Statistics question again

[Gamewizard]

Hi, I have a question about normal distribution. A large population of animals were found to possess a mean

length of 18cm with a standard deviation of 2cm.

 

What percentage of animals are longer than 20cm?

 

so i know you use the formula to standardize the Z. But then from the cumulative table, how do we know which column to look for?

 

and then the next question is, What percentage of animals should be between 15cm and 20.5 cm in length? in the formula, how do you put both figures 15 and 20.5 ? and then the last question is What length should enclose the shortest 60% of the animals ?

please help

 

Also does anyone know a good website for statistics ?

[DJBruce]

Hi, I have a question about normal distribution. A large population of animals were found to possess a mean

length of 18cm with a standard deviation of 2cm.

 

What percentage of animals are longer than 20cm?

 

so i know you use the formula to standardize the Z. But then from the cumulative table, how do we know which column to look for?

 

To use a Z score table you generally go down the left hand side until your Z-score matches the one shown, and then you go across that row until the heading of the column matches the third digit of your Z-score. For example lets say I have Z=1.34, and I want to find P(x<Z). So looking at the table I linked to you go down on the left hand side to the row that says 1.3. Then I go across that row until the the heading of the column is .04. This entry, P(x<Z)=0.4099, is the number you are looking for.

 

and then the next question is, What percentage of animals should be between 15cm and 20.5 cm in length? in the formula, how do you put both figures 15 and 20.5 ? and then the last question is What length should enclose the shortest 60% of the animals ?

please help

 

Calculate the Z-scores for these two lengths. You should then be able to find P(x<Z') and P(x<Z"). Knowing this how would you find the area between these two Z-scores. (Hint: Draw a picture of what P(x<Z') and P(x<Z") represent.)

 

For the last question you need to use an inverse Z-table to find the what Z-score corresponds to P(x<Z)=.6. Once you have this you should be able to slove for the length. Remember:

[math] Z=\frac{x-\mu}{\sigma} [/math]

 

Also does anyone know a good website for statistics ?

 

I am not sure of any great statistics websites, but I do like this website. It is a website for a high school AP program, and it has a lot of really nice handouts and links.

 

 

Also you can avoid tables if you have a graphing calculator as most have built in functions that can calculate these for you.

Edited December 21, 2010 by DJBruce

[ewmon]

Using this table, you split your z value (let's say, 1.78) into two parts, the 1.70 (which you'll find to the left) and the 0.08 (which you'll find at the top). You'll find the cumulative probability (0.9625) of z = 1.78 where the "1.70" row and the "0.08" column intersect. This is for positive z values (as shown in the bell curve diagram). For negative z values (eg, –1.78), can you see what you would do with the 0.9625 value to obtain the cumulative probability for z = –1.78? (Look again at the bell curve diagram for a clue.)

 

[Gamewizard]

To use a Z score table you generally go down the left hand side until your Z-score matches the one shown, and then you go across that row until the heading of the column matches the third digit of your Z-score. For example lets say I have Z=1.34, and I want to find P(x<Z). So looking at the table I linked to you go down on the left hand side to the row that says 1.3. Then I go across that row until the the heading of the column is .04. This entry, P(x<Z)=0.4099, is the number you are looking for.

 

 

 

Calculate the Z-scores for these two lengths. You should then be able to find P(x<Z') and P(x<Z"). Knowing this how would you find the area between these two Z-scores. (Hint: Draw a picture of what P(x<Z') and P(x<Z") represent.)

 

For the last question you need to use an inverse Z-table to find the what Z-score corresponds to P(x<Z)=.6. Once you have this you should be able to slove for the length. Remember:

[math] Z=\frac{x-\mu}{\sigma} [/math]

 

 

 

I am not sure of any great statistics websites, but I do like this website. It is a website for a high school AP program, and it has a lot of really nice handouts and links.

 

 

Also you can avoid tables if you have a graphing calculator as most have built in functions that can calculate these for you.

 

thank you very nicely explained. I have done the first question and I get 1 for that because 2 divide by 2 is 1. Would that be 1.0 or 1.1 in the Z column of cumulative table? and how would I choose the 0.01 column for that ?

thanks (i will do the rest of the questions in the meantime)

 

To use a Z score table you generally go down the left hand side until your Z-score matches the one shown, and then you go across that row until the heading of the column matches the third digit of your Z-score. For example lets say I have Z=1.34, and I want to find P(x<Z). So looking at the table I linked to you go down on the left hand side to the row that says 1.3. Then I go across that row until the the heading of the column is .04. This entry, P(x<Z)=0.4099, is the number you are looking for.

 

 

 

Calculate the Z-scores for these two lengths. You should then be able to find P(x<Z') and P(x<Z"). Knowing this how would you find the area between these two Z-scores. (Hint: Draw a picture of what P(x<Z') and P(x<Z") represent.)

