The manufacturing of a ball bearing is normally distributed with a mean diameter

Question

The manufacturing of a ball bearing is normally distributed with a mean diameter of 22 millimeters and a standard deviation of .016 millimeters. To be acceptable the diameter needs to be between 21.97 and 22.03 millimeters. a. What is the probability that a randomly selected ball bearing will be acceptable ? (Round to tenth of a percent) b. What does the acceptable range of the diameter need to be if you wanted to accept 98% of the ball bearings ? (Round to thousandths)(Type out what TI-83/84 calculator function you used and what numbers were typed in to arrive at your answer)

Explanation / Answer

we are given that mean = 22 and sd = 0.16

and we need to find

P(21.97< X< 22.03) , so we use the z score formula to calculate the values as

Z = (x-mean)/SD

(21.97-22)/0.16) = -0.1875

and (22.03-22/0.16) = 0.1875

now we read the z tables to arrive at the value as

To find the probability of P (0.1875<Z<0.1875), we use the following formula:

P (0.1875<Z<0.1875 )=P ( Z<0.1875 )P (Z<0.1875 )

P ( Z<0.1875 ) can be found by using the following fomula.

P ( Z<a)=1P ( Z<a )

After substituting a=0.1875 we have:

P ( Z<0.1875)=1P ( Z<0.1875 )

We see that P ( Z<0.1875 )=0.5753 so,

P ( Z<0.1875)=1P ( Z<0.1875 )=10.5753=0.4247

At the end we have:

P (0.1875<Z<0.1875 )=0.1506

b)

now we are given the probability as 0.98 , so we first calculate the z score as

2.053

now we use the z score formula again as

Z = (X-Mean)/SD

2.053 = (X - 22)/0.016

X = 2.053*0.016 + 22 = 22.032

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