WeBWorK #9 UPDATE

UPDATE MONDAY 4/15/13:  I found that for some (but not all) students, problem #8 was also having similar trouble.  It is fixed, as of 8:23pm.

Hi everyone,

It was pointed out in class today that some of the problems in WeBWorK #9 were being marked incorrect, even if you follow exactly the steps described in class.  I have tracked down the source of the error, and updated a number of problems in the assignment — they should all be working properly now.

Unless further problems are discovered, the due date for WeBWorK #9 will remain next Tuesday, April 16, at midnight.

If you believe that a problem is still not working correctly (that is, you think your answer is right but WeBWorK is not accepting it) please send me an email and let me know.

– Mr. Reitz

ps.  If you’re interested (GEEK ALERT), the errors arose because we use a table to find z / probability, and the table is limited in the number of decimals it will provide.  Solving these problems on a calculator gives more decimals of accuracy (but requires a more specialized calculator, such as a graphing calculator)

22 thoughts on “WeBWorK #9 UPDATE”

1. some of them still have an issue with the same problem and only a few would work well (depends also with partial question/answers)…

i’ll see what i can do….even though area for z scores on the charts will only use 4 digits after the decimal point…

2. oh i re-checked the answers….and fix it…so all of the questions seem to work well

3. oh i just found that the first part for # 6 is weird:

Because of the relatively high interest rates, most consumers attempt to pay off their credit card bills promptly. However, this is not always possible. An analysis of the amount of interest paid monthly by a bank’s Visa cardholders reveals that the amount is normally distributed with a mean of 29 dollars and a standard deviation of 7 dollars.

A. What proportion of the bank’s Visa cardholders pay more than 32 dollars in interest?
(32-29)/7 = Z

(Z = 0.4285714285714286, Z = .42, area .6628) 1-.6628=.3372

Proportion = .3372
and it says that is incorrect:
0.3372 incorrect

• Hi Anil,
Taking a look through your work, I think you made an error when rounding
Z = 0.4285714285714286
to two decimals. Take another look, and let me know if you’re still stuck.
-Mr. Reitz

• Thanks. i was able to correct it and it worked.

4. questions # 7 to 10 are still weird since i have to find x and it will not be accepted for the corresponding of answers…

most of them have a percent which i use to find the z score and then i find x..though they are not accepted by webwork

• Hi Anil,
To help you out, I would have to know the specific steps you took for one of the problems — if you like, write back and give me some detail on what you did and what numbers you got in each step.
-Mr. Reitz

• Most of them are similar to #7. for the last half of web work 9

# 7.
Suppose the heights of men are normally distributed with mean, = 68 inches, and standard deviation , = 4 inches. Suppose admission to a summer basketball camp requires that a camp participant must be in the top 10 % of men’s heights, what is the minimum height that a camp participant can have in order to meet the camp’s height admission requirement?

i used the conversion for z to x:

10%= 0.1 the closet Z is -1.28 (with area: .1003), so

68+ (-1.28*4)=
68 + -5.12 =
62.88 inches

and webwork responds with:
62.88 incorrect

• i already solved………..the keywords such as “top” meant that the area or percent was at far right of the bell curve and i had to use 1-the area to find the part to the left of it.

5. Can somebody please explain step by step number 10 on the webwork homework. Thank you.

• Hi….oh i had just posted something about it for #10….i had a slightly different question, though it should a bit similar to yours….

• just see the long paragraph below….i hope it helps…

• Let me try.
Part (a) asks you to find x
x = mean + (z) x (standard deviation)

The problem give you:
mean = a
standard deviation = b

Convert the top say 2% to p = 2/100 = .02
Since it is the top 2% it is to the right of the curve. Therefore, subtract from 1. So, 1 – .02 = .9800
.9800 = p’
Use table to find Z. So you find the probability .9800 in the table and that will give you z.

substitute z in the formula above and solve for x.

Part (b) is finding the top say 2% percentage of the population given.

• Let me try.
Part (a) asks you to find x
x = mean + (z) x (standard deviation)

The problem give you:
mean = a
standard deviation = b

Convert the top say 2% to p = 2/100 = .02
Since it is the top 2% it is to the right of the curve. Therefore, subtract from 1. So, 1 – .02 = .9800
.9800 = p’
Use table to find Z. So you find the probability .9800 in the table and that will give you z.

substitute z in the formula above and solve for x.

Part (b) is finding the top say 2% percentage of the population given.

6. the keywords such as “top” meant that the area or percent was at far right of the bell curve and i had to use 1-the area to find the part to the left of it.

THIS was my problem for #10
IQ scores are normally distributed with a mean of 100 and a standard deviation of 15. Mensa is an international society that has one – and only one – qualification for membership: a score in the top 2% of the population on an IQ test.
(a) What IQ score should one have in order to be eligible for Mensa?

the “top” means that 2% is an area to the far right…so the new area should be to the left which would be found with 1-0.02= New area and then find the New Z.
then use the formula for X= U+ [Z] * [σ] with the z of the new area.

(b) In a typical region of 120,000 people, how many are eligible for Mensa?

for this, it is just a percent problem

which you need to find 2% of the population:

[2] / [100] = # / 120000

[2] * [120000] = 240000

[240000] / [100] = 2400.00

try this method and see if it helps.

• oh the area of part (a) for #10 for 2% used was actually 0.0202 from the sheet. with the Z as -2.05

then i used 1-.0202

and the new Z closet to it is 2.05 with the area of .9798

so for X= U+ [Z] * [σ] = 100 + [2.05] * [15]

X = 130.75

7. ok so im still getting 8 wrong maybe im doing something wrong

my problem is The combined math and verbal scores for students taking a national standardized examination for college admission, is normally distributed with a mean of 500 and a standard deviation of 210. If a college requires a student to be in the top 25 % of students taking this test, what is the minimum score that such a student can obtain and still qualify for admission at the college?

im currently doing 1-.25= .75 –> z-score .7486 (.67)
then i plug it into the equation x= m+zo .. x= 500+(.67)(210) and that equals 640.7 yet it is wrong??

• oh i got it i forgot to take off the .7 ha !

8. Proff i know its late but I’m breaking my head with number 4,
I feel like I’m getting a right number but computer doesn’t take it.
Here is a question problem: Using diaries for many weeks, a study on the lifestyles of visually impaired students was conducted. The students kept track of many lifestyle variables including how many hours of sleep obtained on a typical day. Researchers found that visually impaired students averaged 9.53 hours of sleep, with a standard deviation of 2.83 hours. Assume that the number of hours of sleep for these visually impaired students is normally distributed.

I completed question A correctly but having issues with question B

(b) What is the probability that a visually impaired student gets between 6.6 and 10.81 hours of sleep?
What i did was 6.6-9.53/2.83=-1.03= Z=.1515
and 10.81-9.53/2.83=.45=Z=.6736
Last step i did was .6736-.1515= and i keep getting .5221 or if i switch mean and x i get the same answer but “-” computer doesn’t accept neither of the answers

Please let me know if i’m doing the process wrong,
Thank you

• Hi Zinaida,

You were right. I thought I had fixed all of the pesky problems with this assignment, but it seems I didn’t. I have given you credit for this problem – you can check your score online.

I’ll be sure to avoid these problems in future assignments.

Apologies,
Mr. Reitz

• It was a matter of curiosity .. i redid this problem many times, and thought maybe i was doing something incorrect.
Thank you for clearing it out 🙂

Zina