Alright bright folks, I need some help with a statistics problem:

A company produces steel rods. The lengths of the steel rods are normally distributed with a mean of 196.8-cm and a standard deviation of 1.7-cm. For shipment, 18 steel rods are bundled together.

Find P24, which is the average length separating the smallest 24% bundles from the largest 76% bundles.

I keep getting 195.61 and the homework program says its wrong and I'm honestly stumped but I get unlimited attempts

How do I do this

Alright bright folks, I need some help with a statistics problem:

A company produces steel rods. The lengths of the steel rods are normally distributed with a mean of 196.8-cm and a standard deviation of 1.7-cm. For shipment, 18 steel rods are bundled together.

Find P24, which is the average length separating the smallest 24% bundles from the largest 76% bundles.

I keep getting 195.61 and the homework program says its wrong and I'm honestly stumped but I get unlimited attempts

How do I do this

Take 1 rod and stick it up your lecturer's ass

Take 1 rod and stick it up your lecturer's ass

right after i read about statistics and statistic studies, wondering if that's hard and whether i can learn it or not, i stumble across this thread and it completely crashes my hopes and dreams of trying it out

right after i read about statistics and statistic studies, wondering if that's hard and whether i can learn it or not, i stumble across this thread and it completely crashes my hopes and dreams of trying it out

I'm honestly not sure of the formula you'd use (haven't taken stats in a while) but I think it's asking you to look at the midpoints of the bottom 24% and the top 76% (13th and 62nd percentile) and get the difference between them- this should be on the order of a few standard deviations x 18 so I think 195.61 is a little high given that this would mean they're 6.4 standard deviations apart.

I seem to remember something about Z values and a table you can use for this but I haven't done it in about 6 years.

Edit: nvm - looking at the question again its literally just looking for the value at 24%

I'm honestly not sure of the formula you'd use (haven't taken stats in a while) but I think it's asking you to look at the midpoints of the bottom 24% and the top 76% (13th and 62nd percentile) and get the difference between them- this should be on the order of a few standard deviations x 18 so I think 195.61 is a little high given that this would mean they're 6.4 standard deviations apart.

I seem to remember something about Z values and a table you can use for this but I haven't done it in about 6 years.

Edit: nvm - looking at the question again its literally just looking for the value at 24%

http://math.stackexchange.com/

question is confusing but try 195.21?

question is confusing but try 195.21?

It looks like you're on the right track but your value looks slightly off (rounding error?). As you have correctly assessed you want to find x such that P(X<=x) = 0.24 (where X is your random variable following the distribution you described). I'm not sure how you're expected to do this problem (you would need some sort of calculator/computer since I doubt you're gonna compute the inverse error function by hand...). For your sake check your work before looking at the spoilers below. You're only gonna hurt yourself.

Wolfram Alpha:

Show Content

Or on a standard TI calculator (haven't actually tested this one):

Show Content

invNorm(.24, 196.8, 1.7)

Edit: Misread the problem disregard the above

It looks like you're on the right track but your value looks slightly off (rounding error?). As you have correctly assessed you want to find x such that P(X<=x) = 0.24 (where X is your random variable following the distribution you described). I'm not sure how you're expected to do this problem (you would need some sort of calculator/computer since I doubt you're gonna compute the inverse error function by hand...). For your sake check your work before looking at the spoilers below. You're only gonna hurt yourself.

Wolfram Alpha:

[spoiler]http://www.wolframalpha.com/input/?i=24th+percentile+of+normal+distribution+mu%3D196.8+sigma%3D1.7[/spoiler]

Or on a standard TI calculator (haven't actually tested this one):

[spoiler]

invNorm(.24, 196.8, 1.7)

[/spoiler]

Edit: Misread the problem disregard the above

I don't think you can just take the distribution of a single length rod; you'd have to take the distribution for an independent sum of 18 normal distributions, divided by 18, which gets you a new normal distribution with mean 196.8 and standard deviation 1/18*sqrt(18*(1.7)^2).

I suck at stats though, so take this with a huge grain of salt.

EDIT: Plugging that into WA gives me the same value farnsworth suggested.

I don't think you can just take the distribution of a single length rod; you'd have to take the distribution for an independent sum of 18 normal distributions, divided by 18, which gets you a new normal distribution with mean 196.8 and standard deviation 1/18*sqrt(18*(1.7)^2).

I suck at stats though, so take this with a huge grain of salt.

EDIT: Plugging that into WA gives me the same value farnsworth suggested.

Okay folks I have two others that are stumping me

You intend to estimate a population mean with a confidence interval. You believe the population to have a normal distribution. Your sample size is 26.

Find the critical value that corresponds to a confidence level of 98%.

ta/2 = ±

----------------------------------------------------------------------------------

You want to obtain a sample to estimate a population proportion. At this point in time, you have no reasonable estimate for the population proportion. You would like to be 99.9% confident that you estimate is within 0.1% of the true population proportion. How large of a sample size is required?

Okay folks I have two others that are stumping me

You intend to estimate a population mean with a confidence interval. You believe the population to have a normal distribution. Your sample size is 26.

Find the critical value that corresponds to a confidence level of 98%.

ta/2 = ±

----------------------------------------------------------------------------------

You want to obtain a sample to estimate a population proportion. At this point in time, you have no reasonable estimate for the population proportion. You would like to be 99.9% confident that you estimate is within 0.1% of the true population proportion. How large of a sample size is required?

Based on a sample of 200 people, 93% owned cats

The p-value is:

Based on a sample of 50 people, 62% owned cats

The test statistic is:

The critical value is:

Based on a sample of 200 people, 93% owned cats

The p-value is:

Based on a sample of 50 people, 62% owned cats

The test statistic is:

The critical value is:

jdmOkay folks I have two others that are stumping me

You intend to estimate a population mean with a confidence interval. You believe the population to have a normal distribution. Your sample size is 26.

Find the critical value that corresponds to a confidence level of 98%.

ta/2 = ±

Just look in the t table for n-1, 0.02 = 2.485

[quote=jdm]Okay folks I have two others that are stumping me

You intend to estimate a population mean with a confidence interval. You believe the population to have a normal distribution. Your sample size is 26.

Find the critical value that corresponds to a confidence level of 98%.

ta/2 = ±

[/quote]

Just look in the t table for n-1, 0.02 = 2.485