# Statistics Marathon & Questions (1 Viewer)

#### davidgoes4wce

##### Well-Known Member
Statistics Marathon & Questions
This is a marathon thread for statistics. Please aim to pitch your questions for first-year/second-year university level statistics. Excelling & gifted/talented secondary school students are also invited to contribute.

This is also a place to discuss university level statistics. For specific subjects or even higher level stuff, this is probably not the best thread. (Feel to create your own in those instances)

(mod edit 7/6/17 by dan964)

===============================

$The average mass of a man in an office is 85 kg with standard deviation 12 kg. The average mass of a woman in the office is 68 kg with standard deviation 8 kg. The empty lift has a mass of 500 kg. What is the expectation and standard deviation of the total mass of the lift when 3 women and 4 men are inside?$

$For the Expected value. X= men, Y = women$

$E(X)=85, Var(X)=144, E(Y)=68, Var(Y)=64, E(Lift)=500$

$E(4X+3Y+Lift)= 4 \times 85 + 68 \times 3 +500 = 1044$

$I'm having some problems in calculating the standard deviation, I used the formula Var(4X+3Y+Lift)= 4^{2} Var (X)+ 3^{2} Var (Y)$

$The answer turned out to be a standard deviation of 27.7$

Last edited by a moderator:

#### InteGrand

##### Well-Known Member
Re: University Statistics Discussion Marathon

Thought it'd be a good idea to start a marathon for users to post and answer university statistics and probability problems.

$The average mass of a man in an office is 85 kg with standard deviation 12 kg. The average mass of a woman in the office is 68 kg with standard deviation 8 kg. The empty lift has a mass of 500 kg. What is the expectation and standard deviation of the total mass of the lift when 3 women and 4 men are inside?$

$For the Expected value. X= men, Y = women$

$E(X)=85, Var(X)=144, E(Y)=68, Var(Y)=64, E(Lift)=500$

$E(4X+3Y+Lift)= 4 \times 85 + 68 \times 3 +500 = 1044$

$I'm having some problems in calculating the standard deviation, I used the formula Var(4X+3Y+Lift)= 4^{2} Var (X)+ 3^{2} Var (Y)$

$The answer turned out to be a standard deviation of 27.7$
Remember that variance is the square of standard deviation.

They gave us the standard deviations of the male and female masses, so we need to square them to get their variances to use in your line about variance.

Then to get the standard deviation at the end, we'll need to take the square root of what you get for the variance.

Last edited:

#### davidgoes4wce

##### Well-Known Member
Re: University Statistics Discussion Marathon

Remember that variance is the square of standard deviation.

They gave us the standard deviations of the male and female masses, so we need to square them to get their variances to use in your line about variance.

Then to get the standard deviation at the end, we'll need to take the square root of what you get for the variance.
$Yes I did factor that into account I converted SD(X)=12 into Var(X)=144 and SD(Y)=8 into Var(Y)=64$

#### seanieg89

##### Well-Known Member
Re: University Statistics Discussion Marathon

Where are you getting 4X and 3Y from? You are not multiplying single R.V's by a constant, you are summing independent identically distributed R.V's.

(These things aren't actually independent, but this is a good approximation if the number of people in the office is much larger that the number of people in the lift for example.)

When you sum independent (more generally uncorrelated) R.V's, you just sum the variances of the individual R.V's to obtain the variance of the sum. This gives you the correct answer.

(Apologies for anything said above that is incorrect, my stats knowledge is mediocre).

#### seanieg89

##### Well-Known Member
Re: University Statistics Discussion Marathon

Re: my comment on independence.

We can see the extreme case of how this method breaks down if there are exactly 4 men and 3 women in the whole office. Then there is only one way to choose the occupants of the elevator, and so the total mass will have zero variance, regardless of how widely the masses of the individual people vary.

#### davidgoes4wce

##### Well-Known Member
Re: University Statistics Discussion Marathon

I still don't get their answer.....im thinking it should be 53.67

$Based on the rule Var(aX+bY)= a^{2} \ Var(X) +b^{2} \ Var{Y}$

$SD=\sqrt{4^{2} Var (X)+ 3^{2} Var (Y)}$

$SD=\sqrt{4^{2} \times 144+ 3^{2} \times 64}$

$SD =53.67$

#### seanieg89

##### Well-Known Member
Re: University Statistics Discussion Marathon

I still don't get their answer.....im thinking it should be 53.67

$Based on the rule Var(aX+bY)= a^{2} \ Var(X) +b^{2} \ Var{Y}$

$SD=\sqrt{4^{2} Var (X)+ 3^{2} Var (Y)}$

$SD=\sqrt{4^{2} \times 144+ 3^{2} \times 64}$

$SD =53.67$
As I just said, 4X and 3Y are irrelevant to the problem. We have

$\textrm{Var}(\textrm{Total Mass})=\textrm{Var}\left(\sum_{j=1}^4 X_j+ \sum_{j=1}^3 Y_j\right)= 4\cdot\textrm{Var}(X)+3\cdot\textrm{Var}(Y)\\ \\= 4\cdot 144+3\cdot 64 = 768.\\ \\ \sigma=\sqrt{768}=27.713 ( 3 d.p)$

#### seanieg89

##### Well-Known Member
Re: University Statistics Discussion Marathon

Think about rolling a dice and doubling the outcome vs rolling two dice and summing them.

