A company manager for a tire manufacturer is in charge of making sure there is the least amount of defective tires. If the tire manufacturing process is working properly, the average weight of a tire for a 4-door sedan is normally distributed with a mean of 22 pounds and a standard deviation of 0.76 pounds. The manager decides to reject a tire as underweight if it falls more than 1.5 interquartile ranges below the lower quartile of the specified shipment of tires. What proportion of tires will be rejected by this process?

For a standard normal distribution, (with mean 0 and standard deviation 1), the lower and upper quartiles are located at -0.67448 and +0.67448 respectively. Thus the interquartile range (IQR) is 1.34896.

And the manager decides to reject a tire as underweight if it falls more than 1.5 interquartile ranges below the lower quartile of the specified shipment of tires.

1.5 of the Interquartile range = 1.5 × 1.34896 = 2.02344

1.5 of the interquartile range below the lower quartile = (lower quartile) - (1.5 of Interquartile range) = -0.67448 - 2.02344 = -2.69792

The proportion of tires that will fall 1.5 of the interquartile range below the lower quartile = P(x < -2.69792) ≈ P(x < -2.70)

Using data from the normal distribution table

P(x < -2.70) = 0.00347 = 0.347% of the total tires will be rejected as underweight

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JackelineCasarez JackelineCasarez · Answer 2 · 2025-03-26T07:48:37-04:00

The proportion of the tires that would be denied for being underweight through the given process would be:

[tex]0.347[/tex]% of the total tires will be rejected as underweight.

Given that,

Interquartile Range [tex]= 1.5[/tex]

Standard Deviation [tex]= 0.76[/tex]

Considering Mean = 0

and Standard Deviation = 1

Since lower quartile = -0.67448

Upper quartile = +0.67448

IQ range = 1.34896

To find,

The proportion of tires would be rejected due to being underweight through the process would be:

1.5 of Interquartile Range = 1.5 × [tex]1.34896 = 2.02344[/tex]

Now,

1.5 of the IQ range below the lower quartile [tex]= (lower quartile) - (1.5 of Interquartile range)[/tex]

[tex]= -0.67448 - 2.02344[/tex]

[tex]= -2.69792[/tex]

The proportion of tires that would be under 1.5 of the interquartile range below the lower quartile:

[tex]= P(x < -2.69792)[/tex] ≈ [tex]P(x < -2.70)[/tex]

Using data through the Normal Distribution Table,

[tex]P(x < -2.70)[/tex] [tex]= 0.00347[/tex]

[tex]= 0.347[/tex]%

Thus, 0.347% of the total tires would be rejected as underweight.

Learn more about "Proportion" here:

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