Question

Listed below are the amounts of mercury (in parts per million, or ppm) found in tuna sushi sampled at different stores. Find the range, variance, and standard deviation for the set of data. What would be the values of the measures of variation if the tuna sushi contained no mercury?

1.23

0.18

0.62

0.96

0.92

0.65

0.76

Sample variance = ____ ppm2

Answer 1

Solution :

Given sample data are: 1.23 , 0.18 , 0.62 , 0.96 , 0.92 , 0.65 , 0.76

n = 7

Range = maximum - minimum = 1.23 - 0.18 = 1.05

mean = = X / n = 5.32 / 7 = 0.76

sample variance = s² = (X - )² / n-1

= (1.23 - 0.76)² + (0.18 - 0.76)² + (0.62 - 0.76)² + (0.96 - 0.76)² +

(0.92 - 0.76)² + (0.65 - 0.76)² + (0.76 - 0.76)² / 7 - 1

= 0.6546 / 6

= 0.1091

Sample variance = 0.1091 ppm2

sample standard deviation = 0.1091 = 0.3303 ppm2

The measure of variation would be 0 .