Problem 15BSC

In Exercises, find the range, variance, and standard deviation for the given samph data. Include appropriate units (such as “minutes”) in your results. (The same data were used in Section 3-2 where we found measures of center. Here we find measures of variation.) Then answer the given questions.

Mercury in Sushi Listed below are the amounts of mercury (in parts per million, or ppm) found in tuna sushi sampled at different stores in New York City. The study was sponsored by the New York Times, and the stores (in order) are D’Agostino, Eli’s Manhattan, Fairway, Food Emporium, Gourmet Garage, Grace’s Marketplace, and Whole Foods. What would be the values of the measures of variation if the tuna sushi contained no mercury?

Answer :

Step 1 of 1 :

Mercury in Sushi Listed below are the amounts of mercury (in parts per million, or ppm) found in tuna sushi sampled at different stores in New York City. The study was sponsored by the New York Times, and the stores (in order) are D’Agostino, Eli’s Manhattan, Fairway, Food Emporium, Gourmet Garage, Grace’s Marketplace, and Whole Foods.

Given data

0.56 , 0.75 , 0.10 , 0.95 , 1.25 , 0.54 , 0.88

Now we have to find the range, variance, and standard deviation for the given sample data.

Range :

The range is a simple measure of variation in a set of random variables. It is difference between the biggest and smallest random variable.

Range = Maximum value - Minimum value

Given data

0.56 , 0.75 , 0.10 , 0.95 , 1.25 , 0.54 , 0.88

Therefore, the range of the seven random variables 0.56 , 0.75 , 0.10 , 0.95 , 1.25 , 0.54 , 0.88

Range = Maximum value - Minimum value

Range = 1.25 - 0.1

Range = 1.15 mm

Sample variance :

The variance of a sample is defined by slightly different formula:

The sample variance is the square difference of the data value to the mean divided by the number of values.

s2 =

where s2 is the sample variance.

x is the sample mean.

xi is the ith element from the sample and

n is the number of elements in the sample.

= 0.56 + 0.75 + 0.10 + 0.95 + 1.25 + 0.54 + 0.88 = 0.7186

= 0.7186

s2 =

Therefore variance s2 = 0.133 .

The standard deviation is the square root of the sample variance.

s =

s =

s 0.366

If the tuna sushi contained no mercury, all of the measures would be 0 ppm, and the measures of variation would all be 0 as well.