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Daily temperatures for two cities, Salem and Oxford, were recorded for one week.

A 8-column table has 2 rows. Column 1 has entries Salem, Oxford. Column 2 is labeled Sunday with entries 56, 71. Column 3 is labeled Monday with entries 61, 52. Column 4 is labeled Tuesday with entries 55, 58. Column 5 is labeled Wednesday with entries 62, 50. Column 6 is labeled Thursday with entries 58, 89. Column 7 is labeled Friday with entries 61, 60. Column 8 is labeled Saturday with entries 60, 61.

Using statistics, answer the question below:

Which city had the highest variability in temperatures?

1 answer

To determine which city had the highest variability in temperatures, we can calculate the standard deviation of the daily temperatures for both Salem and Oxford. The standard deviation is a measure of dispersion or variability in a set of data.

Temperature Data:

  • Salem: 56, 61, 55, 62, 58, 61, 60
  • Oxford: 71, 52, 58, 50, 89, 60, 61

Calculating the Standard Deviation:

  1. Calculate the mean (average) for each city.

    • For Salem: \[ \text{Mean}_\text{Salem} = \frac{56 + 61 + 55 + 62 + 58 + 61 + 60}{7} = \frac{ 56 + 61 + 55 + 62 + 58 + 61 + 60}{7} \approx 58.71 \]

    • For Oxford: \[ \text{Mean}_\text{Oxford} = \frac{71 + 52 + 58 + 50 + 89 + 60 + 61}{7} = \frac{ 71 + 52 + 58 + 50 + 89 + 60 + 61}{7} \approx 61.14 \]

  2. Calculate the variance for each city.

    • For Salem: \[ \text{Variance}\text{Salem} = \frac{(56 - 58.71)^2 + (61 - 58.71)^2 + (55 - 58.71)^2 + (62 - 58.71)^2 + (58 - 58.71)^2 + (61 - 58.71)^2 + (60 - 58.71)^2}{7} \] \[ \text{Variance}\text{Salem} = \frac{(-2.71)^2 + (2.29)^2 + (-3.71)^2 + (3.29)^2 + (-0.71)^2 + (2.29)^2 + (1.29)^2}{7} \] \[ = \frac{7.3441 + 5.2241 + 13.7641 + 10.8241 + 0.5041 + 5.2241 + 1.6641}{7} \approx \frac{43.63}{7} \approx 6.19 \]

    • For Oxford: \[ \text{Variance}_\text{Oxford}= \frac{(71 - 61.14)^2 + (52 - 61.14)^2 + (58 - 61.14)^2 + (50 - 61.14)^2 + (89 - 61.14)^2 + (60 - 61.14)^2 + (61 - 61.14)^2}{7} \] \[ = \frac{(9.86)^2 + (-9.14)^2 + (-3.14)^2 + (-11.14)^2 + (27.86)^2 + (-1.14)^2 + (-0.14)^2}{7} \] \[ = \frac{97.2996 + 83.5396 + 9.8596 + 124.6596 + 776.6596 + 1.2996 + 0.0196}{7} \approx \frac{1092.39}{7} \approx 155.34 \]

  3. Calculate the standard deviation for each city.

    • For Salem: \[ \text{Std Dev}_\text{Salem} = \sqrt{6.19} \approx 2.49 \]
    • For Oxford: \[ \text{Std Dev}_\text{Oxford} = \sqrt{155.34} \approx 12.47 \]

  4. Comparison: The standard deviation for Salem is approximately 2.49, while the standard deviation for Oxford is approximately 12.47.

Conclusion: Oxford had the highest variability in temperatures, with a standard deviation of approximately 12.47 compared to Salem's standard deviation of approximately 2.49.