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 of Data Set A

The mean formula is $\bar{x}=\frac{\sum_{i = 1}^{n}x_{i}}{n}$. For Data Set A: $\sum_{i=1}^{10}x_{i}=30 + 38+42+42+43+47+51+51+57+59 = 460$, and $n = 10$. So the mean of Data Set A, $\bar{x}_A=\frac{460}{10}=46$. For Data Set B: $\sum_{i = 1}^{10}x_{i}=38+39+40+42+44+46+47+50+51+52=449$, and $n = 10$. So the mean of Data Set B, $\bar{x}_B=\frac{449}{10}=44.9$. Since $46>44.9$, the mean of Data Set A is greater.

Step2: Calculate the range of Data Set A and B

The range formula is $R=\text{max}-\text{min}$. For Data Set A, $\text{max}=59$, $\text{min}=30$, so $R_A=59 - 30=29$. For Data Set B, $\text{max}=52$, $\text{min}=38$, so $R_B=52 - 38 = 14$. Since $29>14$, the range of Data Set A is greater.

Step3: Calculate the median of Data Set A and B

Since $n = 10$ (even), the median formula is $M=\frac{x_{\frac{n}{2}}+x_{\frac{n}{2}+1}}{2}$. For Data Set A, $x_5 = 43$, $x_6=47$, so $M_A=\frac{43 + 47}{2}=45$. For Data Set B, $x_5=44$, $x_6=46$, so $M_B=\frac{44+46}{2}=45$. They are equal.

Step4: Calculate the standard - deviation of Data Set A and B

The standard - deviation formula is $s=\sqrt{\frac{\sum_{i = 1}^{n}(x_{i}-\bar{x})^2}{n - 1}}$. For Data Set A:
First, calculate $\sum_{i = 1}^{10}(x_{i}-46)^2=(30 - 46)^2+(38 - 46)^2+(42 - 46)^2+(42 - 46)^2+(43 - 46)^2+(47 - 46)^2+(51 - 46)^2+(51 - 46)^2+(57 - 46)^2+(59 - 46)^2$
$=(-16)^2+(-8)^2+(-4)^2+(-4)^2+(-3)^2+(1)^2+(5)^2+(5)^2+(11)^2+(13)^2$
$=256+64 + 16+16+9+1+25+25+121+169 = 692$. Then $s_A=\sqrt{\frac{692}{9}}\approx8.7$.
For Data Set B:
First, calculate $\sum_{i = 1}^{10}(x_{i}-44.9)^2=(38 - 44.9)^2+(39 - 44.9)^2+(40 - 44.9)^2+(42 - 44.9)^2+(44 - 44.9)^2+(46 - 44.9)^2+(47 - 44.9)^2+(50 - 44.9)^2+(51 - 44.9)^2+(52 - 44.9)^2$
$=(-6.9)^2+(-5.9)^2+(-4.9)^2+(-2.9)^2+(-0.9)^2+(1.1)^2+(2.1)^2+(5.1)^2+(6.1)^2+(7.1)^2$
$=47.61+34.81+24.01+8.41+0.81+1.21+4.41+26.01+37.21+50.41 = 230.9$. Then $s_B=\sqrt{\frac{230.9}{9}}\approx5.06$. Since $8.7>5.06$, the standard - deviation of Data Set A is greater.

Step5: Calculate the inter - quartile range of Data Set A and B

First, find the first and third quartiles. For Data Set A:
The lower half of Data Set A is $30,38,42,42,43$, the median of the lower half (first quartile $Q_1$) is $42$. The upper half is $47,51,51,57,59$, the median of the upper half (third quartile $Q_3$) is $51$. So $IQR_A=Q_3 - Q_1=51 - 42 = 9$.
For Data Set B:
The lower half of Data Set B is $38,39,40,42,44$, the median of the lower half (first quartile $Q_1$) is $40$. The upper half is $46,47,50,51,52$, the median of the upper half (third quartile $Q_3$) is $50$. So $IQR_B=Q_3 - Q_1=50 - 40 = 10$. Since $9<10$, the inter - quartile range of Data Set A is less than that of Data Set B.

Answer:

mean, range, standard deviation