Step 1:

Given that, the data provided is the shear strength (psi) of the last 10 spot welds.

The data is,

392, 376, 401, 367, 389, 362, 409, 415, 358, 375

 
Step 2:

a.

Assume that the shear strength is normally distributed.

Let X denote the shear strength (psi) of the last 10 spot welds.

Here X follows normal distribution, so the maximum likelihood estimate of true average shear is computed by sample mean $\left( {\hat \mu = \overline X } \right)$ and standard deviation of shear mean is nothing but the sample standard deviation$\left( {\hat \sigma } \right)$.

Now, the following table is used to estimate the average shear strength and standard deviation of shear strength using the maximum likelihood.

The table is,

Xi	Xi-Xbar	(Xi-Xbar)
392	7.6	57.76
376	-8.4	70.56
401	16.6	275.56
367	-17.4	302.76
389	4.6	21.16
362	-22.4	501.76
409	24.6	605.16
415	30.6	936.36
358	-26.4	696.96
375	-9.4	88.36
SumXi=3844		Sum (Xi-Xbar)=3556.4

The sample mean is computed as follows,

\[\begin{array}{c} \bar X = \frac{1}{n}\sum\limits_{i = 1}^n {{X_i}} \\ = \frac{1}{{10}}\sum\limits_{i = 1}^{10} {{X_i}} \\ = \frac{{3.844}}{{10}}\\ = 384.4 \end{array}\]

Therefore, the sample mean is 384.4.

The value of estimated variance ${\hat \sigma ^2}$ is computed as follows,

\[\begin{array}{c} {{\hat \sigma }^2} = \frac{1}{n}\sum\limits_{i = 1}^n {{{\left( {{X_i} - \bar X} \right)}^2}} \\ = \frac{1}{{10}}\sum\limits_{i = 1}^{10} {{{\left( {{X_i} - \bar X} \right)}^2}} \\ = \frac{{3556.4}}{{10}}\\ = 355.64 \end{array}\]

The standard deviation of X is calculated as follows,

\[\begin{array}{c} \hat \sigma = \sqrt {{{\hat \sigma }^2}} \\ = \sqrt {355.64} \\ = 18.86 \end{array}\]

Therefore, the maximum likelihood estimate of the true average shear strength is $\hat \mu = 384.4$ psi and the maximum likelihood estimate of the standard deviation of shear strength is $\hat \sigma = 18.86$ psi.

 
Step 3:

b.

Invariance principle of maximum likelihood estimation:

The invariance principle of maximum likelihood estimator states that, if the mle’s of parameters ${\theta _1},{\theta _2},.....,{\theta _n}$ of a distribution be ${\hat \theta _1},{\hat \theta _2},.....,{\hat \theta _n}$,then the mle for the function of the parameters,

$h\left( {{\theta _1},{\theta _2},.....,{\theta _n}} \right)$ will be $h\left( {{{\hat \theta }_1},{{\hat \theta }_2},.....,{{\hat \theta }_n}} \right)$.

The value of strength is below 95% of all welds is computed as follows:

Let X denote the value of strength is below 95% of all welds.

So,

\[\begin{array}{c} P\left( {X < x} \right) = 0.95\\ P\left( {\frac{{X - \mu }}{\sigma } < \frac{{x - \mu }}{\sigma }} \right) = 0.95\\ P\left( {Z < \frac{{x - \mu }}{\sigma }} \right) = 0.95\\ \phi \left( {\frac{{x - \mu }}{\sigma }} \right) = 0.95 \end{array}\]

Using the standard normal table A.3, the 95% percentile of the standard normal variable Z is 1.645.

\[\phi \left( {1.645} \right) = 0.95\]

Therefore,

\[\begin{array}{c} \phi \left( {1.645} \right) = \phi \left( {1.645} \right)\\ \frac{{x - \mu }}{\sigma } = 1.645\\ x = \mu + 1.645\sigma \end{array}\]

Using the invariance principle of mle,

\[\begin{array}{c} x = \hat \mu + 1.645\hat \sigma \\ = 384.4 + 1.645 \times 18.86\\ = 415.4247\\ \approx 415 \end{array}\]

Therefore, the value of strength is below 95% of all welds is 415.

 
Step 4:

c.

Let X denote the shear strength of the new test spot weld.

Thus,

\[\begin{array}{c} P\left( {X \le 400} \right) = P\left( {\frac{{X - \mu }}{\sigma } \le \frac{{400 - \mu }}{\sigma }} \right)\\ = P\left( {Z \le \frac{{400 - \mu }}{\sigma }} \right) \end{array}\]

Using the invariance principle,

The mle of $\left( {\frac{{400 - \mu }}{\sigma }} \right)$ is $\left( {\frac{{400 - \hat \mu }}{{\hat \sigma }}} \right)$.

So,

\[\begin{array}{c} P\left( {X \le 400} \right) = P\left( {Z \le \frac{{400 - \hat \mu }}{{\hat \sigma }}} \right)\\ = P\left( {Z \le \frac{{400 - 384.4}}{{18.86}}} \right)\\ = \phi \left( {\frac{{400 - 384.4}}{{18.86}}} \right)\\ = \phi \left( {0.8271} \right) \end{array}\]

Using the standard normal table A.3, the value of 0.8271 at left tailed is 0.7959.

\[\phi \left( {0.8271} \right) = 0.7959\]

Thus,

\[P\left( {X \le 400} \right) = 0.7959\]

Therefore, the value of $P\left( {X \le 400} \right)$ is 0.7959.