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: Find the mean lifetime of Brand A.
For Brand A, the observations are
\[ 27,28,30,32,33 \] The sample mean is
\[ \bar{x}_A=\frac{27+28+30+32+33}{5} \] \[ \bar{x}_A=\frac{150}{5}=30 \]

Step 2: Find the sample standard deviation of Brand A.
The deviations from mean are
\[ -3,-2,0,2,3 \] The sum of squares of deviations is
\[ (-3)^2+(-2)^2+0^2+2^2+3^2 \] \[ =9+4+0+4+9=26 \] Since sample variance uses denominator \(n-1\),
\[ s_A^2=\frac{26}{5-1} \] \[ s_A^2=\frac{26}{4}=\frac{13}{2} \] Thus,
\[ s_A=\sqrt{\frac{13}{2}} \]

Step 3: Find the mean lifetime of Brand B.
For Brand B, the observations are
\[ 17,19,20,21,23 \] The sample mean is
\[ \bar{x}_B=\frac{17+19+20+21+23}{5} \] \[ \bar{x}_B=\frac{100}{5}=20 \]

Step 4: Find the sample standard deviation of Brand B.
The deviations from mean are
\[ -3,-1,0,1,3 \] The sum of squares of deviations is
\[ (-3)^2+(-1)^2+0^2+1^2+3^2 \] \[ =9+1+0+1+9=20 \] Thus,
\[ s_B^2=\frac{20}{5-1} \] \[ s_B^2=5 \] So,
\[ s_B=\sqrt{5} \]

Step 5: Compare sample standard deviations.
Since
\[ s_A=\sqrt{\frac{13}{2}}=\sqrt{6.5} \] and
\[ s_B=\sqrt{5} \] we get
\[ s_A>s_B \]

Step 6: Compare coefficients of variation.
The sample coefficient of variation is
\[ CV=\frac{s}{\bar{x}} \] For Brand A,
\[ CV_A=\frac{s_A}{30} = \frac{\sqrt{\frac{13}{2}}}{30} \] For Brand B,
\[ CV_B=\frac{s_B}{20} = \frac{\sqrt{5}}{20} \] To compare, square both positive quantities.
\[ CV_A^2=\frac{\frac{13}{2}}{900} = \frac{13}{1800} \] and
\[ CV_B^2=\frac{5}{400} = \frac{1}{80} \] Now,
\[ \frac{13}{1800}<\frac{1}{80} \] because
\[ 13\times 80=1040<1800 \] Therefore,
\[ CV_A<CV_B \]

Step 7: Final conclusion.
Hence,
\[ CV_A<CV_B \] and
\[ s_A>s_B \] Therefore, the correct option is
\[ \boxed{(B)} \]