91Ó°ÊÓ

The standard deviation is a measurement of a set of values' variation or dispersion. A low standard deviation implies that the values are close to the set's mean, whereas a high standard deviation suggests that the values are dispersed over a larger range.

Allow a random variable\(X\)to represent the diameter of the corks.

Since the diameter of the corks must be between\(2.9\;{\rm{cm}}\)and \(3.1\;{\rm{cm}},\)as a result, whichever machine has a greater \(P\left( {2.9 < X < 3.1} \right),\)is more likely to yield a cork that is acceptable.

Machine first: The mean and standard deviation of cork diameters produced by machine are as follows:

\(\mu = 3\;{\rm{cm}}\)

\(\sigma = 0.1\;{\rm{cm}}\)

Since there's a good chance that the cork diameter will be smaller, between\(2.9\)and\(3.1\;{\rm{cm}}\)is denoted as\(P\left( {2.9 < X < 3.1} \right).\)

As a result, standardisation yields:

\(2.9 < X < 3.1\)

if and only if

\(\frac{{2.9 - 3}}{{0.1}} < \frac{{X - 3}}{{0.1}} < \frac{{3.1 - 3}}{{0.1}}\)

\(\frac{{ - 0.1}}{{0.1}} < \frac{{X - 3}}{{0.1}} < \frac{{0.1}}{{0.1}}\)

\(1 < Z < 1\)

So,

\(P(2.9 < X < 3.1) = P( - 1 < Z < 1)\)

In this case,\(Z\)represents a standard normal distribution rv with edf\(\phi (z)\). Hence

\(\begin{array}{l}P(2.9 < X < 3.1) = P( - 1 < Z < 1)\\ = \phi (1) - \phi ( - 1)\end{array}\)

To get\(\phi (1)\)and\(\phi ( - 1),\)

\(\phi ( - 1) = 0.1587{\rm{ and }}\phi (1) = 0.8413\)

Hence

\(P\left( {2.9 < X < 3.1} \right) = 0.8413 - 0.1587\)

\({\rm{P}}(2.9 < {\rm{X}} < 3.1) = 0.6826\)

Let Z be a continuous rv with df as a proposition\(\phi (z)\). There any\(a\)and\(b\)with\(a < b\),

\(P(a < Z < b) = \phi (b) - \phi (a)\)

Result for machine first\(P(a < Z < b) = \phi (b) - \phi (a)\).

Machine second: The mean and standard deviation of cork diameters produced by machine are as follows:

\(\mu = 3.04\;{\rm{cm}}\)

\(\sigma = 0.02\;{\rm{cm}}\)

Since the likelihood of the cork diameter being between\(2.9\)and\(3.1\;{\rm{cm}}\)is denoted as\(P\left( {2.9 < X < 3.1} \right).\)

As a result, standardisation yields:

\(2.9 < X < 3.1\)

if and only if

\(\frac{{ - 0.14}}{{0.022}} < \frac{{X - 3.04}}{{0.02}} < \frac{{0.06}}{{0.02}}\)

\( - 7 < 2 < 3\)

So,

\(P(2.9 < X < 3.1) = P( - 7 < Z < 3)\)

\(Z\)is a standard normal distribution rv with edr in this case\(\phi (z).\)Hence

\(P(2.9 < X < 3.1) = P( - 1 < Z < 1) = \phi (3) - \phi ( - 7)\)

To get\(\phi (3)\)and\(\phi ( - 7),\)

\(\phi ( - 7) \approx 0{\rm{ and }}\phi (3) = 0.9987\)

Hence

\(P\left( {2.9 < X < 3.1} \right) = 0.9987 - 0\)

\({\rm{P}}(2.9 < {\rm{X}} < 3.1) = 0.9987\)

Therefore, the result of machine second is\({\rm{P}}(2.9 < {\rm{X}} < 3.1) = 0.9987\)