# Solved (Free): The following data represents viscosity measurements from a batch chemical process. The data is assumed to be normally distributed

#### ByDr. Raju Chaudhari

May 31, 2021

The following data represents viscosity measurements from a batch chemical process. The data is assumed to be normally distributed. The batches are checked for conformance to specification. Batches whose viscosity lies between 84 and 96 are acceptable. A random sample of 89 batches are checked.

Viscosity Measure < 80 80-84 84-88 88-92 92-96 96-100 > 100
Frequency 0 7 21 33 23 5 0

i) Calculate the mean, standard deviation, standard error of the mean and coefficient of variation of the data.
ii) Interpret the mean and standard deviation in the context of the problem.

#### Solution

Sr.No. $x$ $f$ $f*x$ $f*x^2$
1 82 7 574 47068
2 86 21 1806 155316
3 90 33 2970 267300
4 94 23 2162 203228
5 98 5 490 48020
Total 89 8002 720932

i) Mean of the data is

 \begin{aligned} \overline{x} &= \frac{1}{\sum f} \sum f*x\\ &= \frac{8002}{89}\\ &=89.9101 \end{aligned}

variance of the data is

 \begin{aligned} s^2 &= \frac{1}{\sum f} \sum f*x^2 -\overline{x}^2\\ &= \frac{720932}{89} - 89.9101^2\\ &=8100.3596 - 8083.8261\\ &=16.5335 \end{aligned}

standard deviation of the data is

 \begin{aligned} s &= \sqrt{s^2}\\ &=\sqrt{16.5335}\\ &=4.0661 \end{aligned}

The coefficient of variation is

 \begin{aligned} cv &= \frac{s}{\overline{x}}\times 100\\ &=\frac{4.0661}{89.9101}\times 100\\ &=4.5224063 \end{aligned}

ii) The sample mean of viscosity from a batch chemical process is $89.9101$ and the standard deviation of viscosity from a batch chemical process is $4.0661$.