Answer
a. The mean temperature for March, June, September, and December in San Francisco, CA is $56.75^{o}$ with a standard deviation of $3.960^{o}$.
b.The mean temperature for March, June, September, and December in Albuquerque, NM is $56.75^{o}$ with a standard deviation of $16.24^o$.
Although the means are identical, the standard deviations' differences show how representative the mean is.
Work Step by Step
a. The mean is calculated by adding all the temperatures together and then dividing by the number of temperatures.
$\bar{x} = \frac{\sum x}{n}$
$\bar{x} = \frac{54+59+62+52}{4}$
$\bar{x} = \frac{227}{4}$
$\bar{x} = 56.75$
The standard deviation is calculated by taking the square root of the sum of the difference of the squared deviations from the mean, divided by the number of temperatures.
$\sigma = \sqrt{\frac{\sum (x-\bar{x})^2}{n}}$
$\sigma = \sqrt{\frac{(54-56.75)^2+(59-56.75)^2+(62-56.75)^2+(52-56.75)^2}{4}}$
$\sigma = \sqrt{\frac{(-2.75)^2+(2.25)^2+(5.25)^2+(-4.75)^2}{4}}$
$\sigma = \sqrt{\frac{(7.5625)+(5.0625)+(27.5625)+(22.5625)}{4}}$
$\sigma = \sqrt{\frac{62.75}{4}}$
$\sigma = \sqrt{15.6875}$
$\sigma = 3.960$
b.The mean is calculated by adding all the temperatures together and then dividing by the number of temperatures.
$\bar{x} = \frac{\sum x}{n}$
$\bar{x} = \frac{46+75+70+36}{4}$
$\bar{x} =56.75$
The standard deviation is calculated by taking the square root of the difference of the sum of the difference of the squared deviations from the mean, divided by the number of temperatures.
$\sigma = \sqrt{\frac{\sum (x-\bar{x})^2}{n}}$
$\sigma = \sqrt{\frac{\sum ({46-56.75})^2+{(75-56.75})^2+{(70-56.75})^2+({36-56.75})^2}{4}}$
$\sigma = \sqrt{\frac{\sum ({-10.75})^2+({18.25})^2+{(13.25})^2+({-20.75})^2}{4}}$
$\sigma = \sqrt{\frac{\sum ({115.5625+333.0625+175.5625+430.5625})}{4}}$
$\sigma = \sqrt{\frac{\ {1054.75}}{4}}$
$\sigma = \sqrt{263.6875}$
$\sigma = 16.2384574$
