Question

two sets of data are shown,
data set a: 30, 38, 42, 42, 43, 47, 51, 51, 57, 59
data set b: 38, 39, 40, 42, 44, 46, 47, 50, 51, 52
choose all the measures which are greater for data set a than for data set b.
□ mean □ range □ median □ standard deviation □ interquartile range

Explanation:

Step1: Calculate the mean

For Data Set A:
$$\bar{x}_A=\frac{30 + 38+42+42+43+47+51+51+57+59}{10}=\frac{460}{10} = 46$$
For Data Set B:
$$\bar{x}_B=\frac{38+39+40+42+44+46+47+50+51+52}{10}=\frac{459}{10}=45.9$$
Since \(46>45.9\), the mean of A is greater than that of B.

Step2: Calculate the range

Range = maximum - minimum
For Data Set A: \(Range_A=59 - 30=29\)
For Data Set B: \(Range_B=52 - 38 = 14\)
Since \(29>14\), the range of A is greater than that of B.

Step3: Calculate the median

For a set of \(n = 10\) (even) data points, the median is the average of the \(\frac{n}{2}\) - th and \((\frac{n}{2}+1)\) - th ordered values.
For Data Set A: The 5 - th value is \(43\) and the 6 - th value is \(47\), \(Median_A=\frac{43 + 47}{2}=45\)
For Data Set B: The 5 - th value is \(44\) and the 6 - th value is \(46\), \(Median_B=\frac{44+46}{2}=45\)
Since \(45 = 45\), the medians are equal.

Step4: Calculate the standard deviation

The formula for the sample standard deviation is \(s=\sqrt{\frac{\sum_{i = 1}^{n}(x_i-\bar{x})^2}{n - 1}}\)
For Data Set A:
\(\sum_{i=1}^{10}(x_i - 46)^2=(30 - 46)^2+(38 - 46)^2+(42 - 46)^2+(42 - 46)^2+(43 - 46)^2+(47 - 46)^2+(51 - 46)^2+(51 - 46)^2+(57 - 46)^2+(59 - 46)^2\)
\(=256+64 + 16+16+9+1+25+25+121+169=702\)
\(s_A=\sqrt{\frac{702}{9}}\approx8.82\)
For Data Set B:
\(\sum_{i = 1}^{10}(x_i-45.9)^2=(38 - 45.9)^2+(39 - 45.9)^2+(40 - 45.9)^2+(42 - 45.9)^2+(44 - 45.9)^2+(46 - 45.9)^2+(47 - 45.9)^2+(50 - 45.9)^2+(51 - 45.9)^2+(52 - 45.9)^2\)
\(=62.41+47.61+34.81+15.21+3.61+0.01+1.21+16.81+26.01+37.21=244.9\)
\(s_B=\sqrt{\frac{244.9}{9}}\approx5.23\)
Since \(8.82>5.23\), the standard deviation of A is greater than that of B.

Step5: Calculate the inter - quartile range (IQR)

For Data Set A:
The first quartile \(Q_1\) (average of 2 - nd and 3 - rd values of the lower half). The lower half is \(30,38,42,42,43\), \(Q_{1A}=\frac{38 + 42}{2}=40\)
The third quartile \(Q_3\) (average of 2 - nd and 3 - rd values of the upper half). The upper half is \(51,51,57,59\), \(Q_{3A}=\frac{51+57}{2}=54\)
\(IQR_A=Q_{3A}-Q_{1A}=54 - 40 = 14\)
For Data Set B:
The lower half is \(38,39,40,42,44\), \(Q_{1B}=\frac{39+40}{2}=39.5\)
The upper half is \(47,50,51,52\), \(Q_{3B}=\frac{50 + 51}{2}=50.5\)
\(IQR_B=Q_{3B}-Q_{1B}=50.5-39.5 = 11\)
Since \(14>11\), the inter - quartile range of A is greater than that of B.

Answer:

mean, range, standard deviation, interquartile range