Hi Dylan,
This question essentially asks all permutations of the classic equation in elementary statistics: z= (x-mu)/sigma. Let's get to it.
(a) This is looking at a single observation, so the equation is:
z=(x-mu)/sigma
Breaking this down:
x=weight we're interested in testing
mu=mean weight
sigma=standard deviation
Therefore, for this problem:
x=152
mu=185
sigma=29
z=(152-185)/29
z= -.1.14
Now, go to your z-table and search for -1.1 in the row and 0.04 in the column. This gives you the probability that z< -1.14, which is equivalent to P(x<152) So:
P(Z< -.1.14)=P(X<152)=0.1271
But we were asked for greater. This is where you apply the complement rule or, as I call it, the one minus trick:
P(X>a)=1-P(X<a), so here:
P(X>152)=1-P(X<152)=1-P(Z< -1.14)=
1-0.1271=
P=0.8729
(b) Now, we have to modify the classic equation a bit since we are working with a sample size. Equation is now:
z=(x-mu)/SE
SE=standard error, which has its own calculation and formula
SE=sigma/sqrt(n)
Let's break this down:
sigma=standard deviation
n=sample size
Therefore:
SE=sigma/sqrt(n)
sigma=29
n=27
SE=29/sqrt(27)
SE=5.58
Now, we can plug this back into our initial equation:
z=(x-mu)/SE
x=152
mu=185
SE=5.58
z=(152-185)/5.58
z= -5.91
Now, if you look at the z-table, this value is off the scale. That means:
P(Z< -5.91) is effectively 0, so low it can't be practically displayed on a table.
Again, we are asked for greater, so:
P(Z> -5.91)=P(X>152)= 1-P(Z< -5.91)= 1.00
It is almost certain that mean weight of a 27-man sample will exceed 152 lbs.; statistical software will provide a more exact answer.
(c) It's hazardous. If a 27-man sample is virtually guaranteed to weigh over 152 lbs., and that's your safe mean weight, you will likely have problems with that elevator. Maybe make it women's only if women weigh less?
I hope this helps.
