# Thread: Assignment 3 help..!!! so hard......!!!!

1. ## Assignment 3 help..!!! so hard......!!!!

i just finish the home work........can any one check my work correct me if i didnt get it right, i only done the questions that i can solve.....and can any one do question #4,9,10,12 (b,c) ,13,14,15,16 <- those are the one i dont know how to solve .........

2. where is the second page of assignment three?

3. Originally Posted by Quick
where is the second page of assignment three?

4. That's the first page of assignment 2...

5. Originally Posted by Quick
That's the first page of assignment 2...

no is not.....look at the top left corner....it say assignment 3 page 2......however can u do ur best....and ASAP

by the one.....how come u keep saying assig 2 ? i dont get it....?

6. Question 4.

(a) If for two events $\displaystyle A$ and $\displaystyle B$

$\displaystyle P(B|A)=P(B)$,

then as:

$\displaystyle P(B)=P(B|A)+P(B| \mathrm{not }A)$

we have:

$\displaystyle P(B| \mathrm{not }A)=0$.

(b) A dice is rolled twice. The first roll is a 2, what is the probability
that the sum of the two rolls is greater than 6?

$\displaystyle P(a+b>6|a=2)=P(b=5)+P(b=6)$

that is the probability of the sum being greater than 6 given the first roll
is a 2, is the sum of the probabilities that the second roll is a 5 and that
the second roll is 6. So:

$\displaystyle P(a+b>6|a=2)=1/6+1/6=2/6=1/3$

RonL

7. #16:

$\displaystyle \hat{p}=.62\;\ and \;\ \hat{q}=.38$

From the table, the z-score for a 92% CI is 1.75.

$\displaystyle E=1.75\sqrt{\frac{(.62)(.38)}{100}}\approx{.085}$

.62+.085=.705

.62-.085=.535

.535<p<.705

You can say with 92% confidence that the proportion of people who buy SNO flakes is between 53.5% and 70.5%.

8. Sonymd23 do not beg. If you question is not answered do not start posting totally useless posts begging for answers. Wait.
-=USER WARNED=-

9. Originally Posted by galactus
#16:

$\displaystyle \hat{p}=.62\;\ and \;\ \hat{q}=.38$

From the table, the z-score for a 92% CI is 1.75.

$\displaystyle E=1.75\sqrt{\frac{(.62)(.38)}{100}}\approx{.085}$

.62+.085=.705

.62-.085=.535

.535<p<.705

You can say with 92% confidence that the proportion of people who buy SNO flakes is between 53.5% and 70.5%.
how did u get the 1.75?

10. how did u get the 1.75?
Virus.BagelJQ.Dll

11. Originally Posted by sonymd23
how did u get the 1.75?
Since you are looking for a 92% CI the upper limit must cut off 4% of
the area under the normal density above the limit (the lower limit will cut off
the other 4% below, to leave the 92% you require). So the area under the
normal density curve below the upper limit is 96% which is what you look up
in a standard normal table to get 1.75.

RonL

12. Originally Posted by sonymd23
how did u get the 1.75?
From the standard normal distribution table, aka the z-table.

I calculated 92% CI by looking up .96 in the table and seeing what z-score sorresponded.

You should see your teacher.

13. ^ thnx every one......however i dont get number #13........

explain, using a sketch graph, why the continuity correction is needed for discrete data. use an example to illustrate your explanation.

14. When you use a continuous normal distribution to approximate a binomial probability, you need to move 0.5 units to the left and right of the midpoint. This is so you can include all possible x-values in the interval.

This is called a continuity correction

For instance, let's convert the following binomial to normal distribution.

The probability of getting between 200 and 250 successes, inclusive.

The discrete midpoint values are 200, 201, 202,....., 250. The interval for the

normal distribution is 199.5<x<250.5

15. ^ thnx........but do u have a graph ?

beucase if i dont have a graph.....then the teacher will take marks off.........

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