91影视

Internal pressure describes how a system's internal energy varies as it expands or contracts at a constant temperature.

\(X\)has a normal distribution with

\(\mu = 30mm\)

\(\sigma = 7.8mm\)

(a)

The standardised score is the average of all students' scores value\({\bf{x}}\)lowered by the standard deviation and then divided by the mean.

\(z = \frac{{x - \mu }}{\sigma } = \frac{{20 - 30}}{{7.8}} \approx - 1.28\)

\(P(X \le 20) = P(Z \le - 1.28) = 0.1003 = 10.03\% \)

\(P\left( {X < 20} \right) = P\left( {Z < - 1.28} \right) = 0.1003 = 10.03{\rm{ }}\% \)

The probability is \(P\left( {X < 20} \right) = P\left( {Z < - 1.28} \right) = 0.1003 = 10.03\% \).

(b)

The probabilities to the left of z-scores are listed in the normal probability table in the appendix. likelihood):

\(z = 0.67\)

\(\begin{aligned} x &= \mu + z\sigma \\ &= 30 + 0.67(7.8)\\ &= 30 + 5.226\\ &= 35.226\end{aligned}\)

Thus, \(35.226\;{\rm{mm}}\) is the \(75th\) percentile.

(c)

The probabilities to the left of z-scores are listed in the normal probability table in the appendix.

The\(15th\)percentile has the property that\(15\% \)of the data values lie below it The corresponding z-score is then the z-score in the normal probability table that corresponds with a probability of\(15\% \)or\(0.15\)(or the closest probability):

\(z = - 1.04\)

Because the z-score measures the number of standard deviations that a value is distant from the mean, the equivalent values are the mean multiplied by the product of the z-score and the standard deviation:

\(\begin{aligned}x &= \mu + z\sigma \\ &= 30 - 1.04(7.8)\\ &= 30 - 8.112\\ &= 21.888\end{aligned}\)

Thus, \(21.888\;{\rm{mm}}\)is the \(15th\) percentile.

(d)

The probabilities to the left of z-scores are listed in the normal probability table in the appendix.

The middle\(80\% \)has the property that\(10\% \)of the data values lie below its lower boundary and\(90\% \)of the data values lie below its upper boundary. The corresponding z-score is then the z-score in the normal probability table that corresponds with a probability of\(10\% \)\(,90\% \)or\(0.10/0.90\)(or the closest probability):

\(z = \pm 1.28\)

Because the z-score measures the number of standard deviations that a value is distant from the mean, the equivalent values are the mean multiplied by the product of the z-score and the standard deviation:

\(\begin{aligned}x &= \mu + z\sigma \\ &= 30 - 1.28(7.8)\\ &= 30 - 9.984\\ &= 20.016\end{aligned}\)

\(\begin{aligned}x &= \mu + z\sigma \\ &= 30 + 1.28(7.8)\\ &= 30 + 9.984\\ &= 39.984\end{aligned}\)

So, the values that separate the middle \(80\% \) of the defect length distribution from the smallest \(10\% \) and the largest \(10\% \) are \(20.016\;{\rm{mm}}\)and \(39.984\;{\rm{mm}}.\)