1. ## Normal distribution question

The tensile strength of a certain metal component is normally distributed with a mean of 10,000 kilograms per square centimeter and a standard deviation of 100 kilograms per square centimeter. Measurements are recorded to the nearest 50 kilograms per square centimeter.

What is proportion of these proponents exceed 10,150 kilograms per square centimeter in tensile strength?

$\displaystyle p(X>10,150)$

$\displaystyle p(Z>\frac{10,150-10,000}{100}) = p(Z>1.50) = p(Z<-1.50) =$$\displaystyle 0.0668\approx 6.7\%$

In the book the answer is 0.0401. Working backwards:

$\displaystyle p(Z< z)=0.0401$

$\displaystyle \frac{x-10,000}{100}=0.0401$

$\displaystyle x=10,750$

Why is that? I see that the problem states that measurements are recorded to the nearest 50 kilograms per centimeter but why is x=10,750? I would have thought it'd equal to either 10,150 or 10,200, not their average. Is this always the case with similarly worded problems where it's stated that x is rounded?

2. $\displaystyle P\left(X \geq 10,175\right) = .0400592030528$

"exceed 10,150"

So it can't be equal to 10,150, Since its round to the nearest 50, then 10,200, 24 would round down, 25 would round up. so 10,150 + 25 = 10,175, round up would be 10,200.

EDIT: quote from "mr fantastic", "I cannot stand answers to questions like these expressed as a percentage. You should be giving a number correct to four decimal places that lies between 0 and 1."

.0400592030528 = .0401

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# the tensile strength of a certain metal component is normally distributed with a mean of

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