Question

a sets of data are given in the table below
set a: 20, 28, 59, 38, 54, 46, 49
set b: 28, 45, 58, 32, 47, 54, 51, 32, 48, 27
which statements are true? choose 2 that apply
a. set a has a higher mean than set b
b. set a has a higher median than set b
c. set a has a larger standard - deviation than set b
d. set a and set b have the same mean

Explanation:

Step1: Calculate mean of set A

Mean of set A, $\bar{x}_A=\frac{20 + 28+59+38+54+46+49}{7}=\frac{294}{7} = 42$

Step2: Calculate mean of set B

Mean of set B, $\bar{x}_B=\frac{28+44+58+51+32+48+47}{7}=\frac{308}{7}=44$

Step3: Arrange set A in ascending order

$20,28,38,46,49,54,59$. Median of set A (since $n = 7$ is odd) is the 4 - th value, so median of set A is $46$.

Step4: Arrange set B in ascending order

$32,44,47,48,51,58,28$. Median of set B (since $n = 7$ is odd) is the 4 - th value, so median of set B is $48$.

Step5: Calculate standard - deviation formula

The formula for the sample standard deviation is $s=\sqrt{\frac{\sum_{i = 1}^{n}(x_i-\bar{x})^2}{n - 1}}$.
For set A:
$\sum_{i=1}^{7}(x_i - 42)^2=(20 - 42)^2+(28 - 42)^2+(38 - 42)^2+(46 - 42)^2+(49 - 42)^2+(54 - 42)^2+(59 - 42)^2$
$=(-22)^2+(-14)^2+(-4)^2+4^2+7^2+12^2+17^2$
$=484 + 196+16 + 16+49+144+289=1194$
$s_A=\sqrt{\frac{1194}{6}}\approx\sqrt{199}\approx14.1$.
For set B:
$\sum_{i = 1}^{7}(x_i - 44)^2=(28 - 44)^2+(44 - 44)^2+(58 - 44)^2+(51 - 44)^2+(32 - 44)^2+(48 - 44)^2+(47 - 44)^2$
$=(-16)^2+0^2+14^2+7^2+(-12)^2+4^2+3^2$
$=256+0 + 196+49+144+16+9=670$
$s_B=\sqrt{\frac{670}{6}}\approx\sqrt{111.67}\approx10.6$.

Answer:

C. Set A has a larger standard deviation than set B.
D. Set A and set B have the same mean (This is false as $\bar{x}_A = 42$ and $\bar{x}_B=44$).
A. Set A has a higher mean than set B (This is false as $\bar{x}_A = 42$ and $\bar{x}_B = 44$).
B. Set A has a higher median than set B (This is false as median of set A is $46$ and median of set B is $48$).
So the correct answers are C.