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

[lake]

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?

 

[stapel]

Re: Rivets:

...the shear strength (in pounds) is modeled by N(800,50).

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!

Eliz.

 

[galactus]

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

 

[lake]

Re: Rivets:

...the shear strength (in pounds) is modeled by N(800,50).

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!

Eliz.

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.

 

[galactus]

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%