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neonofarm [45]
3 years ago
11

The standard diameter of a golf ball is 42.67 mm. A golf ball factory does quality control on the balls it manufactures. Golf ba

lls are randomly measured to ensure the correct size. One day, an inspector decides to stop production if the discrepancy in diameter is more than 0.002 mm. What is the range of acceptable values?
42.668 mm to 42.670 mm
42.668 mm to 42.672 mm
42.670 mm to 42.672 mm
42.670 mm to 42.674 mm
Mathematics
1 answer:
sergiy2304 [10]3 years ago
4 0

THE ANSWER IS A:

This is the function:

f(x)=|42.67-x|.

42.670 -.002 =42.668

42.668 mm to 42.670

Explanation:

Production will be stopped if the difference in diameter is greater than 0.002 mm.

This difference can mean that the golf ball has a diameter more than 0.002 larger than 42.67 mm, or a diameter more than 0.002 smaller than 42.67. This is the reason for using an absolute value function.

 Absolute value represents the distance a number is from 0. For this function, we would want only the values where the function is less than 0.002.

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Answer:

(−0.103371 ; 0.063371) ;

No ;

( -0.0463642, 0.0063642)

Step-by-step explanation:

Shift 1:

Sample size, n1 = 30

Mean, m1 = 10.53 mm ; Standard deviation, s1 = 0.14mm

Shift 2:

Sample size, n2 = 25

Mean, m2 = 10.55 ; Standard deviation, s2 = 0.17

Mean difference ; μ1 - μ2

Zcritical at 95% confidence interval = 1.96

Using the relation :

(m1 - m2) ± Zcritical * (s1²/n1 + s2²/n2)

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Lower boundary :

-0.02 - 0.0833710 = −0.103371

Upper boundary :

-0.02 + 0.0833710 = 0.063371

(−0.103371 ; 0.063371)

B.)

We cannot conclude that gasket from shift 2 are on average wider Than gasket from shift 1, since the interval contains 0.

C.)

For sample size :

n1 = 300 ; n2 = 250

(10.53-10.55) ± 1.96*sqrt(0.14^2/300 + 0.17^2/250)

Lower boundary :

-0.02 - 0.0263642 = −0.0463642

Upper boundary :

-0.02 + 0.0263642 = 0.0063642

( -0.0463642, 0.0063642)

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