Question

the diameter of a cylindrical shaft door hinge is normally distributed with a population mean of 12.8mm and standard deviation of 0.35mm. The specifications on the shaft are 13 ± 0.15 mm. what proportion of the shafts conforms to the specifications?

Answer 1

Answer:

sorry

Step-by-step explanation:

I don't know

Answer 2

28.57% of the shafts conforms to the specifications.

Step-by-step explanation:

We are given that the diameter of a cylindrical shaft door hinge is normally distributed with a population mean of 12.8 mm and standard deviation of 0.35 mm.

Also, the specifications on the shaft are 13 ± 0.15 mm.

Let X = diameter of a cylindrical shaft door hinge

SO, X ~ Normal($\mu=12.8,\sigma^{2} =0.35^{2}$)

The z-score probability distribution of normal distribution is given by;

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

where, $\mu$ = population mean = 12.8 mm

            $\sigma$ = standard deviation = 0.35 mm

Now, we are given that the specifications on the shaft are 13 ± 0.15 mm which that the the specification are between (13-0.15) = 12.85 mm and (13+0.15) = 13.15 mm

So, the proportion of the shafts conforms to the specifications is given by = P(12.85 mm < X < 13.15 mm)

 P(12.85 mm < X < 13.15 mm) = P(X < 13.15 mm) - P(X $\leq$ 12.85 mm)

 P(X < 13.15 mm) = P( $\frac{X-\mu}{\sigma}$ < $\frac{13.15-12.8}{0.35}$ ) = P(Z < 1) = 0.84134

 P(X $\leq$ 12.85 mm) = P( $\frac{X-\mu}{\sigma}$ $\leq$ $\frac{12.85-12.8}{0.35}$ ) = P(Z $\leq$ 0.14) = 0.55567

The above probabilities are calculated using z table by looking at the critical values of x = 1 and x = 0.14 which gives an probability area of 0.84134 and 0.55567 respectively.

Therefore, P(12.85 mm < X < 13.15 mm) = 0.84134 - 0.55567 = 0.28567

Hence, 28.57% of the shafts conforms to the specifications.

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