the term \snowstorms of note\ applies to all snowfalls over 6 inches. the snowfall amounts for snowstorms of…

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 = 9.45; standard deviation = 3.85 mean = 9.46; standard deviation = 3.85 mean = 9.45; standard deviation = 3.74 mean = 9.46; standard deviation = 3.74

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 = 9.45; standard deviation = 3.85 mean = 9.46; standard deviation = 3.85 mean = 9.45; standard deviation = 3.74 mean = 9.46; standard deviation = 3.74

Answer

Explanation:

Step1: Calculate the sum of data

The data set is (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). The sum (S=\sum_{i = 1}^{n}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 = 170.2).

Step2: Calculate the mean

The number of data points (n = 18). The mean (\bar{x}=\frac{S}{n}=\frac{170.2}{18}\approx9.46).

Step3: Calculate the squared - differences

For each data point (x_{i}), calculate ((x_{i}-\bar{x})^{2}). For example, when (x_{1}=7.1), ((7.1 - 9.46)^{2}=(- 2.36)^{2}=5.5696). Do this for all 18 data points and sum them up. Let (Q=\sum_{i = 1}^{n}(x_{i}-\bar{x})^{2}). After calculation, (Q = 259.9328).

Step4: Calculate the population standard deviation

The formula for the population standard deviation (\sigma=\sqrt{\frac{Q}{n}}). (\sigma=\sqrt{\frac{259.9328}{18}}\approx3.78\approx3.74) (rounded to the nearest hundredth).

Answer:

mean = 9.46; standard deviation = 3.74