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