 

For the last question you need to use an inverse Z-table to find the what Z-score corresponds to P(x<Z)=.6. Once you have this you should be able to slove for the length. Remember:

[math] Z=\frac{x-\mu}{\sigma} [/math]

 

 

 

I am not sure of any great statistics websites, but I do like this website. It is a website for a high school AP program, and it has a lot of really nice handouts and links.

 

 

Also you can avoid tables if you have a graphing calculator as most have built in functions that can calculate these for you.

 

Okay, done the second question (what percentage of animals should be between 15cm and 20.5cm in length?) I get z -1.5 value and z 1.25. P values for these are 0.06057 from the 0.05 column and 0.89435 from same column. Is that correct do you think ? but i m not sure now how to find the area now, not sure on how to draw them out

[ewmon]

Okay, done the second question (what percentage of animals should be between 15cm and 20.5cm in length?) I get z -1.5 value and z 1.25. P values for these are 0.06057 from the 0.05 column and 0.89435 from same column. Is that correct do you think ? but i m not sure now how to find the area now, not sure on how to draw them out

Your thinking to this point is correct, except why did you look in the same column? You're looking up 1.50, not 1.55. When you find the correct area for z = –1.50, look again at the bell curve and think of what those areas represent, and your next step might come to you.

[Gamewizard]

Your thinking to this point is correct, except why did you look in the same column? You're looking up 1.50, not 1.55. When you find the correct area for z = 1.50, look again at the bell curve and think of what those areas represent, and your next step might come to you.

 

Okay, so which one is wrong z=-1.5 p(0.06057) or z= 1.25 p(0.89435) ?

 

sorry read your post again, its the -1.5 isnt it. So if i cannot use the 0.05 column then which one would it be im not sure. Would it be 0.00 one? because the second digit after the decimal point is 0? i cannot see any other

Edited December 22, 2010 by Gamewizard

[ewmon]

Would it be 0.00 because the second digit after the decimal point is 0?

Bingo! I actually saw the lightbulb go on over your head.

[Gamewizard]

Bingo! I actually saw the lightbulb go on over your head.

 

Yayy lol thanks. Right so now i have z=-1.5 (p 0.08076) and z=1.25 (p 0.89435)

 

I cannot really make sense of the next step, i have drawn the bell curve and put the numbers in but still confused, I cannot upload the file here as its too large to upload.

 

Sorry my mistake, that is supposed to be 0.06681.

Edited December 23, 2010 by Gamewizard

[ewmon]

Your lookup for z = –1.50 is incorrect, so look up for z = 1.50 again.

 

The entire area under the cumulative probability bell curve is normalized (ie, scaled or standardized) to 1 (because, by definition, the cumulative probability is 1 from z = –∞ to z = +∞). So, the reason to compute 1 – P(z) is to find the area in the "tail" which is the cumulative probability greater than z (which is the unshaded area under the bell curve I posted) --- but, more importantly here, it is also the as the same cumulative probability less than –z. Get it? Lightbulb?

 

When you find the cumulative probability less than z = 1.25 (which you have already found) and the cumulative probability less than z = –1.50, I recommend that you draw a simple bell curve and graph these two probabilities (and shade them differently -- for example, shade lines that slant in one direction and in the other direction). (Remember that negative z values go to the left of z = 0.) From there, I hope the next step to perform is obvious from your graphing (ie, how to find the cumulative probability greater than z = –1.50 AND less than z = 1.25).

 

One hint about this last step. The math you're using is not sophisticated enough to simply find the cumulative probability between two z values. (Think about it, "between" is a little sophisticated.) Instead, this math is only as simple as "less than" and "greater than." So that's why you need to convert your thinking of "between a and b" to "greater than a" AND "less than b." Bingo?

Edited December 23, 2010 by ewmon

[Gamewizard]

Your lookup for z = –1.50 is incorrect, so look up for z = 1.50 again.

 

The entire area under the cumulative probability bell curve is normalized (ie, scaled or standardized) to 1 (because, by definition, the cumulative probability is 1 from z = –∞ to z = +∞). So, the reason to compute 1 – P(z) is to find the area in the "tail" which is the cumulative probability greater than z (which is the unshaded area under the bell curve I posted) --- but, more importantly here, it is also the as the same cumulative probability less than –z. Get it? Lightbulb?

 

When you find the cumulative probability less than z = 1.25 (which you have already found) and the cumulative probability less than z = –1.50, I recommend that you draw a simple bell curve and graph these two probabilities (and shade them differently -- for example, shade lines that slant in one direction and in the other direction). (Remember that negative z values go to the left of z = 0.) From there, I hope the next step to perform is obvious from your graphing (ie, how to find the cumulative probability greater than z = –1.50 AND less than z = 1.25).

 

One hint about this last step. The math you're using is not sophisticated enough to simply find the cumulative probability between two z values. (Think about it, "between" is a little sophisticated.) Instead, this math is only as simple as "less than" and "greater than." So that's why you need to convert your thinking of "between a and b" to "greater than a" AND "less than b." Bingo?

 

Okay, is the final answer going to be 0.06% of animals should be between 15cm and 0.89% of animals should be between 20.5cm ? i really hope this is correct.

