Question

1. the term \snowstorms of note\ applies to all snowfalls over 6 inches. the total amounts of snowstorms of note in utica, new york, over a four - year period are given: 3.1, 3.2, 3.5, 5.4, 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 sample standard deviation for these data? a) mean = 9.46; standard deviation = 3.74 b) mean = 9.46; standard deviation = 3.85 c) mean = 9.45; standard deviation = 3.85 11. what is the sample variance for the data given: 5, 10, 7, 12, 0, 20, 15, 22, 8, 2 a) 52.77 b) 47.47 c) 81 d) 102.01 12. for a population of five university students with gpas of 2.5, 2.3, 1.7, 1.4, and 1.1, find the population standard deviation. a) 0.22 b) 0.41 c) 0.35 d) 0.53

Explanation:

Step1: Calculate the mean for question 10

The data set for question 10 is \(3.1,3.2,3.5,5.4,14.4,8.5,6.1,6.8,7.7,21.5,6.7,9.0,8.4,7.0,11.5,14.1,8.5,8.6\). The formula for the mean \(\bar{x}=\frac{\sum_{i = 1}^{n}x_{i}}{n}\), where \(n = 18\) and \(\sum_{i=1}^{18}x_{i}=3.1 + 3.2+3.5+5.4+14.4+8.5+6.1+6.8+7.7+21.5+6.7+9.0+8.4+7.0+11.5+14.1+8.5+8.6=170.2\). So \(\bar{x}=\frac{170.2}{18}\approx9.46\).

Step2: Calculate the sample - standard deviation for question 10

The formula for the sample - standard deviation \(s=\sqrt{\frac{\sum_{i = 1}^{n}(x_{i}-\bar{x})^{2}}{n - 1}}\). First, calculate \((x_{i}-\bar{x})^{2}\) for each \(x_{i}\), sum them up: \(\sum_{i = 1}^{18}(x_{i}-\bar{x})^{2}=(3.1 - 9.46)^{2}+(3.2 - 9.46)^{2}+\cdots+(8.6 - 9.46)^{2}=238.7928\). Then \(s=\sqrt{\frac{238.7928}{18 - 1}}\approx3.74\).

Step3: Calculate the sample variance for question 11

The data set for question 11 is \(5,10,7,12,0,20,15,22,8,2\). First, find the mean \(\bar{x}=\frac{5 + 10+7+12+0+20+15+22+8+2}{10}=\frac{101}{10}=10.1\). The formula for the sample variance \(s^{2}=\frac{\sum_{i = 1}^{n}(x_{i}-\bar{x})^{2}}{n - 1}\). Calculate \((x_{i}-\bar{x})^{2}\) for each \(x_{i}\), sum them up: \(\sum_{i=1}^{10}(x_{i}-10.1)^{2}=(5 - 10.1)^{2}+(10 - 10.1)^{2}+\cdots+(2 - 10.1)^{2}=474.9\). Then \(s^{2}=\frac{474.9}{10 - 1}\approx52.77\).

Answer:

1. a) mean = 9.46; standard deviation = 3.74
2. a) 52.77
3. No data provided for question 12 to calculate.