Chapter 14, Problem 1E

To determine

To find:

The mean, median, mode, and standard deviation using built-in functions.

Answer to Problem 1E

Solution:

The required answers are stated as follows.

Explanation of Solution

The given data is,

$\begin{array}{l}\text{Prod}\text{.}\text{line}\text{A:15}\text{.94}\text{15}\text{.98}\text{15}\text{.94}\text{16}\text{.16}\text{15}\text{.86}15.86\text{15}\text{.90}\text{15}\text{.88}\\ \text{Prod}\text{.}\text{line}\text{B:15}\text{.96}\text{15}\text{.94}\text{16}\text{.02}\text{16}\text{.10}\text{15}\text{.92}\text{16}\text{.00}\text{15}\text{.96}\text{16}\text{.02}\end{array}$

The formula to calculate the mean of set of data is,

$\begin{array}{rll}\text{Mean}\left(\overline{x}\right)& =& \frac{x_1+x_2+\mathrm{...}+x_n}{n}\\ & =& \frac{{\displaystyle \sum _{i1}^n{x}_i}}{n}\end{array}$

Here, $x_i$ is the $i^{\text{th}}$ data value and $n$ is the number of data values in the data set.

Substitute $\text{15}\text{.94,}\text{15}\text{.98}\text{15}\text{.94},\mathrm{...},15.88$ for $x_1,x_2,x_3,\mathrm{...},x_8$ respectively in the formula.

Here number of data in the given data set is $8$, substitute $8$ for $n$ in the formula.

$\begin{array}{rll}\text{Mean}\left(\overline{x}\right)& =& \frac{x_1+x_2+\mathrm{...}+x_n}{n}\\ & =& \frac{\text{15}\text{.94}+\text{15}\text{.98}+\text{15}\text{.94}+\text{16}\text{.16}+\text{15}\text{.86}+15.86+\text{15}\text{.90}+\text{15}\text{.88}}{8}\\ & \approx & 15.94\end{array}$

Mean of A is $15.94$.

Substitute $\text{15}\text{.96, 15}\text{.94, 16}\text{.02,}\mathrm{......}\text{16}\text{.02}$ for $x_1,x_2,x_3,\mathrm{...},x_8$ respectively in the formula.

$\begin{array}{rll}\text{Mean}\left(\overline{x}\right)& =& \frac{x_1+x_2+\mathrm{...}+x_n}{n}\\ & =& \frac{\text{15}\text{.96}+\text{15}\text{.94}+\text{16}\text{.02}+\text{16}\text{.10}+\text{15}\text{.92}+\text{16}\text{.00}+\text{15}\text{.96}+\text{16}\text{.02}}{8}\\ & \approx & 15.99\end{array}$

Mean of B is $15.99$.

The formula to calculate the median of set of ordered data is,

$\text{Median}=\{\begin{array}{l}{\left(\frac{n+1}{2}\right)}^{\text{th}}\text{value},n=\text{odd}\\ \frac{1}{2}\left[{\left(\frac{n}{2}\right)}^{\text{th}}\text{value}+{\left(\frac{n+1}{2}\right)}^{\text{th}}\text{value}\right],n=\text{even}\end{array}$

$n$ is the number of data values in the data set.

Arrange the given data in increasing order.

$\begin{array}{l}\text{Prod}\text{.}\text{line}\text{A:15}\text{.86}15.86\text{15}\text{.88}\text{15}\text{.90}\text{15}\text{.94}\text{15}\text{.94}\text{15}\text{.98}\text{16}\text{.16}\\ \text{Prod}\text{.}\text{line}\text{B:15}\text{.92}\text{15}\text{.94}\text{15}\text{.96}\text{15}\text{.96}\text{16}\text{.00}\text{16}\text{.02}\text{16}\text{.02}\text{16}\text{.10}\end{array}$

Number of values in data set is $8$, it is even.

Substitute $8$ for $n$ in the formula for even number of data set A.

$\begin{array}{rll}\text{Median}& =& \frac{1}{2}\left[{\left(\frac{n}{2}\right)}^{\text{th}}\text{value}+{\left(\frac{n}{2}+1\right)}^{\text{th}}\text{value}\right]\\ & =& \frac{1}{2}\left[{\left(\frac{8}{2}\right)}^{\text{th}}\text{value}+{\left(\frac{8}{2}+1\right)}^{\text{th}}\text{value}\right]\\ & =& \frac{1}{2}\left[4^{\text{th}}\text{value}+5^{\text{th}}\text{value}\right]\\ & =& \frac{1}{2}\left[15.90+15.94\right]\end{array}$

Simplify further,

$\begin{array}{rll}\text{Median}& =& \frac{1}{2}\left[15.90+15.94\right]\\ & =& 15.92\end{array}$

The median of A is $15.92$.

Substitute $8$ for $n$ in the formula for even number of data set B.

$\begin{array}{rll}\text{Median}& =& \frac{1}{2}\left[{\left(\frac{n}{2}\right)}^{\text{th}}\text{value}+{\left(\frac{n}{2}+1\right)}^{\text{th}}\text{value}\right]\\ & =& \frac{1}{2}\left[{\left(\frac{8}{2}\right)}^{\text{th}}\text{value}+{\left(\frac{8}{2}+1\right)}^{\text{th}}\text{value}\right]\\ & =& \frac{1}{2}\left[4^{\text{th}}\text{value}+5^{\text{th}}\text{value}\right]\\ & =& \frac{1}{2}\left[15.96+16.00\right]\end{array}$

Simplify further,

$\begin{array}{rll}\text{Median}& =& \frac{1}{2}\left[15.96+16.00\right]\\ & =& 15.98\end{array}$

The median of B is $15.98$.

The most frequently occurred data is the mode.

Here, each of the data $\text{15}\text{.86}$ in A occur 2 times.

Mode of the given data set A are $\text{15}\text{.86}$.

Here, each of the data $\text{15}\text{.96}$ in B occur 2 times.

Mode of the given data set B are $\text{15}\text{.96}$.

Since, the mean value of set B is closer to diameter $16\text{mm}$. Hence, set B is better than set A.

MATLAB Code:

A = [15.94, 15.98, 15.94, 16.16, 15.86, 15.86, 15.90, 15.88];

B = [15.96, 15.94, 16.02, 16.10, 15.92, 16.00, 15.96, 16.02];

mean_A = mean(A)

mean_B = mean(B)

median_A = median(A)

median_B = median(B)

mode_A = mode(A)

mode_B = mode(B)

stdev_A = std(A)

stdev_B = std(B)

Save the MATLAB script with the name, Chapter14_54793_14_1E.m in the current folder. Execute the script by typing the script name at the command window to generate output.

Result:

Therefore, the required program is stated above.