Hello,

I am doing my homework and just have a gut feeling that I didn't do a few problems correct. Any help would be greatly appreciated.

1. An architect designing the men’s gymnasium at Slippery Rock University wants to make the interior doors high enough to allow 95% of the men at least a 1-foot clearance. Assuming that the height of men are normally distributed with mean of 70 inches and standard deviation of 3 inches, how high must the architect make the doors?

For my answer, I got 75.88. I got this by:

70 + 1.960(3)/sqrt of 1



2. The weights of packages for a certain high-volume product received at a department store have a mean of 200 pounds and standard deviation of 15 pounds. What is the probability that 16 packages received at random and loaded on an elevator will exceed the safety limit of the elevator, specified at 3300 pounds?

My answer was .8372 chance the packages will exceed the safety limit.
I got that by:
3300/16= 206.25
z-score: (206.25-200)/15 = .4166666667
1-1.628= .8372


3. An automatic fluid filling machine has been adjusted so that it pours an average (mean) of 16.25 ounces into bottles whose contents are supposed to be 16 ounces. The amounts of fill in individual bottles are normally distributed with a standard deviation of 0.5 ounce.

(a) If a random sample of 9 bottles is selected, what is the probability that the sample-mean amount of fill will be less than 16 ounces?
I got that there is a .0668 chance.
I got that by 16-16.25/(0.5/3)= -1.5 = .0668

(b) In order to lower this probability of under-fill, the machine was planned to be re-adjusted so that its setting targets a higher mean. What mean should be targeted so that the probability of the sample mean (of 9 bottles) being less than 16 ounce is 0.01
I got the target mean should be 16.387 ounces.

16-16.387/(.5/3)= =2.322=.012

1. An architect designing the men's gymnasium at Slippery Rock University wants to make the interior doors high enough to allow 95% of the men at least a 1-foot clearance. Assuming that the height of men are normally distributed with mean of 70 inches and standard deviation of 3 inches, how high must the architect make the doors?

For my answer, I got 75.88. I got this by:

70 + 1.960(3)/sqrt of 1

It's really much easier if you post one question at a time so we don't get confused over what's going on.

It's one-tailed -- it's 1.645, not 1.96.

Why the square root thing? Not that it changes this, but where's it coming from? Multiplying by 3 gives you the distance from the mean, then add it. You're right at the spot for 95%. There's no division there.

The answer you get is the spot where 95% of the men are at, or lower. (i.e. no more than 95% are taller than that.) But the door itself needs to allow for one foot clearance. You didn't allow for that.