Question

a company that manufactures rivets believes the shear strength of the rivets they manufacture follows a normal model with a mean breaking strength of 950 pounds and a standard deviation of 40 pounds

a) what percentage of rivets selected at random will break when tested under a 900 pound load?
b) you're trying to improve the rivets and want to examine some that fail. use a simulation to estimate how many rivets you might need to test in order to find three that fail at 900 pounds or below.

Answer 1

Answer on Question #55447 – Math – Statistics and Probability

a company that manufactures rivets believes the shear strength of the rivets they manufacture follows a normal model with a mean breaking strength of 950 pounds and a standard deviation of 40 pounds

a) what percentage of rivets selected at random will break when tested under a 900 pound load?

b) you're trying to improve the rivets and want to examine some that fail. use a simulation to estimate how many rivets you might need to test in order to find three that fail at 900 pounds or below.

Solution

a) Let $X$ follow a normal model with a mean of 950 pounds and a standard deviation of 40 pounds. Let $Z$ follow a normal model with a mean of 0 pound and a standard deviation of 1 pound.

Then $Z = \frac{X - 950}{40}$.

Calculate

$$P(X < 900) = P\left(\frac{X - 950}{40} < \frac{900 - 950}{40}\right) = P\left(Z < \frac{900 - 950}{40}\right) = P(Z < -1.25)$$

From z-table obtain that

$$P(Z < -1.25) = 0.1056,$$

hence

$$P(X < 900) = 0.1056$$

10.56% of rivets selected at random will break when tested under a 900 pound load.

b) Proportion of rivets that fail at 900 pounds or below is $0.1056 = \frac{3}{n}$.

So sample size is determined by

$$n = \frac{3}{0.1056} = 28.4 \text{ round up to } 29.$$

Answer: a) 10.56%; b) 29.

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