Question

Problem 2 Poly(3hydroxybut yrate) (PHB), a semicrystallne polymer that is fully biodegrad- able and biocompatible, is obtained from renewable resources. From a sus tainability perspective, PHB offers many attactive properties though i is more expensive to produce than standard psties. The following data con tains the melting point (C") values for 12 specimens of the polymer: 80. 181.7 180.9 181.6 182.6 181.6 181.3 182.1 1821 180.3 181.7 180.5 ( 8 paintsl Give the 2) (2 points) Caleulate the sa mple mean of he s 3) (4 points) Caleulate the sample st andard deviation

Answer 1

First of all let's write all the numbers in following list:-

{180.5, 181.7, 180.9 , 181.6 , 182.6, 181.6, 181.3, 182.1, 182.1, 180.3, 181.7, 180.5}

Now for finding the five number summary of the data , first we will sort (in Ascending order) the data and re-write in the list as following:-

{180.3,180.5,180.5,180.9,181.3,181.6,181.6,181.7,181.7,182.1,182.1,182.6}

1) Now we will find five number summary of the given data

Minimum :- 180.3 C^o

Now we will find Median (or Quartile Q2)...

Here in the given data total data points n = 12 which is even number.. So for finding median we will take an average of 6^th and 7^th data point, i.e.

$\small \frac{181.6 + 181.6}{2} = {181.6 \ C^o = Median}$

For finding Quartile Q1, we will find median of upper half i.e. data points upto n = 6 , so upper half data points in list as following.

{180.3,180.5,180.5,180.9,181.3,181.6} as here n=6 , for finding median of this data we will take average of 3^rd and 4^th data point

$\small \frac{180.5 + 180.9}{2} = 180.7 \ C^o = Quartile \ Q1$

Similarly, for finding Quartile Q3, we will find median of lower half i.e. data points from n=7 to n = 12 , so lower half data points in list as following.

{181.6,181.7,181.7,182.1,182.1,182.6} as here n=6, for finding median of this data we will take average of 3^rd and 4^th data point

$\small \frac{181.7 + 182.1}{2} = {181.9 \ C^o = Quartile \ Q3}$

Maximum :- 182.6 C^o

So,

Minimum: 180.3 C^o
Quartile Q1: 180.7 C^o
Median: 181.6 C^o
Quartile Q3: 181.9 C^o
Maximum: 182.6 C^o

2) Sample mean can be calculated as following...

$\small \frac{180.3+180.5+180.5+180.9+181.3+181.6+181.6+181.7+181.7+182.1+182.1+182.6}{12}$

$\small = {181.4083 \ C^o = Mean}$

So ,

Sample Mean: μ = 181.4083 C^o

3) Sample standard deviation for sample data can be found using following formula :-

$\small s = \sqrt{\frac{1}{N-1} \sum_{i=1}^N (x_i - \mu)^2}$

$\small s=\sqrt{\frac{(180.3-\mu)^2+(180.5-\mu)^2+(180.5-\mu)^2+(180.9-\mu)^2+(181.3-\mu)^2+(181.6-\mu)^2+(181.6-\mu)^2+(181.7-\mu)^2+(181.7-\mu)^2+(182.1-\mu)^2+(182.1-\mu)^2+(182.6-\mu)^2} {12-1} }$

$\small s = 0.7242$

So,

sample standard deviation s = 0.7242 C^o