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The mean diameter of holes produced by a drilling machine bit is 4.05 mm and the stan dard deviation of the diameters is 0.0028 mm For twenty holes drilled using this machine, determine, correct to the nearest whole num ber, how many are likely to have diame ters of between (a) 4.048 and 4.0553 mm and (b) 4.052 and 4.056 mm, assuming the diameters are normally distributed.

Sagot :

The total number of holes will be (a)15 and (b) 4 when the drilling machine is used whose standard deviation is 0.0028mm.

The square root of the variance is used to calculate the standard deviation, a statistic that expresses how widely distributed a dataset is in relation to its mean.

The sum of all values divided by the total number of values constitutes a dataset's mean.

The mean of the holes are given as 4.05 mm([tex]\mu=4.08[/tex])

The standard deviation is given as 0.0028 mm([tex]\rho=0.0028[/tex])

Now we will use the z-score and probability distribution to calculate the frequency of the holes to have diameter between 4.048 and 4.0553.

Here [tex]Pr(4.048\leq X\leq 4.0053)[/tex] ,so we need to compute the z-values

[tex]Z_1=\frac{X_1-\mu}{\rho} \\Z_1=\frac{4.048-4.05}{0.0028} \\Z_1=-0.7143[/tex]

Similarly:

[tex]Z_2=1.8929[/tex]

Therefore from the above statements we can say that

[tex]Pr(-0.7143\leq X\leq 1.8929)[/tex]

Solving for probability distribution we get

[tex]Pr(4.048\leq X\leq 4.0053)=0.7333[/tex]

Total number of holes=0.7333 × 20= 14.666..

Therefore the number of holes is 15

Similarly

The z-value corresponding to 4.052 mm is given by 0.71

The z-value corresponding to 4.056 mm is given by:2.14

The probability of the diameter being between 4.052 mm and 4.056 mm is 0.4838 – 0.2611 =0.2227

The number likely to have a diameter between 4.052 mm and 4.056 mm = 0.2227 × 20

= 4.454

= 4, correct to nearest whole number

To learn more about standard deviation:

https://brainly.com/question/24298037

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