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Statistics / Applied Statistics and Probability for Engineers 6 / Chapter 4.12 / Problem 202SE

Applied Statistics and Probability for Engineers | 6th Edition | ISBN: 9781118539712 | Authors: Douglas C. Montgomery, George C. Runger

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The size of silver particles in a photographic emulsion is

Chapter 4, Problem 202SE

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QUESTION:

Problem 202SE

The size of silver particles in a photographic emulsion is known to have a log normal distribution with a mean of 0.001 mm and a standard deviation of 0.002 mm.

(a) Determine the parameter values for the lognormal distribution.

(b) What is the probability of a particle size greater than 0.005 mm?

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QUESTION:

Problem 202SE

The size of silver particles in a photographic emulsion is known to have a log normal distribution with a mean of 0.001 mm and a standard deviation of 0.002 mm.

(a) Determine the parameter values for the lognormal distribution.

(b) What is the probability of a particle size greater than 0.005 mm?

ANSWER:

Solution

Step 1 of 2

a) We have to find the parameters of the lognormal distribution

Let X be the size of the silver particles

Here X is log normal random variable with mean 0.001 mm and standard deviation 0.002 mm

Then $E(X)=e{\ }^{\theta{}+\frac{\omega{}{\ }^{2}}{2}}=0.001$

$V(X)=e{\ }^{2\theta{}+\omega{}{\ }^{2}}(e{\ }^{\omega{}{\ }^{2}}-1)=(0.002){\ }^{2}$

Let $x=e{\ }^{\theta{}}$ and $y=e{\ }^{\omega{}{\ }^{2}}$

Now substitute in above equations

Then $E(X)=e{\ }^{\theta{}+\frac{\omega{}{\ }^{2}}{2}}=e{\ }^{\theta{}}e{\ }^{\omega{}{\ }^{2}/2}=x\sqrt{y}=0.001$

Now square on both the sides

Now $x{\ }^{2}y=(0.001){\ }^{2}$

And $V(X)=e{\ }^{2\theta{}+\omega{}{\ }^{2}}(e{\ }^{\omega{}{\ }^{2}}-1)=x{\ }^{2}y(y-1)=(0.002){\ }^{2}$

Now substitute $x{\ }^{2}y=(0.001){\ }^{2}$ in the above equation

$(0.001){\ }^{2}(y-1)=(0.002){\ }^{2}$ $\$

$y-1=\frac{(0.002){\ }^{2}}{(0.001){\ }^{2}}$

$y-1=2$

$y=3$

Now from $x\sqrt{y}=0.001$

$x\sqrt{3}=0.001$

$x=0.001/\sqrt{3}$

$x=0.00058$

Now from $x=e{\ }^{\theta{}}$

$0.00058=e{\ }^{\theta{}}$

$\theta{}=ln(0.00058)$

$\theta{}=-7.45$

And from $y=e{\ }^{\omega{}{\ }^{2}}$

$3=e{\ }^{\omega{}{\ }^{2}}$

$\omega{}{\ }^{2}=ln(3)$

$\omega{}{\ }^{2}=1.099$

Hence the parameters of a lognormal distribution are $\theta{}=-7.45\ and\ \ \ \omega{}{\ }^{2}=1.099$

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