91Ó°ÊÓ

The standard deviation is a statistic that calculates the square root of the variance and measures the dispersion of a dataset relative to its mean. By calculating each data point's divergence from the mean, the standard deviation is calculated as the square root of variance.

Let the bolt thread length be denoted by RV X and the mean and standard deviation of its distribution to \(\mu \)and \(\sigma \)respectively.

(a) The probability that the thread length of a randomly selected bolt is within \(1.5\)SDs of its mean value is denoted as\(P(\mu - 1.5\sigma < X < \mu + 1.5\sigma )\). Standardizing gives

\(\mu - 1.5\sigma < X < \mu + 1.5\sigma \)

if and only if

\(\begin{array}{l}\frac{{(\mu - 1.5\sigma ) - \mu }}{\sigma } < \frac{{X - \mu }}{\sigma } < \frac{{(\mu + 1.5\sigma ) - \mu }}{\sigma }\frac{{ - 1.5\sigma }}{\sigma }\\ < \frac{{X - \mu }}{\sigma } < \frac{{1.5\sigma }}{\sigma } - 1.5 < Z < 1.5\end{array}\)

Thus

\(P(\mu - 1.5\sigma < X < \mu + 1.5\sigma ) = P( - 1.5 < Z < 1.5)\)

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

\(P(\mu - 1.5\sigma < X < \mu + 1.5\sigma ) = \phi (1.5) - \phi ( - 1.5)\)

To get \(\phi (1.5)\)and \(\phi ( - 1.5)\), we check Appendix Table A.3, from there

\(\phi ( - 1.5) = 0.0668{\rm{ and }}\phi (1.5) = 0.9332\)

Hence

\(\begin{array}{l}P(\mu - 1.5\sigma < X < \mu + 1.5\sigma ) = 0.9332 - 0.0668\\P(\mu - 1.5\sigma < X < \mu + 1.5\sigma ) = 0.8664\end{array}\)

Proposition: Let $X$ be a continuous rv with cdf \(\phi (x)\). Then for any a and b with a<b,

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

(b) First, we calculate the probability that the thread length of a randomly selected bolt is within \(2.5\)SDs of its mean value is denoted as \(P(\mu - 2.5\sigma < X < \mu + 2.5\sigma )\). Standardizing gives

\(\mu - 2.5\sigma < X < \mu + 2.5\sigma \)

if and only if

\(\begin{array}{l}\frac{{(\mu - 2.5\sigma ) - \mu }}{\sigma } < \frac{{X - \mu }}{\sigma } < \frac{{(\mu + 2.5\sigma ) - \mu }}{\sigma }\frac{{ - 2.5\sigma }}{\sigma }\\ < \frac{{X - \mu }}{\sigma } < \frac{{2.5\sigma }}{\sigma } - 2.5 < Z < 2.5\end{array}\)

Thus

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

\(P(\mu - 2.5\sigma < X < \mu + 2.5\sigma ) = \phi (2.5) - \phi ( - 2.5)\)

To get \(\phi (2.5)\)and\(\phi ( - 2.5)\), we check Appendix Table A.3, from there

\(\phi ( - 2.5) = 0.0062{\rm{ and }}\phi (2.5) = 0.9938\)

Hence

\(\begin{array}{l}P(\mu - 2.5\sigma < X < \mu + 2.5\sigma ) = 0.9938 - 0.0062\\P(\mu - 2.5\sigma < X < \mu + 2.5\sigma ) = 0.9876\end{array}\)

By using definition of the probability of the complement of an event, we can say that:

\(\begin{array}{l}P(Farther\,than\,2.5\,{\rm{SDs}}\,from\,mean) = 1 - P(\mu - 2.5\sigma < X < \mu + 2.5\sigma )\\P(Farther\,than\,2.5\,{\rm{SDs}}\,from\,mean) = 1 - 0.9876\\P(Farther\,than\,2.5\,{\rm{SDs}}\,from\,mean) = 0.0124\end{array}\)

Definition: If \(\bar A\) is the complement of an event A, then

\(P(\bar A) = 1 - P(A)\)

(c) Observing the results from previous two parts, we can conclude following result:

\(P(\mu - n\sigma < X < \mu + n\sigma ) = \phi (n) - \phi ( - n)\)

Using this result, we can say that

\(\begin{array}{c}P({\rm{ Between }}1{\rm{ and }}2{\rm{ SDs from mean }}) = P({\rm{ Within }}2{\rm{ SDs from mean }}) - P({\rm{ Within }}1{\rm{ SDs from mean }})\\ = P(\mu - 2\sigma < X < \mu + 2\sigma ) - P(\mu - \sigma < X < \mu + \sigma )\\ = (\phi (2) - \phi ( - 2)) - (\phi (1) - \phi ( - 1))\\ = (0.9772 - 0.0228) - (0.8413 - 0.1587)\\ = 0.9544 - 0.6826\\P({\rm{ Between }}1{\rm{ and }}2{\rm{ SDs from mean }}) = 0.2718\end{array}\)