Question:
The following data represent the lifetimes (in months) of a sample of eleven bulbs:
The following data represent the lifetimes (in months) of a sample of eleven bulbs:
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When every observation is multiplied by a constant \(c\), the mean, median, and standard deviation are also multiplied by \(c\), while variance is multiplied by \(c^2\).
Updated On: Jun 4, 2026
- The sample mean is greater than the sample median
- The mean deviation about the point \(15\) equals \(3\)
- The range of the data is twice the interquartile range of the data
- The sample standard deviation as well as the sample median are scaled by \(2\) if every data point is scaled by \(2\)
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The Correct Option is A, B, D
Solution and Explanation
Step 1: Arrange the data in ascending order.
The given data are
\[ 15,10,12,10,17,12,14,20,13,22,15 \] Arranging in ascending order,
\[ 10,10,12,12,13,14,15,15,17,20,22 \]
Step 2: Check option (A).
The sample mean is
\[ \bar{x}=\frac{15+10+12+10+17+12+14+20+13+22+15}{11} \] \[ \bar{x}=\frac{160}{11} \] \[ \bar{x}\approx 14.545 \] Since there are \(11\) observations, the median is the \(6\)th observation in the ordered data.
\[ \text{Median}=14 \] Thus,
\[ \bar{x}>14 \] Hence, option (A) is true.
Step 3: Check option (B).
Mean deviation about the point \(15\) is
\[ \frac{1}{11}\sum_{i=1}^{11}|x_i-15| \] Now,
\[ |15-15|+|10-15|+|12-15|+|10-15|+|17-15|+|12-15| \] \[ +|14-15|+|20-15|+|13-15|+|22-15|+|15-15| \] \[ =0+5+3+5+2+3+1+5+2+7+0 \] \[ =33 \] Therefore,
\[ \text{Mean deviation}=\frac{33}{11}=3 \] Hence, option (B) is true.
Step 4: Check option (C).
The range is
\[ 22-10=12 \] For the ordered data,
\[ Q_1=12 \] and
\[ Q_3=17 \] So, the interquartile range is
\[ IQR=Q_3-Q_1=17-12=5 \] Twice the interquartile range is
\[ 2IQR=10 \] But the range is \(12\), not \(10\).
Hence, option (C) is false.
Step 5: Check option (D).
If every observation is multiplied by \(2\), then every measure of location such as median is also multiplied by \(2\).
Also, every deviation from the mean is multiplied by \(2\), so the sample standard deviation is also multiplied by \(2\).
Hence, option (D) is true.
Step 6: Final conclusion.
The true statements are
\[ \boxed{(A),\,(B)\text{ and }(D)} \]
The given data are
\[ 15,10,12,10,17,12,14,20,13,22,15 \] Arranging in ascending order,
\[ 10,10,12,12,13,14,15,15,17,20,22 \]
Step 2: Check option (A).
The sample mean is
\[ \bar{x}=\frac{15+10+12+10+17+12+14+20+13+22+15}{11} \] \[ \bar{x}=\frac{160}{11} \] \[ \bar{x}\approx 14.545 \] Since there are \(11\) observations, the median is the \(6\)th observation in the ordered data.
\[ \text{Median}=14 \] Thus,
\[ \bar{x}>14 \] Hence, option (A) is true.
Step 3: Check option (B).
Mean deviation about the point \(15\) is
\[ \frac{1}{11}\sum_{i=1}^{11}|x_i-15| \] Now,
\[ |15-15|+|10-15|+|12-15|+|10-15|+|17-15|+|12-15| \] \[ +|14-15|+|20-15|+|13-15|+|22-15|+|15-15| \] \[ =0+5+3+5+2+3+1+5+2+7+0 \] \[ =33 \] Therefore,
\[ \text{Mean deviation}=\frac{33}{11}=3 \] Hence, option (B) is true.
Step 4: Check option (C).
The range is
\[ 22-10=12 \] For the ordered data,
\[ Q_1=12 \] and
\[ Q_3=17 \] So, the interquartile range is
\[ IQR=Q_3-Q_1=17-12=5 \] Twice the interquartile range is
\[ 2IQR=10 \] But the range is \(12\), not \(10\).
Hence, option (C) is false.
Step 5: Check option (D).
If every observation is multiplied by \(2\), then every measure of location such as median is also multiplied by \(2\).
Also, every deviation from the mean is multiplied by \(2\), so the sample standard deviation is also multiplied by \(2\).
Hence, option (D) is true.
Step 6: Final conclusion.
The true statements are
\[ \boxed{(A),\,(B)\text{ and }(D)} \]
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