespan.
• Consistency (CV): Brand A ($25.48\%$) vs. Brand B ($10.37\%$). Brand B has a lower coefficient of variation, indicating more consistent performance.

Based on both a higher average lifespan and greater consistency, Brand B is better.

Here are the results:
Brand A:

• Mean: $\boxed{\text{1583.6 hours}}$
• Median: $\boxed{\text{1811 hours}}$
• Mode: $\boxed{\text{None}}$
• Standard Deviation: $\boxed{\text{403.55 hours}}$
• Coefficient of Variation: $\boxed{\text{25.48\%}}$

Brand B:

• Mean: $\boxed{\text{1715.8 hours}}$
• Median: $\boxed{\text{1787 hours}}$
• Mode: $\boxed{\text{None}}$
• Standard Deviation: $\boxed{\text{177.99 hours}}$
• Coefficient of Variation: $\boxed{\text{10.37\%}}$

Which brand is better?
$\boxed{\text{Brand B}}$