Question

The diameters of bolts produced in a machine shop are normally distributed with a mean of 5.725.72 millimeters and a standard deviation of 0.070.07 millimeters. Find the two diameters that separate the top 5%5% and the bottom 5%5%. These diameters could serve as limits used to identify which bolts should be rejected. Round your answer to the nearest hundredth, if necessary.

Answer 1

Answer:

Top 5% is 5.84 milliters and the bottom 5% is 5.60 millimeters.

Step-by-step explanation:

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$, the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

$\mu = 5.72, \sigma = 0.07$

Top 5%:

X when Z has a pvalue of 0.95. So X when Z = 1.645

$Z = \frac{X - \mu}{\sigma}$

$1.645 = \frac{X - 5.72}{0.07}$

$X - 5.72 = 1.645*0.07$

$X = 5.84$

Bottom 5%:

X when Z has a pvalue of 0.05. So X when Z = -1.645

$Z = \frac{X - \mu}{\sigma}$

$-1.645 = \frac{X - 5.72}{0.07}$

$X - 5.72 = -1.645*0.07$

$X = 5.60$

Top 5% is 5.84 milliters and the bottom 5% is 5.60 millimeters.

Answer 2

Answer:

The two diameters that separate the top 5% and the bottom 5% are 5.84 and 5.60 respectively.

Step-by-step explanation:

We are given that the diameters of bolts produced in a machine shop are normally distributed with a mean of 5.72 millimeters and a standard deviation of 0.07 millimeters.

Let X = diameters of bolts produced in a machine shop

So, X ~ N($\mu=5.72,\sigma^{2} = 0.07^{2}$)

The z score probability distribution is given by;

         Z = $\frac{X-\mu}{\sigma}$ ~ N(0,1)

where, $\mu$ = mean diameter = 5.72 millimeter

            $\sigma$ = standard deviation = 0.07 millimeter

Now, we have to find the two diameters that separate the top 5% and the bottom 5%.

• Firstly, Probability that the diameter separate the top 5% is given by;

       P(X > x) = 0.05

       P( $\frac{X-\mu}{\sigma}$ > $\frac{x-5.72}{0.07}$ ) = 0.05

       P(Z > $\frac{x-5.72}{0.07}$ ) = 0.05

So, the critical value of x in z table which separate the top 5% is given as 1.6449, which means;

                 $\frac{x-5.72}{0.07}$  = 1.6449

               ${x-5.72} = 1.6449 \times {0.07}$

                        $x$   = 5.72 + 0.115143 = 5.84

• Secondly, Probability that the diameter separate the bottom 5% is given by;

       P(X < x) = 0.05

       P( $\frac{X-\mu}{\sigma}$ < $\frac{x-5.72}{0.07}$ ) = 0.05

       P(Z < $\frac{x-5.72}{0.07}$ ) = 0.05

So, the critical value of x in z table which separate the bottom 5% is given as -1.6449, which means;

                 $\frac{x-5.72}{0.07}$  = -1.6449

               ${x-5.72} = -1.6449 \times {0.07}$

                        $x$   = 5.72 - 0.115143 = 5.60

Therefore, the two diameters that separate the top 5% and the bottom 5% are 5.84 and 5.60 respectively.