Question

Two different brands, namely Brand A and Brand B, of mobile batteries are tested and the following data on their lifetimes in months are obtained:

• A. \(CV_A<CV_B\) and \(s_A<s_B\)
• B. \(CV_A<CV_B\) and \(s_A>s_B\)
• C. \(CV_A>CV_B\) and \(s_A<s_B\)
• D. \(CV_A>CV_B\) and \(s_A>s_B\)

Hint:

The sample standard deviation measures absolute variation, while the coefficient of variation compares relative variation by dividing the standard deviation by the mean.

Answer 1

The Correct Option is B

Step 1: Get the two means.
Brand A: $27,28,30,32,33$ averages to $30$. Brand B: $17,19,20,21,23$ averages to $20$.

Step 2: Find the standard deviations.
A has squared deviations summing to $26$, so $s_A^2=\frac{26}{4}=6.5$. B has them summing to $20$, so $s_B^2=\frac{20}{4}=5$. Hence $s_A=\sqrt{6.5}>\sqrt5=s_B$.

Step 3: Compare the coefficients of variation.
$CV=s/\bar x$. Compare squares: $CV_A^2=\frac{6.5}{900}=\frac{13}{1800}$, $CV_B^2=\frac{5}{400}=\frac1{80}$.

Step 4: Decide the order.
Since $13\times80=1040<1800$, we get $CV_A^2<CV_B^2$, so $CV_A<CV_B$.

Step 5: Conclude.
$CV_A<CV_B$ and $s_A>s_B$, option (B).
\[ \boxed{(B)} \]