The first RV only takes even values and takes each value in {2,4,...10,12} with equal likelihood.
The second RV can take any value in {2,3,...,11,12} and takes the values close to 7 much more often than it takes the values far from 7.

We have an analogous situation in your elevator question and you are mixing up the above two.

Last edited:

#### leehuan

##### Well-Known Member
Re: University Statistics Discussion Marathon

I'm starting probability next lecture. Should I post all of my questions here?

#### InteGrand

##### Well-Known Member
Re: University Statistics Discussion Marathon

I'm starting probability next lecture. Should I post all of my questions here?
I guess that's up to you. If you'd prefer having your own questions in specific threads, you could make separate threads for them. Depends on whether you'd prefer that or prefer having them in one thread that also includes Q's from other people.

#### seanieg89

##### Well-Known Member
Re: University Statistics Discussion Marathon

Here is a exercise that some of you might enjoy from a probability course I am grading:

$Let X_j be independent uniformly distributed random variables on [0,1]. Prove that for every a>1/2, there exists a corresponding b>0 such that \mathcal{P}(\frac{1}{n}\sum_{j=1}^n X_j >a)< e^{-nb} for all positive integers n.\\ \\ You may not use the central limit theorem for the purpose of this exercise. (Although feel free to post a CLT based proof as well if you wish.)$

#### leehuan

##### Well-Known Member
Re: University Statistics Discussion Marathon

One of the few questions I got out by myself. (Good revision for you I suppose david)

$A genetic experiment on cell division can give rise to at most 2n cells. The probability distribution of the number of cells X recorded is\\ P(X=k)=\frac{\theta^k(1-\theta)}{1-\theta^{2n+1}} for 0\le k \le 2n\\ where \theta is a constant with 0<\theta<1$

$What is the probability that an odd number of cells is recorded?$

#### davidgoes4wce

##### Well-Known Member
Re: University Statistics Discussion Marathon

Does anyone in here have access to the solutions manual for Business Statistics by Selvanathan, E., Selvanathan, S. and Keller, G? If there are any students that studied that course before that still have the weekly tutorials from their respective universities or tutors or lecturers that have access to the solutions (as I am trying to do as many questions as possible to get my standards up). If there is a price and terms you have for them, please message me.

#### BlueGas

##### Well-Known Member
Re: University Statistics Discussion Marathon

Probably an easy question compared to the questions you guys post but I need help with this.

For the class of 2010, the average score on the Writing portion of the SAT (Scholastic Aptitude Test) is 492 with a standard deviation of 111. Find the mean and standard deviation of the distribution of mean scores if we take random samples of 1100 scores at a time and compute the sample means.

#### BlueGas

##### Well-Known Member
Re: University Statistics Discussion Marathon

Probably an easy question compared to the questions you guys post but I need help with this.

For the class of 2010, the average score on the Writing portion of the SAT (Scholastic Aptitude Test) is 492 with a standard deviation of 111. Find the mean and standard deviation of the distribution of mean scores if we take random samples of 1100 scores at a time and compute the sample means.

Bump.

#### davidgoes4wce

##### Well-Known Member
Re: University Statistics Discussion Marathon

Not sure if you have a solution with you but I think the mean would still be the same = 492, the standard deviation would it be 3.35 (to 2 decimal places)?

#### BlueGas

##### Well-Known Member
Re: University Statistics Discussion Marathon

Not sure if you have a solution with you but I think the mean would still be the same = 492, the standard deviation would it be 3.35 (to 2 decimal places)?
How did you get that solution?

#### InteGrand

##### Well-Known Member
Re: University Statistics Discussion Marathon

How did you get that solution?
$\noindent The mean of the sample mean is just the original mean. The standard deviation becomes \frac{\sigma}{\sqrt{n}}. These are standard facts about sample means.$

#### BlueGas

##### Well-Known Member
Re: University Statistics Discussion Marathon

I attempted this next question and wanted to make sure if I'm right.

Test whether males are less likely than females to support a ballot initiative, if 24 % of a random sample of 52 males plan to vote yes on the initiative and 31 % of a random sample of 52 females plan to vote yes.

P1 is for males who said yes and P2 is for females who said yes

P1 = 0.24
P2 = 0.31
Pooled proportion = 0.275

Test Statistic = -0.80
P-value = 0.212

and the conclusion is do not reject the null hypothesis as the p-value as greater than the significance level of 5%

#### davidgoes4wce

##### Well-Known Member
Re: University Statistics Discussion Marathon

$Pooled Estimate of p$

$\hat{p}= \frac{\hat p_1 \ n_1 + \hat p_{2} \ n_2}{n_{1}+n_{2}}$

$\hat{p}=\frac{0.24 \times 52 + 0.31 \times 52}{52+52}=0.275$

$SE_{0}=\sqrt{\frac{\hat p (1-\hat p)}{n_1} +\frac{ \hat p (1-\hat p) }{n_2}}$

$SE_{0}=\sqrt{\frac{(0.275) \ (0.725)}{52} + \frac{(0.275) \ (0.725)}{52} }= 0.0883$

$z=\frac{( \hat p_1 - \hat p_2)-0}{SE_O} =\frac{(0.24-0.31)-0}{0.0883}=-0.8$

$Microsoft Excel calculation in order to get the p-value:$

$Since, z-calculated lies in the non-rejection region, we accept the null hypothesis that there aren't any significant differences between the two genders.$

Last edited: