Ask Me Help Desk - Statistics Homework Help (Z-Scores, etc.)

Hi, I am looking for help with my statistics homework.
Question:
A National Bank presently has a total of 5,000 checking accounts. The bank has found that the average daily balance on all its personal checking accounts is $550, with a standard deviation of $150. The average daily balances have been found to be normally distributed.

Questions I have completed, please tell me if these are right or wrong:

1. Based upon the population data, what percentage of customers carry average daily balances above $800?

Answer:
Find z score, (800-550)/150 = 1.667
1-.9525=.0475; 4.75%

2. Based upon the population data, what percentage of customers carry average balances below $200?
Answer: Find z-score (200-550)/150= 2.333 = 99%

3. Based upon the population data, which percentage of customers carry average daily balances between $200 and $700?
Answer: (200-550)/150= 2.333 = 1-.9901 = .0099
(700-550)/150= 1 = 1-.8413= .1587
.1587-.0099 = .1488; 14.88%

4. Determine the mean average account balance based upon the sample data:
(1) 539 (2) 575 (3) 700 (4) 714 (5) 546 (6) 523 (7) 569 (8) 565 (9) 531 (10) 402 (11) 535 (12) 386 (13) 810) (14) 550 (15) 290 (16) 800 (17) 545 (18) 559 (19) 461 (20) 560

Added, you get: 1160, divided by 20, the mean equals 558.

Questions that I need help with are:

1. If you were to draw another random sample of 20 observations from the same population, what is the likelihood (probability) that you would find a larger sample mean than the value you calculated in question 1?

2. The bank has conducted an analysis that reveals approx. 5% of its accounts have an averge daily balance below $304. In a randomly drawn sample of 2,000 accounts, what percentage of accounts could be expected to have an average daily balance below $304?

3. Based upon the population data, what is the daily balance level that you would expect only 25 percent of the customers to exceed?

4. If the bank is considering a minimum account balance of $127, how many customers would be expected to have a balance below this value?

5. If the bank chose to model the distribution below, what is the probability of a customer carrying an average daily balance below $751?

Daily Balance: Percent of Customers:
$1-150 .38%
$151-250 1.90
$251-350 6.90
$351-450 15.96
$451-550 24.86
551-650 24.86
751-850 15.96
851-950 1.90
951+ .38

Any assistance in helping on how to set these problems up would be very helpful. Thank you.

1. From my table, I'm getting 0.9522 instead of the 0.9525 that you got :confused:

2. Well, 99.02% is the number that have above $200, while you were asked to find the percentage having below that.

3. Again, I got 0.9902 instead of 0.9901. But that's not the main issue. Between $200 and $700 envelops a big majority, including the mean amount. This obviously means that your percentage should be big, very big.

From a picture, the shaded part represents the percentage asked for:
http://www.p1cture.it/images/b1g25g0acgc8cb2cd61f.png

Does that appear to be 14% to you? This is to show you how sketches of the normal distribution greatly helps you.

4. Good, except that you typed 1160 instead of 11160 :p

Okay, next part now.

1. Okay, for that one, I had to get my book and read a chapter I never studied before.

But from what I get, you have this:
For $X\sim N(\mu , \sigma^2)$ , the mean $\bar{X}$ of a sample taken from the population is normally distributed as $\bar{X} \sim N(\mu , \frac{\sigma^2}{n})$ (mu is the mean, n is the sample size)

And now, you can find the probability that $\bar{X}$ is greater than 558.

2. Do there same here with n = 2000.

3. This one is easy. You go from the probability to the z-score and work back to the value of x.

4. This one is easy too. Just like the ones you did in the first part, like 1. topmost.

5. This one is easy too... just add all the percentages below 751... seems you missed the line for 651-750 though. If such and such percent of the population fall in a range, this means that the probability of any person falling within that range is equal to the percentage you were given.

Now, if you were given something like 752 instead of 751, you could work out some form of interpolation (assuming that the amounts in the range are uniformly distributed) and estimate the probability.

Thanks for your help.

Here's the answers I have worked out so far.
1. 4.78%
2. 1-.9902= .98%
3. 83.14%
Found by finding z scores, 2.333= .9901 and zscore of 700= 1 - .8413
1-.9901= .0099
1-.8413=.1587

.1587+.0099= .1686; 1-.1686= 83.14%

4. 558... yes, I did goof off on my typing! Sorry!

Questions Needed Help With:
1. Z score= (550-558)/(120/sqrt(20)) = -.2385, 1-.5948=.4052, 5-.4052=9.48%

2..?. I'm still getting stuck here. I am using (304-550)/(150/sqrt(2000)) = -73.343... isn't this z-score too large?

3. . 67= (x-550)/150... Solve for x, $650.50 is the daily balance level that you would expect only 25% to exceed is $650.50.

4.(127-550)/150= 2.82, z score of 2.82=.9976; 1-.9976= .0024; .0024(5000)= 12 customers would be below this value.

5. I did miss the value for 651-750; the percentage at this rate is 15.96%.
To get my answer, I took 100-6.9-1.9-.38, to get 90.82% as an answer.

I have another question, similar to #3, the question is: "Based upon the population data, what is the daily balance level that you would expect 90 percent of the customers to exceed?"

So far, I have figured that 1.28=(x-550)/150... if figured correctly, the answer would be 742=x... However, I can't see 90% of customers to exceed $742, if only 25% are exceeding $650.50. It doesn't add up to me.

Please guide me in the right direction. Thank you.

Quote:

Originally Posted by collegegirl08

Thanks for your help.

Here's the answers I have worked out so far.
1. 4.78%
2. 1-.9902= .98%
3. 83.14%
Found by finding z scores, 2.333= .9901 and zscore of 700= 1 - .8413
1-.9901= .0099
1-.8413=.1587

.1587+.0099= .1686; 1-.1686= 83.14%

For this one, a shorter, easier one would be 0.8413 - 0.0099 = 0.8314

Anyway, as long as you understand what you are doing, that doesn't matter much. The only thing, is that one can be short on time sometimes. If you don't understand how the way I just did works, post back.

Quote:

4. 558... yes, I did goof off on my typing! Sorry!

:)

Quote:

Questions Needed Help With:
1. Z score= (550-558)/(120/sqrt(20)) = -.2385, 1-.5948=.4052, 5-.4052=9.48%

Okay, I really don't understand what you did here. Here is how my work would be:

$P(\bar{X} > 558) = P(z > \frac{558 - 550}{\sqrt{1125}}) = P(z > 0.239)$

Remember: $\frac{\sigma^2}{n} = \frac{150^2}{20} = 1125$

Then, $P(z > 0.239) = 1 - P(z < 0.239) = 1 - 0.5945 = 0.4055 = 40.55%$

Quote:

2..?. I'm still getting stuck here. I am using (304-550)/(150/sqrt(2000)) = -73.343... isn't this z-score too large?

Okay, sorry for that one, I thought it was again a question asking for the mean, but it makes sense. If you take a sample of 2000, but there are only 250 (5% of 5000) who have less than $304, the mean of the sample cannot be 304 with the 1500 greater than 304. Now, to the new method (this is new to me too, sorry again :o):

I can't seem to find anything in my book about such an example, but my guess is that you have to use proportions with the percentage given.

In 5000, you expect to have 5% of accounts less than $304.
In 2000, you expect to have (5/5000)*2000 = 2% of accounts less than $304.

Quote:

3. . 67= (x-550)/150... Solve for x, $650.50 is the daily balance level that you would expect only 25% to exceed is $650.50.

My book says that $\phi(0.75) = 0.674$ while you got 0.67. I really don't know the accuracy of your z-table :rolleyes: But this said, my final answer would become $651.10

Quote:

4.(127-550)/150= 2.82, z score of 2.82=.9976; 1-.9976= .0024; .0024(5000)= 12 customers would be below this value.

Good! :)

Quote:

5. I did miss the value for 651-750; the percentage at this rate is 15.96%.
To get my answer, I took 100-6.9-1.9-.38, to get 90.82% as an answer.

Except that the only percentages you have above 751 are:

Code:

751-850 15.96 851-950 1.90 951+ 0.38

This said, I get 81.76%

Quote:

I have another question, similar to #3, the question is: "Based upon the population data, what is the daily balance level that you would expect 90 percent of the customers to exceed?"

So far, I have figured that 1.28=(x-550)/150... if figured correctly, the answer would be 742=x... However, I can't see 90% of customers to exceed $742, if only 25% are exceeding $650.50. It doesn't add up to me.

Please guide me in the right direction. Thank you.

I get 1.282 in my book for 90% (again accuracy issue :p)

And you are thinking correctly! This cannot be $742.30 (that's the number I got), but if you look at my drawing, it will be right at the other end, to the left of the mean.

742.30 is 192.30 units from the mean, 550, right?
Then, go 192.30 units to the left of 550, that is 550 - 193.30 = $357.7

This is the balance 90% of the people will exceed.

I hope my answers helped you despite the clumsy number 2 :o

Thanks so much for your help!

You're welcome! :)

If you need any more help, I'll be glad to help you again! Well, as long as it's within my reach :o