[ewmon]

is the final answer going to be 0.06% of animals should be between 15cm and 0.89% of animals should be between 20.5cm?

Uh, nope.

 

Sorry, let me re-phrase the logic involved because all the cumulative probabilities are for "less than" a value of z (ie, P[Z<z], as shown in my diagram).

 

You need to convert your thinking to "less than z = 1.25" BUT NOT "less than z = –1.50".

"less than z = 1.25" means P[Z<1.25], which you correctly found to equal 0.89435.

 

"less than z = –1.50" means P[Z<-1.50], which you know means 1 – P[Z<1.50].

So, look up P[Z<1.50] in the table, and subtract it from 1.

 

Hopefully, when you draw the bell curve and shade the areas, you'll see how to calculate "less than z = 1.25" BUT NOT "less than z = –1.50". The value of P[Z<a] refers to the area under the curve from –∞ to a, like this:

 

 

The graph below is different because the shaded area shows the final answer in graphic form.

 

 

If you know the areas representing P[Z<1.25] and P[Z<–1.50], then what must you do to find the answer?

 

And be careful converting probabilities into percentages; a probability of 1.00 equals 100%, a probability of 0.50 equals 50%, etc.

 

Merry Christmas.

[Gamewizard]

Uh, nope.

 

Sorry, let me re-phrase the logic involved because all the cumulative probabilities are for "less than" a value of z (ie, P[Z<z], as shown in my diagram).

 

 

"less than z = 1.25" means P[Z<1.25], which you correctly found to equal 0.89435.

 

"less than z = 1.50" means P[Z<-1.50], which you know means 1 P[Z<1.50].

So, look up P[Z<1.50] in the table, and subtract it from 1.

 

Hopefully, when you draw the bell curve and shade the areas, you'll see how to calculate "less than z = 1.25" BUT NOT "less than z = 1.50". The value of P[Z<a] refers to the area under the curve from ∞ to a, like this:

 

 

The graph below is different because the shaded area shows the final answer in graphic form.

 

 

If you know the areas representing P[Z<1.25] and P[Z<1.50], then what must you do to find the answer?

 

And be careful converting probabilities into percentages; a probability of 1.00 equals 100%, a probability of 0.50 equals 50%, etc.

 

Merry Christmas.

 

Oh i think i m getting it now. So i already know that 6.681% of animals should be less than 15cm long, and 89.435% should be less than 20.5cm long, right, so to find the answer i should minus them together isnt it ? 89.435-6.681= 82.754 final answer ?

Edited December 28, 2010 by Gamewizard

[ewmon]

Oh i think i m getting it now. So i already know that 6.681% of animals should be less than 15cm long, and 89.435% should be less than 20.5cm long, right, so to find the answer i should minus them together isnt it ? 89.435-6.681= 82.754 final answer ?

Bingo!

[Gamewizard]

Bingo!

 

Yayyyyy thanks so much for your help !! I have another statistics question, which I think I'll have to make a new thread for, its about ANOVA, have you done that before ?

Edited December 29, 2010 by Gamewizard

[Gamewizard]

And where can I find an inverse z table for the last question ? thanks

[ewmon]

And where can I find an inverse z table for the last question?

the last question is What length should enclose the shortest 60% of the animals ?

Just convert the percent to probability, then use the P[Z<z] table in reverse, and then, knowing the mean and SD, convert the z into the length.

[Gamewizard]

Just convert the percent to probability, then use the P[Z<z] table in reverse, and then, knowing the mean and SD, convert the z into the length.

 

 

Oh thanks. Sorry but I need to ask something else on a slightly diffrent question. Why are diffrent t-percentiles used? i mean for example in one question there is 97.5 percentile used, and in another question 99.9, how do we know which percentile column to look in for diffrent questions?

[ewmon]

What's the whole question regarding t-percentiles?

[Gamewizard]

What's the whole question regarding t-percentiles?

 

If the mean mercury concentration in fish caught in a particular lake is below 0.5ppm then the state of florida would class that lake as acceptable. On the basis of the data provided in the case study how should the state classify each lake ?

so there are basically two lakes, East Lake and Lake Iamonia.

 

East lake mean= 1.1765 sd=0.2802

Lake Iamonia mean= 0.598 sd= 0.2836

 

formula used = t= mean-value(0.5)/s/square root of n

 

 

The t-value for the first lake is 15.27, and the total population is 40(fishes)

so the answer given by my teacher is t39(0.025) = 3.307< 15.27. So hes looked at the 99.9 % column in the t-table. For the next lake, hes looked at the 97.5% column, and so his answer is t29(0.025)=2.045

 

What I do not understand is why is he looking at diffrent percentiles? and how would I know which percentile column to look for in diff questions?

[ewmon]

Someone else needs to answer this. I've used Student t tests to compare populations, but this is unfamiliar to me.

[Gamewizard]

Someone else needs to answer this. I've used Student t tests to compare populations, but this is unfamiliar to me.

 

 

Oh right okay I will try to post this as new thread then