91Ó°ÊÓ

The standard deviation, defined as the square root of the variance, is a statistic that represents the dispersion of a dataset relative to its mean. By determining each data point's departure relative to the mean, the standard deviation is calculated as the square root of variance.

Let the Rockwell hardness of a particular alloy be represented by a rv X then its mean and standard deviation are given as:

\(\begin{array}{l}\mu = 70\\\sigma = 3\end{array}\)

(a) Since a specimen is acceptable only if its hardness is between\(67{\rm{ }}and{\rm{ }}75\).

Hence the probability that a randomly chosen specimen has an acceptable hardness can be represented as \(P(67 < X < 75)\). Standardizing gives:

\(67 < X < 75\)

if and only if

\(\begin{array}{l}\frac{{67 - 70}}{3} < \frac{{X - 70}}{3} < \frac{{75 - 70}}{3}\frac{{ - 3}}{3}\\ < \frac{{X - 70}}{3} < \frac{5}{3} - 1 < Z < 1.67\end{array}\)

Thus

\(P(67 < X < 75) = P( - 1 < Z < 1.67)\)

Here, Z is a standard normal distribution rv with cdf \(\phi (z)\). Hence, we can write

\(P(67 < X < 75) = \phi (1) - \phi ( - 1)\)

To get \(\phi (1)\)and\(\phi ( - 1)\), we check Appendix Table A.3, from there

\(\phi ( - 1) = 0.1587{\rm{ and }}\phi (1.67) = 0.9525\)

Hence,

\(\begin{array}{l}P(67 < X < 75) = 0.9525 - 0.1587\\{\rm{P}}(67 < {\rm{X}} < 75) = 0.7938\end{array}\)

Proposition: Let Z be a continuous rv with\({\mathop{\rm ddf}\nolimits} \phi (z)\). Then for any a and b with a<b,

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

(b) Since the acceptable range of hardness is \((70 - c,70 + c)\). Hence the probability that a randomly chosen specimen has an acceptable hardness can be represented as \(P(70 - c < X < 70 + c)\). Standardizing gives:

\(70 - c < X < 70 + c\)

if and only if

\(\begin{array}{l}\frac{{(70 - c) - 70}}{3} < \frac{{X - 70}}{3} < \frac{{(70 + c) - 70}}{3}\frac{{ - c}}{3}\\ < \frac{{X - 70}}{3} < \frac{c}{3} - \frac{c}{3} < Z < \frac{c}{3}\end{array}\)

Thus

\(P(70 - c < X < 70 + c) = P\left( { - \frac{c}{3} < Z < \frac{c}{3}} \right)\)

Here $Z$ is a standard normal distribution rv with cdf \(\phi (z)\). Hence, we can write

\(P(70 - c < X < 70 + c) = \phi \left( {\frac{c}{3}} \right) - \phi \left( { - \frac{c}{3}} \right)\)

Now due to the symmetry of standard normal distribution curve, we know that

\(\phi \left( { - \frac{c}{3}} \right) = 1 - \phi \left( {\frac{c}{3}} \right)\)

Hence,

\(P(70 - c < X < 70 + c) = 2\phi \left( {\frac{c}{3}} \right) - 1\)

But it is already given in the question that this probability is equal to \(0.95\), Hence

\(\begin{array}{c}2\phi \left( {\frac{c}{3}} \right) - 1 = 0.95\phi \left( {\frac{c}{3}} \right)\\ = \frac{{1 + 0.95}}{2}\phi \left( {\frac{c}{3}} \right) = 0.975\end{array}\)

We check Appendix Table A.3 if \(\phi (z)\)is equal to \(0.975\)for any z-value, from there

\(\begin{array}{l}\frac{c}{3} = 1.96\\c = 5.88\end{array}\)

(c) Let Y denote the number of acceptable specimens out of \(10\), then Y is binomially distributed. i.e.,

\(Y \approx {\mathop{\rm Bin}\nolimits} (10,p)\)

Where the sample size is \(10\) and p denotes the probability of success or in this case it denotes the probability that a randomly selected specimen out of the sample is acceptable.

From part(a):

\(p = 0.7938\)

Then using the fundamental property of Binomial distribution, we can write:

\(\begin{array}{l}E(Y) = n \cdot p\\E(Y) = 10 \cdot (0.7938)\\E(Y) = 7.938\end{array}\)

(d) Let Y be the rv that denotes the number among the ten specimens with hardness less than\(73.84\), then Y is binomially distributed with \(n = 10\)and p denotes the probability that a selected specimen has hardness less than \(73.84\).

p is nothing but \(P(X < 73.84)\). Standardizing gives:

\(X < 73.84\)

if and only if

\(\begin{array}{l}\frac{{X - 70}}{3} < \frac{{73.84 - 70}}{3}\frac{{X - 70}}{3} < \frac{{3.84}}{3}\\Z < 1.28\end{array}\)

Thus

\(P(X < 73.84) = P(Z < 1.28)\)

Here Z is a standard normal distribution rv with cdf \(\phi (z)\). Hence, we can write

\(P(X < 73.84) = \phi (1.28)\)

To get \(\phi (1.28)\), we check Appendix Table A.3, from there

\(\phi (1.28) = 0.8997\)

Hence

\(p = P(X < 73.84) = 0.8997\)

Proposition: Let Z be a continuous rv with \({\mathop{\rm cdf}\nolimits} \phi (z)\). Then for any a and b with a<b,

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

Step 6

Now let us recall the following result related to Binomial distribution: Let Y be a binomial rv based on n trials with success probability p. Then if the binomial probability histogram is not too skewed, Y has approximately a normal distribution with \(\mu = np\)and\(\sigma = \sqrt {np(1 - p)} \). In particular, for \(y = \)a possible value of Y,

\(P(Y \le y) = {\mathop{\rm Bin}\nolimits} (y,n,p) = \phi \left( {\frac{{y + 0.5 - \mu }}{\sigma }} \right)\)

Since we already know that

\(Y \approx {\mathop{\rm Bin}\nolimits} (10,0.8997)\)

Then it's mean and standard deviation are given as:

\(\begin{array}{l}\mu = 10(0.8997) = 8.997\\\sigma = \sqrt {10(0.8997)(1 - 0.8997)} = 0.9499\end{array}\)

Then the probability that at most eight of ten independently selected specimens have a hardness of less than \(73.84\)can be denoted as \(P(Y \le 8)\)

\(\begin{array}{l}P(Y \le 8) = \sum\limits_{y = 0}^8 {{{(0.8997)}^y}} {(1 - 0.8997)^{1 - y}}\\P(Y \le 8) = 0.2651\end{array}\)