Question

There are two machines available for cutting corks intended for use in wine bottles. The first produces corks with diameters that are normally distributed with mean 3 cm and standard deviation0.08 cm. The second machine produces corks with diameters that have a normal distribution with mean 3.04 cm and standard deviation0.03 cm. Acceptable corks have diameters between 2.9 cm and 3.1 cm.

What is the probability that the first machine produces an acceptable cork? (Round your answer to four decimal places.)

What is the probability that the second machine produces an acceptable cork? (Round your answer to four decimal places.)

Answer 1

Concepts and reason

The concept of probability generating function of normal distribution will be used. The random variable X will be converted into z-Score by changing normal distribution into standard normal distribution and then computing the probability for the provided range.

Fundamentals

Consider X be the random variable representing the number of corks.

$X \sim {\rm{Normal }}\left( {\mu ,{\sigma ^2}} \right)$

Probability of random variable X can be calculated as follows:

$P\left( {a < X < b} \right) = P\left( {\frac{{a - \mu }}{\sigma } < \frac{{X - \mu }}{\sigma } < \frac{{b - \mu }}{\sigma }} \right)$

Where,

$\begin{array}{l}\\X{\rm{ is the random variable representing number of corks}}\\\\\mu {\rm{ is population mean}}\\\\\sigma {\rm{ population standard deviation}}\\\\\left( {a,b} \right){\rm{ is the range of acceptable corks}}\\\end{array}$

Consider X be the random variable representing the number of corks.

$X \sim {\rm{Normal }}\left( {\mu = 3,\sigma = 0.08} \right)$

The corks produced by first machine are normally distributed with a,

$\begin{array}{c}\\{\rm{Mean }}\left( \mu \right) = 3\\\\{\rm{Standard deviation }}\left( \sigma \right) = 0.08\\\end{array}$

The corks produced by first machine are normally distributed with a,

$\begin{array}{c}\\{\rm{Mean }}\left( \mu \right) = 3.04\\\\{\rm{Standard deviation }}\left( \sigma \right) = 0.03\\\end{array}$

According to question, the acceptable diameter range is,

$\left( {2.9 < X < 3.1} \right)$

Calculate the probability of corks meeting the diameter range as follows:

$\begin{array}{c}\\P\left( \begin{array}{l}\\{\rm{That corks from first machine }}\\\\{\rm{are meeting the diameter range}}\\\end{array} \right) = P\left( {2.9 < X < 3.1} \right)\\\\ = P\left( {\frac{{2.9 - \mu }}{\sigma } < \frac{{X - \mu }}{\sigma } < \frac{{3.1 - \mu }}{\sigma }} \right)\\\\ = P\left( {\frac{{2.9 - 3}}{{0.08}} < z < \frac{{3.1 - 3}}{{0.08}}} \right)\\\end{array}$

$\begin{array}{c}\\ = P\left( { - 1.25 < z < 1.25} \right)\\\\ = P\left( {z > 1.25} \right) - P\left( {z < - 1.25} \right)\\\\ = 0.8944 - 0.1056\\\\ = 0.7887\\\end{array}$

Calculate the probability of corks meeting the diameter range as follows:

$\begin{array}{c}\\P\left( \begin{array}{l}\\{\rm{That corks from second machine }}\\\\{\rm{are meeting the diameter range}}\\\end{array} \right) = P\left( {2.9 < X < 3.1} \right)\\\\ = P\left( {\frac{{2.9 - \mu }}{\sigma } < \frac{{X - \mu }}{\sigma } < \frac{{3.1 - \mu }}{\sigma }} \right)\\\\ = P\left( {\frac{{2.9 - 3.04}}{{0.03}} < z < \frac{{3.1 - 3.04}}{{0.03}}} \right)\\\end{array}$

$\begin{array}{c}\\ = P\left( { - 4.67 < z < 2} \right)\\\\ = P\left( {z > 2} \right) - P\left( {z < - 4.67} \right)\\\\ = 0.9722 - 0\\\\ = 0.9722\\\end{array}$

Ans:

Probability of corks from first machine meeting the diameter range is 0.7887.

Probability of corks from second machine meeting the diameter range is 0.9722.