Question

the data shows the number of pages murray and brian read each day last week. murray: 48, 40, 42, 50, 36, 44, 41 brian: 72, 31, 28, 47, 38, 43, 42 which sentences about the data are true? select all that apply. a) murrays mean is greater than brians. b) murray and brian have identical means. c) murrays mad is 3.71 when rounded to the nearest hundredth. d) brians mad is less than double murrays mad. e) murrays data is skewed by the presence of an outlier.

Explanation:

Step1: Calculate Murray's mean

$\text{Mean}_M=\frac{48 + 40+42+50+36+44+41}{7}=\frac{301}{7}=43$

Step2: Calculate Brian's mean

$\text{Mean}_B=\frac{72+31+28+47+38+43+42}{7}=\frac{301}{7}=43$

Step3: Calculate Murray's MAD

1. Find the absolute - deviation of each data point from the mean:

$|48 - 43| = 5$, $|40 - 43| = 3$, $|42 - 43| = 1$, $|50 - 43| = 7$, $|36 - 43| = 7$, $|44 - 43| = 1$, $|41 - 43| = 2$

1. Calculate the mean of these absolute - deviations:

$\text{MAD}_M=\frac{5 + 3+1+7+7+1+2}{7}=\frac{26}{7}\approx3.71$

Step4: Calculate Brian's MAD

1. Find the absolute - deviation of each data point from the mean:

$|72 - 43| = 29$, $|31 - 43| = 12$, $|28 - 43| = 15$, $|47 - 43| = 4$, $|38 - 43| = 5$, $|43 - 43| = 0$, $|42 - 43| = 1$

1. Calculate the mean of these absolute - deviations:

$\text{MAD}_B=\frac{29+12+15+4+5+0+1}{7}=\frac{66}{7}\approx9.43$
$2\times\text{MAD}_M = 2\times3.71 = 7.42$, and $\text{MAD}_B=9.43>7.42$

1. Check for outliers in Murray's data:

The inter - quartile range (IQR) method:
First, order the data: $36,40,41,42,44,48,50$
The median (Q2) is 42.
The lower half is $36,40,41$, and the median of the lower half (Q1) is 40.
The upper half is $44,48,50$, and the median of the upper half (Q3) is 48.
$IQR = Q3 - Q1=48 - 40 = 8$
Lower fence: $Q1-1.5\times IQR=40-1.5\times8 = 40 - 12 = 28$
Upper fence: $Q3 + 1.5\times IQR=48+1.5\times8 = 48 + 12 = 60$
There are no outliers in Murray's data.

Answer:

B. Murray and Brian have identical means.
C. Murray's MAD is 3.71 when rounded to the nearest hundredth.