Question

Use the normal approximation to the binomial to find that probability for the specific value of X.
n=30, p=0.6, X=17
A) 0.14
B) 0.57
C) 0.34
D) 0.6

Answer 1

Do you know the formulas for the mean and the standard deviation of the normal approximation to the binomial distribution?

Answer 2

I think so. n*p is the mean, and the standard deviation is the square root of the mean? :/

Answer 3

sorry, i'm just rusty because i'm making up this test and we learned the stuff about a week ago.. i appreciate the help tho!!

Answer 4

\[mean=n \times p\]
\[standard\ deviation=\sqrt{npq}\]
where q =1 - p
So can you find the values of the mean and the standard deviation and post them please?

Answer 5

sure.. one second my friend

Answer 6

mean=18 and standard dev.= 2.68 ??

Answer 7

is that with the chart?

Answer 8

Good work. Now you need to find the z-score for X = 17 to enable you to use the standard normal distribution table. Do you know how to find the z-score?
You need to use the formula\[z=\frac{X-\mu}{\sigma}\]

Answer 9

wait.. how am i finding the z-score?

Answer 10

nvm lol

Answer 11

-0,37313433

Answer 12

You are reasonably correct. However I have just read the question more carefully and realise that the probability of a discrete value of X (17) is required. Please wait a few minutes.

Answer 13

would it be c) 0.34??

Answer 14

We are changing from a discrete distribution (binomial) to a continuous distribution (normal). We need to find the following probability:\[P(16\le X \le18)\]
To do this we need to find two z-scores each of which has a correction term 0.5 caused by the change from a discrete to a continuous distribution.
\[z _{\alpha}=\frac{16-18-0.5}{2.683}=you\ can\ calculate\]
\[z _{\beta}=\frac{18-18+0.5}{2.683}=you\ can\ calculate\]
Please post the calculated values. Then we can go to the next and final step.

Answer 15

-o.93179277 and 0.18635855

Answer 16

Excellent! Now we need to use a standard normal distribution table to find the two probability values corresponding to these z-scores. I suggest that you use a table found by going to this link
http://www.math.bgu.ac.il/~ngur/Teaching/probability/
and selecting 'normal.pdf'
Please post the two values.

Answer 17

0.1762 and 0.5753

Answer 18

Finally we need to find the difference between these two values of probability:\[P(X=17)=0.5753-0.1762\approx 0.4\]
This value is not one of the given choices. Therefore I will calculate by using the binomial distribution and compare the result. Please wait.

Answer 19

The binomial distribution calculation gives P(X = 17) = 0.136.
Please wait a few minutes and I will recheck the normal approximation method.

Answer 20

dont worry about it man, youve done enough. im sure ill get some partial credit and imma bout to go to bed. you think you could help me with one last question that might be easier? if its another tough one, then dont worry about it buddy

Answer 21

the average diameter of sand dollars on a certain island is 4.50 centimeters with a standard deviation of 1.00 centimeters. if 9 sand dollars are chosen at random for a collection, find the probability that the average diameter of those sand dollars is more than 4.10 centimeters. assume that the variable is normally distributed.

Answer 22

a. 0.615 b. 0.820 c. 0.385 d. 0.885

Answer 23

just let me know what you want to do, because im getting tired lol.. i really appreciate all of your help though!!!!!

Answer 24

I will look at this last one. It should be quick :)

Answer 25

ok thanks a lot!!

Answer 26

im getting off in about 3 mins kropot. so if its taking you a while, don't worry about it buddy!!

Answer 27

Sorry I can't be sure it will be quick :(
Have a good sleep!

Answer 28

don't be sorry at all. you too, good night!!!!!

Answer 29

Regarding the first question, the correct answer is obtained if the values of the two z-scores are calculated as follows:
\[z _{\alpha}=\frac{16-18+0.5}{2.683}=-0.56\]
\[z _{\beta}=\frac{18-18-0.5}{2.683}=-0.186\]
If you use the table at the link that I gave previously, find the two probability values, and then find the difference:
\[P(X=17)=P _{\beta}-P _{\alpha}\]