#1

Does anyone out there know how to solve problems like this:

"A cop does a survey of the speeds of travelling cars. The area in which she conducts her survey has a 50km/h speed limit. He finds a mean of 56km/h and a standard deviation of 8km/h.

Calculate the percentage of people driving between 50 and 60km/h?"

How would I solve that?

Thanks.

"A cop does a survey of the speeds of travelling cars. The area in which she conducts her survey has a 50km/h speed limit. He finds a mean of 56km/h and a standard deviation of 8km/h.

Calculate the percentage of people driving between 50 and 60km/h?"

How would I solve that?

Thanks.

#2

Well you can't unless it says how many people were surveyed.

-Jayke

-Jayke

#3

Her..he?

#4

42.

#5

try trial and error of numbers until u come across the standard deviation. then thats ur answer/

ask for a graphics calculator

ask for a graphics calculator

#6

Well, if the question is how many people get pulled over between the ages of 16 and 22 then the answer is all of them.

#7

try trial and error of numbers until u come across the standard deviation. then thats ur answer/

ask for a graphics calculator

How do I look for the standard deviation though? I can't put it in as '0,' that comes up with 'MATH ERROR' and if I put it in as 8 (what it should be) then I still have to find the area which is what I'm trying to find.

#8

**** im supposed to know this for my exams in may too

find the variance, and wotk the formula back till you get n or something

ya im screwed

find the variance, and wotk the formula back till you get n or something

ya im screwed

#9

42

#10

42

Thanks, but can you please tell me how you got it?

#11

Can't you graph the distribution function and do an integral from 50 to 60 mph? That'll give you the proportion of people in that speed range.

#12

42 isn't really the answer to that question.

it IS the answer to EVERYTHING on the other hand.

unless it is, I'm **** at maths LOL

it IS the answer to EVERYTHING on the other hand.

unless it is, I'm **** at maths LOL

#13

Can't you graph the distribution function and do an integral from 50 to 60 mph? That'll give you the proportion of people in that speed range.

Maybe, but I have no idea how to do that and I'll probably forget. I just really need someone to tell me how they got 42. Or to tell me how to do Inverse Normal Problems. I don't know how to find the area, I know the standard deviation and mean because we are told it, but I have no idea how to find the area. The question I posted was just an example of the questions I've got to know how to answer.

#14

Thanks, but can you please tell me how you got it?

This is how I did it, but it's probably not the orthodox way.

One SD to the left/right of the mean is 34.1%. I divided that by the SD (8) giving 4.2625. Then: 6*4.2625=25.575 (left) and 4*4.2625=17.05 (right), then 25.575+17.05=42.625.

Someone else will have a better way I'm sure, but that's the best I can do after a year of not doing stats. (Disclaimer: I don't actually know if it's right...)

*Last edited by Kiwi Ace at Apr 7, 2008,*

#15

no 42 isn't the correct answer...its the answer to why we are here though

#16

This is how I did it, but it's probably not the orthodox way.

One SD to the left/right of the mean is 34.1%. I divided that by the SD (8) giving 4.2625. Then: 6*4.2625=25.575 (left) and 4*4.2625=17.05 (right), then 25.575+17.05=42.625.

Someone else will have a better way I'm sure, but that's the best I can do after a year of not doing stats. (Disclaimer: I don't actually know if it's right...)

Hmm..Yeah, I get ya , but do you know the way that you're supposed to do it?

#17

Well I don't know the answer however I think TS needs to read The Hitchhikers Guide To The Galaxy.

#18

normalcdf (50,60,56,8)

consult your graphic display calculator.

consult your graphic display calculator.

#19

Well I don't know the answer however I think TS needs to read The Hitchhikers Guide To The Galaxy.

I will, as well as all of Douglas Adams' other works, as long as someone can answer my question .

#20

Hmm..Yeah, I get ya , but do you know the way that you're supposed to do it?

Give me two secs, I'm just looking it up for ya.

#21

Give me two secs, I'm just looking it up for ya.

Ooh, thanks mate

#22

Ooh, thanks mate

Ok, I've brushed some dust of my stats knowledge so I'll give it a shot.

Firstly, you know the standardising formula z=(x-u)/s right? (s being sigma).

We want P(50<X<60) with a u=56 and s=8. I'd draw your standard normal curve here and mark in your values and shade the areas you want.

We standardise each side, (50-56)/8=-0.75 for the LHS and (60-56)/8=0.5 for the RHS.

Now we have P(-0.75<Z<0.5), so we turn to our standard normal table and look each up, 0.75 is .2734 and 0.5 is .1915. Add them together, 0.2734+0.1915=0.4649.

Times by 100, 46.5%.

Or something like that, don't take it as gospel, I haven't done much stats in the past year.

*Last edited by Kiwi Ace at Apr 7, 2008,*

#23

Does anyone out there know how to solve problems like this:

"A cop does a survey of the speeds of travelling cars. The area in which she conducts her survey has a 50km/h speed limit. He finds a mean of 56km/h and a standard deviation of 8km/h.

Calculate the percentage of people driving between 50 and 60km/h?"

How would I solve that?

Thanks.

Ok, well the speed of the cars is normally distributed, and so the question is asking for the probability of a random car travelling between 50 and 60 km/h.

Now, you should have been told at some point how to do questions like this, so I'm not going to explain every step.

