How do I find the variance and what is the answer to the question?

[AoKai]

Ive been stuck at this question for over 5 hours now trying to figure out what the answer is. Please help!

We have observed 250 cars on a motorway with speed limit of 90 km/h

Speed number of cars
75 32
85 56
90 36
97 23
105 57
120 46

A, calculate the average speed and standard deviation

What is the answer to the question, how do I find the variance?

 

[chisigma]

Ive been stuck at this question for over 5 hours now trying to figure out what the answer is. Please help!

We have observed 250 cars on a motorway with speed limit of 90 km/h

Speed number of cars
75 32
85 56
90 36
97 23
105 57
120 46

A, calculate the average speed and standard deviation

What is the answer to the question, how do I find the variance?

Wellcome on MHB AoKai!... let's start from the basic definitions: do You know what is a discrete random variable and what is its expected value [or mean value or even average value...]?... Kind regards $\chi$ $\sigma$

 

[AoKai]

Wellcome on MHB AoKai!... let's start from the basic definitions: do You know what is a discrete random variable and what is its expected value [or mean value or even average value...]?... Kind regards $\chi$ $\sigma$

Yes I do what those are, I've just been stuck on this particular question for about 5 hours now trying to figure out how to do it.

Ive been using this formula given by my teacher.

Sum ( (speed-mean)^2 * number of cars) / 249

But I'm not getting the correct answer. Apparently the variance is 213, but the variance I keep getting is 464. I am not sure if I've been misscalculating so I ran through the numbers over and over and I still keep getting 464 as my variance. How do I find the right answer?

 

[chisigma]

Yes I do what those are, I've just been stuck on this particular question for about 5 hours now trying to figure out how to do it.

Ive been using this formula given by my teacher.

Sum ( (speed-mean)^2 * number of cars) / 249

But I'm not getting the correct answer. Apparently the variance is 213, but the variance I keep getting is 464. I am not sure if I've been misscalculating so I ran through the numbers over and over and I still keep getting 464 as my variance. How do I find the right answer?

All right!... You have a random variable that can be $v_{1}= 75$, $v_{2}= 85$, $v_{3}= 90$, $v_{4}= 97$, $v_{5}= 105$, $v_{6}= 120$, each with probability $p_{1}= \frac{32}{250}$, $p_{2}= \frac{56}{250}$, $p_{3}= \frac{36}{250}$, $p_{4}= \frac{23}{250}$, $p_{5}= \frac{57}{250}$, $p_{6}= \frac{46}{250}$. By definition the expected value is...

$\displaystyle E \{v\}= \mu = \sum_{k=1}^{6} v_{k}\ p_{k} = 96,544\ (1)$

Now let's define variance that is the expected value $\displaystyle E \{(v - \mu)^{2} \}$. Are You able to proceed?... Kind regards $\chi$ $\sigma$

 

[AoKai]

All right!... You have a random variable that can be $v_{1}= 75$, $v_{2}= 85$, $v_{3}= 90$, $v_{4}= 97$, $v_{5}= 105$, $v_{6}= 120$, each with probability $p_{1}= \frac{32}{250}$, $p_{2}= \frac{56}{250}$, $p_{3}= \frac{36}{250}$, $p_{4}= \frac{23}{250}$, $p_{5}= \frac{57}{250}$, $p_{6}= \frac{46}{250}$. By definition the expected value is...

$\displaystyle E \{v\}= \mu = \sum_{k=1}^{6} v_{k}\ p_{k} = 96,544\ (1)$

Now let's define variance that is the expected value $\displaystyle E \{(v - \mu)^{2} \}$. Are You able to proceed?... Kind regards $\chi$ $\sigma$

Yes. I've already found the mean. How do I find the variance?

 

[chisigma]

Yes. I've already found the mean. How do I find the variance?

Is...

$\displaystyle E \{(v - \mu)^{2}\} = \sigma^{2} = \sum_{k=1}^{6} (v_{k} - \mu)^{2}\ p_{k}\ (1)$

... where $\mu$, the $v_{k}$ and the $p_{k}$ have been previously found...The numerical evaluation of (1) is left to You... Kind regards $\chi$ $\sigma$

 

[chisigma]

The p.d.f. proposed by AoKai is very suggestive for the reason that the following histogram highlights...

http://d16cgiik7nzsna.cloudfront.net/8b/91/i82940299._szw380h285_.jpg

The mean value of speed that has been computed is v = 96.544... but in the diagram v=97 has the lowest probability!... this [apparent] paradox can be solved if we consider that the test have been done in two intervals of time, one interval in which police were present and the other interval in which police were not present...

This is what i call 'camel hump histogram' and is to be considered as indicating some sort of 'big fraud'. Several years ago [... in ther year 2007...] I made a statistical analysis about the so called 'white electoral ballots' [the ballots that are left white by the voting people...] of the political elections in Italy of the year 2006, where the winning coalition won with a margin of only 25,000 votes. The result of the analysis was that the statistican distribution of the white ballots of election districts with a large number of voters was 'camel bump' and that meant that in some districts, because of absece of control from the political opponents, several white ballots had become valid votes(Punch)...

Unfortunately I published the result of my work in a math forum with a well defined political label... the result of course had been that I was banned almost immediately! (Wasntme) ... may be that to be clever doesn't mean to be lucky! (Emo) ...

Kind regards

$\chi$ $\sigma$