Find the range, variance, and standard deviation for each of the two samples, then compare the two sets of results.
-When investigating times required for drive-through service, the following results (in seconds) were obtained. $ \begin{array}{lllllllll}\text { Restaurant A } & 120 & 123 & 153 & 128 & 124 & 118 & 154 & 110\end{array} $
$ \begin{array}{lllllllll}\text { Restaurant B } & 115 & 126 & 147 & 156 & 118 & 110 & 145 & 137\end{array} $
A)
$\begin{array} { l l l l l l l l l } \text { Restaurant A } & 120 & 123 & 153 & 128 & 124 & 118 & 154 & 110 \end{array}$
Restaurant B: $46 \mathrm { sec } ; 285.64 \mathrm { sec } ^ { 2 } ; 16.90 \mathrm { sec }$
There is more variation in the times for restaurant A.
B) Restaurant A: $46 \mathrm { sec } ; 260.79 \mathrm { sec } ^ { 2 } ; 16.15 \mathrm { sec }$
Restaurant B: $44 \mathrm { sec } ; ^ { 285 } .64 \mathrm { sec } ^ { 2 } ; 16.90 \mathrm { sec }$
There is more variation in the times for restaurant $B$.
C) Restaurant A: $44 \mathrm { sec } _ { ; } 260.79 \mathrm { sec } ^ { 2 } ; 16.15 \mathrm { sec }$
Restaurant B: $46 \mathrm { sec } ^ { ; } ; 285.64 \mathrm { sec } ^ { 2 } ; 16.90 \mathrm { sec }$
There is more variation in the times for restaurant B.
D) Restaurant A: $44 \mathrm { sec } ^ { * } 260.79 \mathrm { sec } ^ { 2 } ; 16.15 \mathrm { sec }$
Restaurant B: $46 \mathrm { sec } ^ { ; } ; 245.64 \mathrm { sec } ^ { 2 } ; 15.67 \mathrm { sec }$
There is more variation in the times for restaurant A.

Question

Quizplus · Accepted Answer

To find the range, we subtract the smallest value from the largest value. 
For restaurant A, range = 154 - 110 = 44 seconds. 
For restaurant B, range = 156 - 110 = 46 seconds.

To find the variance and standard deviation, we need to use formulas that require us to find the mean of each set. 
For restaurant A: 
mean = (120+123+153+128+124+118+154+110) / 8 = 128.5 
variance = [(120-128.5)^2 + (123-128.5)^2 + ... + (110-128.5)^2] / 8 = 260.79 seconds squared 
standard deviation = square root of variance = 16.15 seconds

For restaurant B: 
mean = (115+126+147+156+118+110+145+137) / 8 = 134.5 
variance = [(115-134.5)^2 + (126-134.5)^2 + ... + (137-134.5)^2] / 8 = 285.64 seconds squared 
standard deviation = square root of variance = 16.90 seconds

Comparing the results, we see that the range is greater for restaurant B, but the variance and standard deviation are greater for restaurant A. Therefore, the best choice is C, as it correctly identifies that there is more variation in the times for restaurant B based on the range, while there is more variation in the times for restaurant A based on the variance and standard deviation.

Find the Range, Variance, and Standard Deviation for Each of the Two