So, first you standardise the distribution to get the CRV (continuous random variable) Z=(X-56)/8 (here, X is the CRV with normal distribution with mean 56 and standard deviation 8). So when X=50, Z=-3/4 and when X=60, Z=1/2

You then need to find normal tables (there'll be plenty on the internet) to find P(-3/4<Z<1/2). Express this as a percentage and you'll have your answer.

#24

Ok, well the speed of the cars is normally distributed, and so the question is asking for the probability of a random car travelling between 50 and 60 km/h.

Now, you should have been told at some point how to do questions like this, so I'm not going to explain every step.

So, first you standardise the distribution to get the CRV (continuous random variable) Z=(X-56)/8 (here, X is the CRV with normal distribution with mean 56 and standard deviation 8). So when X=50, Z=-3/4 and when X=60, Z=1/2

You then need to find normal tables (there'll be plenty on the internet) to find P(-3/4<Z<1/2). Express this as a percentage and you'll have your answer.

Sweet, I'm right then.

#25

Sweet, I'm right then.

Ha, yeah

#26

TS, if you've doing 7th form stats and have a Casio FX-9750G calculator, go to Stat, then Dist, then Norm, then Ncd, fill it out, and it will give you the probability.

#27

TS, if you've doing 7th form stats and have a Casio FX-9750G calculator, go to Stat, then Dist, then Norm, then Ncd, fill it out, and it will give you the probability.

Nah, I'm doing 6th form stats, but I do have a Casio FX-9750G calculator. Yeah, I plugged in those numbers. I went:

Lower: 50

Upper: 60

Standard Deviation: 8

Mean: 56

And got the answer 0.46483, however, that's a probability. Can I multiply that by 100 and call it a percentage or would that be wrong?

#28

Nah, I'm doing 6th form stats, but I do have a Casio FX-9750G calculator. Yeah, I plugged in those numbers. I went:

Lower: 50

Upper: 60

Standard Deviation: 8

Mean: 56

And got the answer 0.46483, however, that's a probability. Can I multiply that by 100 and call it a percentage or would that be wrong?

Yep that's right, but they'll probably want the working porcupineoris and I showed, but the calculator is a good check.

#29

Nah, I'm doing 6th form stats, but I do have a Casio FX-9750G calculator. Yeah, I plugged in those numbers. I went:

Lower: 50

Upper: 60

Standard Deviation: 8

Mean: 56

And got the answer 0.46483, however, that's a probability. Can I multiply that by 100 and call it a percentage or would that be wrong?

Yeah, thats fine. Probabilities, percentages, ratios, fractions are all the same idea just expressed slightly differently. Also, unless i suggest you get used to the proper way of doing it (without a calculator!), it'll help you a lot more, and its not that difficult.

#30

Ok, well the speed of the cars is normally distributed, and so the question is asking for the probability of a random car travelling between 50 and 60 km/h.

Now, you should have been told at some point how to do questions like this, so I'm not going to explain every step.

So, first you standardise the distribution to get the CRV (continuous random variable) Z=(X-56)/8 (here, X is the CRV with normal distribution with mean 56 and standard deviation 8). So when X=50, Z=-3/4 and when X=60, Z=1/2

You then need to find normal tables (there'll be plenty on the internet) to find P(-3/4<Z<1/2). Express this as a percentage and you'll have your answer.

Okay, I'm with you until the last bit. When I plug those numbers in:

Lower: -0.75

Upper: 0.5

Std. Deviation: 8

Mean: 56

I get 1.343 x 10^-12, nowhere near 46%, what am I doing wrong?

Yep that's right, but they'll probably want the working porcupineoris and I showed, but the calculator is a good check.

Yeah, I figured as much .

Yeah, thats fine. Probabilities, percentages, ratios, fractions are all the same idea just expressed slightly differently. Also, unless i suggest you get used to the proper way of doing it (without a calculator!), it'll help you a lot more, and its not that difficult.

Okay cool. Yeah, I wanna learn how to do it properly so I understand it better, thanks .

#31

Okay, I'm with you until the last bit. When I plug those numbers in:

Lower: -0.75

Upper: 0.5

Std. Deviation: 8

Mean: 56

I get 1.343 x 10^-12, nowhere near 46%, what am I doing wrong?

Just forget about your calculator for that. Once you have standardised your X values (50 and 60) and gotten Z values (-0.75 and 0.5), with the formula (x-u)/s, you need to look up 0.75 and 0.5 on a Z-table like this http://www.isixsigma.com/library/content/zdistribution.asp

(there should be one in your book)

Then you find the probabilities for each and add them together. (the Z table gives the probabilities, here 0.2734 and 0.1915)

*Last edited by Kiwi Ace at Apr 7, 2008,*

#32

Just forget about your calculator for that. Once you have standardised your X values (50 and 60) and gotten Z values (-0.75 and 0.5), with the formula (x-u)/s, you need to look up 0.75 and 0.5 on a Z-table like this http://www.isixsigma.com/library/content/zdistribution.asp

(there should be one in your book)

Then you find the probabilities for each and add them together. (the Z table gives the probabilities, here 0.2734 and 0.1915)

Okay sweet. The formula to get the Z values is X-mean/std. deviation right?

#33

Okay sweet. The formula to get the Z values is X-mean/std. deviation right?

Yep that's the key formula to remember in your course, everything builds off it. Anyway, I'm going to bed now, good luck mate!

#34

Yep that's the key formula to remember in your course, everything builds off it. Anyway, I'm going to bed now, good luck mate!

Okay, awesome. Yeah, me too .

Thanks for all the help bro, you're the man!!

#35

find a bell curve use that