Hey, I'm new here and somewhat new to statistics. I have been doing a bit of homework lately, and have been stuck on a relatively simple problem and would like some insight as to whether or not I am doing it right.


Problem: Shoppers enter Hamilton Place Mall at an average of 120 per hour.
Q1. What is the probability that exactly 5 shoppers will enter the mall between noon and
12:05 p.m.?
Q2: What is the probability that at least 5 shoppers will enter the mall between noon and
12:05 p.m.?


Here is what I have done/concluded:
For Q1:
120 per hour = 10 every 5 min (2 every 1 min).

f(x) = 10^5 * 2.71828 ^-10 / 5! = .03783.

I am not sure if I am using the proper procedure or how to state the results if it is correct, or what the difference between Q1 and Q2 would be to get the solution...

Also, I am not even sure how to begin this one:


Problem: A professor at a local community college noted that the grades of his students were
normally distributed with a mean of 74 and a standard deviation of 10. The professor has
informed us that 6.3 percent of his students received A's while only 2.5 percent of his
students failed the course and received F's.
Q1. What is the minimum score needed to make an A?
Q2. What is the maximum score among those who received an F?


Any help would be appreciated, thanks.

For Q1 you should look into poisson distribution check this one for a good example of how to calculate

http://mathforum.org/library/drmath/view/56606.html

For Q2 you need to look into a standard distribution table to calculate what z-score is associated with 6.3 percent, from there you should be able to solve for the score needed to achieve the grade, since you know both E(X) and SD(X)

Thanks, I will be looking into that within the hour.

Also, by "Q1" do you mean the first problem I have posted, or the actual 'Q1' for the first problem?
(And "Q2," the second problem, or the second part of the first problem...?)

I have two separate problems posted.

with q1 i meant problem 1, q2 is problem 2. Missed that there were several questions per problem

For Q2 you need to look into a standard distribution table to calculate what z-score is associated with 6.3 percent, from there you should be able to solve for the score needed to achieve the grade, since you know both E(X) and SD(X)

I have looked up on my charts and I am getting .063 on a Z Table to be .5293.

With that, and this formula:
z = X - MEAN / STANDARD DEV.

I plug in: .5293 = X? - 74 / 10 -> 79.293 = X,
but with the second part of the problem, i am using 2.3%:
.51 = X? - 74 / 20 -> 79.1 = X.

These numbers are too closely related, and they don't make sense for the problem. Either I am using the wrong formula or on the wrong track entirely...

Thanks.

you doing the right thing just in reverse. You want to find the z-score that represents 6.3 per cent left of the normal distribution right hand side.

that is P(Z)=1-0.063 -> P(Z)=0.937

from a standard distribution table this represents a z-score of around 1.53

put this in your formula Z=(X-E(X))/SD(X) solve for X -> X=Z*SD(X)+E(X)

which gives

1.53*10+74=89.3 that is you need to get 89.3 or higher to receive an A

the intuition from this puzzle is that 6.3 per cent represents 1.53 standard deviations.

Now do the same for getting an F

I appreciate the help because I felt I was on the right track for the most part, but my numbers just were not making sense with the problem. I have already submitted the homework, but I'd like to understand why this is working the way it is...

Here was my initial solution:

a) z = X – 74 / 10 -> z @ 6.3% = -1.53.
-1.53 = X – 74 / 10 -> solve for X: X= 58.7
b) z = X – 74 / 10 -> z @ 2.5% = -1.96
-1.96 = X – 74 / 10 -> solve for X: X = 54.4


However, doing it your method for F's (I can see why it works for A's), here was my second (and probably more accurate) conclusion:

a) z = 1 – 0.063 = 0.937 -> 1.53
1.53 × 10 + 74 = 89.3 for A
b) z = 1 + 0.025 = -0.975 - > 1.96 <- leaving it + gives me an answer that is 93.6, which doesn't make sense, because you need an 89.3 to get an A...

-1.96 x 10 + 74 = 54.4 for F?


Again, thanks for the help.

you need to remember on which side of the bell curve you are investigating.

For A, which is above the mean, you are on the right hand side, hence a positive Z

For F, which is below the mean, you are on the left hand side, hence a negative Z.

The intuition however is how many SD away from the mean the threshold for X percent lies. i.e. for an A, which means that you belong to the top 6.3 percent. On a bell curve, the threshold that encapsulates everything but the top 6.3 per cent lies 1.53SD above the mean. Z-score is therefore a measure of how many SD away from the mean that is associated with a probability percentage.