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Blood pressure values are often reported to the nearest5 mmHg (100, 105, 110, etc.). Suppose the actual bloodpressure values for nine randomly selected individuals are

118.6 127.4 138.4 130.0 113.7 122.0 108.3131.5 133.2

  1. What is the median of the reportedblood pressure values?
  2. Suppose the blood pressure of the second individual is 127.6 rather than 127.4 (a small change in a single value). How does this affect the median of the reported values? What does this say about the sensitivity of the median to rounding or grouping in the data?

Short Answer

Expert verified

a.The median value is 125 mmHg.

b. The median value is 130 mmHg. The median is sensitive to rounding or grouping in small data.

Step by step solution

01

Given information

The blood pressure values are provided.

The size of the sample is 9.

02

Step 2:Compute the median

a.

Let x represents the blood pressure values.

The reported blood pressure values are computed by rounding off the actual blood pressure values.

The data for the reported blood are,

120

125

140

130

110

120

110

135

130

To compute the median, first, arrange the data in ascending order,

110

110

120

120

125

130

130

135

140

For the odd number of observations, the median value is computed as,

\(\begin{aligned}\tilde x &= {\left( {\frac{{n + 1}}{2}} \right)^{th}}ordered\;value\\ &= {\left( {\frac{{9 + 1}}{2}} \right)^{th}}ordered\;value\\ &= {\left( 5 \right)^{th}}ordered\;value\end{aligned}\)

Thus, the median value of x is 125 mmHg.

03

Comment on the change in median value

b.

It is given that the blood pressure of the second individual is 127.6 rather than 127.4.

The reported blood pressure values are computed by rounding off the actual blood pressure values.

The data for the reported blood are,

120

130

140

130

110

120

110

135

130

To compute the median, first arrange the data in ascending order,

110

110

120

120

130

130

130

135

140

For the odd number of observations, the median value is computed as,

\(\begin{aligned}\tilde x &= {\left( {\frac{{n + 1}}{2}} \right)^{th}}ordered\;value\\ &= {\left( {\frac{{9 + 1}}{2}} \right)^{th}}ordered\;value\\ &= {\left( 5 \right)^{th}}ordered\;value\end{aligned}\)

Thus, the median value of x is 130 mmHg.

Therefore, the median value increased by 5. Also, the median value is highly sensitive to rounding or grouping in the data.

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