Question

the term \snowstorms of note\ applies to all snowfalls over 6 inches. the snowfall amounts for snowstorms of note in utica, new york, over a four - year period are as follows: 7.1, 9.2, 8.0, 6.1, 14.4, 8.5, 6.1, 6.8, 7.7, 21.5, 6.7, 9.0, 8.4, 7.0, 11.5, 14.1, 9.5, 8.6. what are the mean and population standard deviation for these data, to the nearest hundredth? mean = 9.46; standard deviation = 3.85 mean = 9.46; standard deviation = 3.74 mean = 9.45; standard deviation = 3.85 mean = 9.45; standard deviation = 3.74

Explanation:

Step1: Calculate the mean

The mean $\bar{x}=\frac{\sum_{i = 1}^{n}x_{i}}{n}$, where $x_{i}$ are the data - points and $n$ is the number of data - points.
$n = 18$, and $\sum_{i=1}^{18}x_{i}=7.1 + 9.2+8.0 + 6.1+14.4+8.5+6.1+6.8+7.7+21.5+6.7+9.0+8.4+7.0+11.5+14.1+9.5+8.6=170.3$.
$\bar{x}=\frac{170.3}{18}\approx9.46$.

Step2: Calculate the population standard deviation

The formula for the population standard deviation $\sigma=\sqrt{\frac{\sum_{i = 1}^{n}(x_{i}-\bar{x})^{2}}{n}}$.
First, calculate $(x_{i}-\bar{x})^{2}$ for each $x_{i}$:
$(7.1 - 9.46)^{2}=(-2.36)^{2}=5.5696$, $(9.2 - 9.46)^{2}=(-0.26)^{2}=0.0676$, $(8.0 - 9.46)^{2}=(-1.46)^{2}=2.1316$, $(6.1 - 9.46)^{2}=(-3.36)^{2}=11.2896$, $(14.4 - 9.46)^{2}=(4.94)^{2}=24.4036$, $(8.5 - 9.46)^{2}=(-0.96)^{2}=0.9216$, $(6.1 - 9.46)^{2}=(-3.36)^{2}=11.2896$, $(6.8 - 9.46)^{2}=(-2.66)^{2}=7.0756$, $(7.7 - 9.46)^{2}=(-1.76)^{2}=3.0976$, $(21.5 - 9.46)^{2}=(12.04)^{2}=144.9616$, $(6.7 - 9.46)^{2}=(-2.76)^{2}=7.6176$, $(9.0 - 9.46)^{2}=(-0.46)^{2}=0.2116$, $(8.4 - 9.46)^{2}=(-1.06)^{2}=1.1236$, $(7.0 - 9.46)^{2}=(-2.46)^{2}=6.0516$, $(11.5 - 9.46)^{2}=(2.04)^{2}=4.1616$, $(14.1 - 9.46)^{2}=(4.64)^{2}=21.5296$, $(9.5 - 9.46)^{2}=(0.04)^{2}=0.0016$, $(8.6 - 9.46)^{2}=(-0.86)^{2}=0.7396$.
$\sum_{i = 1}^{18}(x_{i}-\bar{x})^{2}=5.5696+0.0676 + 2.1316+11.2896+24.4036+0.9216+11.2896+7.0756+3.0976+144.9616+7.6176+0.2116+1.1236+6.0516+4.1616+21.5296+0.0016+0.7396 = 253.82$.
$\sigma=\sqrt{\frac{253.82}{18}}\approx3.74$.

Answer:

mean = 9.46; standard deviation = 3.74