Rivets believed to have shear strength modeled by....

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lake

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A company that manufactures rivets believes the shear strength (in pounds) is modeled by N(800,50).

About what percent of these rivets would you expect to fall below 900 pounds?
 
Re: Rivets:

lake said:
...the shear strength (in pounds) is modeled by N(800,50).
Click to expand...
What is the meaning of "N(800, 50)"? Does this mean "a normal distribution, with a mean of mu = 800 and a standard deviation of sigma = 50"? Or something else?

What have you tried? How far have you gotten? Where are you stuck?

Please be complete. Thank you! :D

Eliz.
 
If 800 is the mean and 50 the SD, just apply the 68-95-99.7 rule. It's straightforward.
 
Re: Rivets:

stapel said:
lake said:
...the shear strength (in pounds) is modeled by N(800,50).
Click to expand...
What is the meaning of "N(800, 50)"? Does this mean "a normal distribution, with a mean of mu = 800 and a standard deviation of sigma = 50"? Or something else?

What have you tried? How far have you gotten? Where are you stuck?

Please be complete. Thank you! :D

Eliz.
Click to expand...


The meaning is that N(880,50) means that 800 is the mean and 50 is the standard deviation. I'm not really stuck, I got 97.7%. I was just trying to see if that was the correct answer.
 
You're correct. Using the rule, it's 97.5%.

Because 50+34+13.5=97.5

More accurate is \(\displaystyle \frac{900-800}{50}=2\)

Which corresponds to .9772 in the z-table. 97.72%
 
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