# Thread: Help with probability question

1. ## Help with probability question

Hi all I'm currently taking Statistics 1 and I'm stuck on a homework problem that I've been trying to figure out for a while...hopefully one of you guys can enlighten me on how to do it..heres the question:

The weight of corn chips dispensed into a 10-ounce bag by the dispensing machine has been identified as possessing a normal distribution with a mean of 10.5 ounces and a standard deviation of .2 ounce. Suppose 100 bags of chips are randomly selected. Find the probability that the mean weight of these 100 bags exceeds 10.45 ounces.

Any help will be greatly appreciated. Thanks

2. Originally Posted by lilyungn
Hi all I'm currently taking Statistics 1 and I'm stuck on a homework problem that I've been trying to figure out for a while...hopefully one of you guys can enlighten me on how to do it..heres the question:

The weight of corn chips dispensed into a 10-ounce bag by the dispensing machine has been identified as possessing a normal distribution with a mean of 10.5 ounces and a standard deviation of .2 ounce. Suppose 100 bags of chips are randomly selected. Find the probability that the mean weight of these 100 bags exceeds 10.45 ounces.

Any help will be greatly appreciated. Thanks
The mean of a sample follows a normal distribution ..... Read this: Sampling Distribution of the Mean

Use this distribution to calculate $\Pr(\overline{X} > 10.45)$.

3. I got this 0.40129367 i'm not sure if i did it right...

4. Originally Posted by lilyungn
I got this 0.40129367 i'm not sure if i did it right...
Show all your working that led to this answer and someone will review it.

5. actually I got 0.598706 since I dont have my TI-83 I used this site to fill in the required fields...Z table - Normal Distribution

6. Originally Posted by lilyungn
actually I got 0.598706 since I dont have my TI-83 I used this site to fill in the required fields...Z table - Normal Distribution

7. sorry im stuck the professor let's us use the TI-83 to solve the problems so I'm not to sure how to solve it by hand...can you help me out

8. I gave it a shot and this is what I got...Since 10.5 - 10.45 = 0.05 which is (1/4) of a standard deviation you're to look for the z-score of 0.25 and add it to 0.5.

0.5 + 0.0943 = 0.50943 is that right?

9. Originally Posted by lilyungn
sorry im stuck the professor let's us use the TI-83 to solve the problems so I'm not to sure how to solve it by hand...can you help me out
By working I mean saying what mean and what standard deviation you're using.

And if you're using a TI-83 you don't need to convert to a z-score (I suggest you carefully review how to use the TI-83 to solve normal distribution problems).

10. Is this right?

Mean(100 bags) = 10.5 ounces
Std deviation(100 bags) = 0.2 / sqrt(100) = 0.02 ounces

P(x-bar > 10.45) = P(Z > -0.05/0.02) = P(Z>-2.5) ~= 99.38%

11. Originally Posted by lilyungn
Is this right?

Mean(100 bags) = 10.5 ounces
Std deviation(100 bags) = 0.2 / sqrt(100) = 0.02 ounces

P(x-bar > 10.45) = P(Z > -0.05/0.02) = P(Z>-2.5) ~= 99.38%
Yes, but I would write it as 0.9938 (probability is a number that lies between 0 and 1 inclusive).

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### The weight of corn chips dispensed into a 19 - ounce bag by the dispensing machine has been identified as possessing a normal distribution with a mean of 19.5 ounces and a standard deviation of 0.1 ounce. Suppose 100bags of chips are randomly selected. Fi

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