Answer to Question #277707 in Statistics and Probability for Owence

Question

Find the mean, sample standard deviation, quartiles, box and whisker plot, and stem and leaf diagram, of the following body temperature (°F)

97.1 97.8 98.0 98.9 99.5 96.3

98.4 98.5 98.1 100.8 98.6 98.2

99.0 99.3 98.8 97.6 97.4 99.0

97.4 96.4 98.0 98.1 97.8 98.5

98.7 98.8 98.2 97.6 98.2 98.8

Answer 1

1.

Mean $=\frac{97.1+97.8+98.0+98.9+99.5+96.3}{6}=97.93$

Standard deviation = $\sqrt{\frac{\sum(x-\bar x)^2}{n-1}}$

From the given data, the standard deviation is:

$S=1.16$

The data: 96.3, 97.1, 97.8, 98.0, 98.9, 99.5

First Quartile(Q1) $=((n+1)/4)^{th}$ term $=7/4=1^{st}\hspace{3pt}term=96.9$

Second Quartile(Q2)$==((n+1)/2)^{th}\hspace{3pt} term=97.9$

Third Quartile(Q3) $=(3(n+1)/4)^{th}\hspace{3pt}term=99.05$

Box and Whisker Plot:

Stem and Leaf Diagram:

2.

Mean $=\frac{98.4+98.5+98.1+100.8+98.6+98.2}{6}=98.77$

Standard deviation = $\sqrt{\frac{\sum(x-\bar x)^2}{n-1}}$

From the given data, the standard deviation is:

$S=1.01$

The data: 98.1,98.2,98.4,98.5,98.6,100.8

First Quartile(Q1) $=((n+1)/4)^{th}$ term $=7/4=1^{st}\hspace{3pt}term=98.175$

Second Quartile(Q2)$==((n+1)/2)^{th}\hspace{3pt} term=98.45$

Third Quartile(Q3) $=(3(n+1)/4)^{th}\hspace{3pt}term=99.15$

Box and Whisker Plot:

Stem and Leaf Diagram:

3.

Mean $=98.52$

Standard deviation = $\sqrt{\frac{\sum(x-\bar x)^2}{n-1}}$

From the given data, the standard deviation is:

$S=0.81$

The data: 97.4,97.6,98.8,99.0,99.0,99.3

First Quartile(Q1) $=((n+1)/4)^{th}$ term $=7/4=1^{st}\hspace{3pt}term=97.55$

Second Quartile(Q2)$==((n+1)/2)^{th}\hspace{3pt} term=98.9$

Third Quartile(Q3) $=(3(n+1)/4)^{th}\hspace{3pt}term=99.075$

Box and Whisker Plot:

Stem and Leaf Diagram:

4.

Mean $=97.7$

Standard deviation = $\sqrt{\frac{\sum(x-\bar x)^2}{n-1}}$

From the given data, the standard deviation is:

$S=0.73$

The data: 96.4,97.4,97.8,98.0,98.1,98.5

First Quartile(Q1) $=((n+1)/4)^{th}$ term $=7/4=1^{st}\hspace{3pt}term=97.15$

Second Quartile(Q2)$==((n+1)/2)^{th}\hspace{3pt} term=97.9$

Third Quartile(Q3) $=(3(n+1)/4)^{th}\hspace{3pt}term=98.2$

Box and Whisker Plot:

Stem and Leaf Diagram:

5.

Mean $=98.38$

Standard deviation = $\sqrt{\frac{\sum(x-\bar x)^2}{n-1}}$

From the given data, the standard deviation is:

$S=0.47$

The data: 97.6,98.2,98.2,98.7,98.8,98.8

First Quartile(Q1) $=((n+1)/4)^{th}$ term $=7/4=1^{st}\hspace{3pt}term=98.05$

Second Quartile(Q2)$==((n+1)/2)^{th}\hspace{3pt} term=98.45$

Third Quartile(Q3) $=(3(n+1)/4)^{th}\hspace{3pt}term=98.8$

Box and Whisker Plot:

Stem and Leaf Diagram: