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 $4.0661$.
Further Reading
- 5 Number Summary Calculator
- Statistics
- Descriptive Statistics
- Probability Theory
- Probability Distribution
- Hypothesis Testing
- Confidence interval
- Sample size determination
- Non-parametric Tests
- Correlation Regression
- Statistics Calculators
