Answer:
Step-by-step explanation:
Mean is the ratio of sum of the dataset to the sample size. Mathematically:
[tex]\overline x = \frac{\sum Xi}{N}[/tex]
Xi are the individual periods
N is the sample size
[tex]\sum Xi = 1.06+1.31+1.28+0.99+1.48+1.37+0.98+1.31+1.59+1.55\\\sum Xi = 12.92[/tex]
N = 10
Substitute
[tex]\overline x = \frac{12.92}{10}\\\overline x = 1.292[/tex]
hence the mean of the samples is 1.292
For the standard deviation:
[tex]\sigma = \sqrt{\frac{\sum (x-\overline x)^2}{N} } \\[/tex]
[tex]\sum (x-\overline x)^2 = (1.06-1.292)^2+ (1.31-1.292)^2+ (1.28-1.292)^2+ (10.99-1.292)^2+ (1.48-1.292)^2+ (1.37-1.292)^2+ (0.98-1.292)^2+ (1.31-1.292)^2+ (1.59-1.292)^2+ (1.55-1.292)^2\\\sum (x-\overline x)^2 = 0.43996[/tex]
Substitute into the formula:
[tex]\sigma = \sqrt{\frac{0.43996}{10} }} \\\sigma = \sqrt{{0.043996}}} \\\sigma = 0.2098[/tex]
Hence the standard deviation is 0.2098
