# Thread: University Statistics Discussion Marathon

1. ## Re: University Statistics Discussion Marathon

Id rule out A straight up as well, which would love me down to B or C.

2. ## Re: University Statistics Discussion Marathon

Statistical validity refers to whether a statistical study is able to draw conclusions that are in agreement with statistical and scientific laws. This means if a conclusion is drawn from a given data set after experimentation, it is said to be scientifically valid if the conclusion drawn from the experiment is scientific and relies on mathematical and statistical laws.

id also refer out C, and would go for B in this question.

4. ## Re: University Statistics Discussion Marathon

Originally Posted by davidgoes4wce
Third option

5. ## Re: University Statistics Discussion Marathon

Damn I would have gone for 4th..........

6. ## Re: University Statistics Discussion Marathon

any idea how you would explain it?

$\sum_{i=1} ^n (x_i -1)^2$

$\sum_{i=1} ^n x_i ^2-2x_i+1$

I don't see how you then can get to n-1

7. ## Re: University Statistics Discussion Marathon

Originally Posted by davidgoes4wce
any idea how you would explain it?

$\sum_{i=1} ^n (x_i -1)^2$

$\sum_{i=1} ^n x_i ^2-2x_i+1$

I don't see how you then can get to n-1
$\noindent Use the sample variance formula: S^{2} = \frac{1}{n-1}\sum_{i = 1}^{n}\left(x_{i} - \overline{x}\right)^{2}, where S^{2} is the sample variance and \overline{x} is the sample mean. Note that if the (sample) standard deviation equals the variance, since one is the square of the other, we must have that the sample variance is S^{2} = 1 (it can't be 0 since we are told it is positive). We are told the mean is equal to this too, so the mean is also \overline{x} = 1. Hence we have 1 = \frac{1}{n-1}\sum_{i=1}^{n}\left(x_{i} - 1\right)^{2}, from the sample variance formula. Now rearranging this yields the result.$

8. ## Re: University Statistics Discussion Marathon

Originally Posted by davidgoes4wce
Damn I would have gone for 4th..........
$\noindent An example of such a dataset: n = 2, with x_{1} = \frac{2-\sqrt{2}}{2}, x_{2} = \frac{2 + \sqrt{2}}{2}.$

9. ## Re: University Statistics Discussion Marathon

The company Atherton uses a particular machine to fill 1 kg bags with sugar. The weight X of the machine processed bags follows a normal distribution with mean $\mu$ and standard deviation $\sigma$ =0.24. The machine mean weight can be set by the operator to any value between 0 kg and 50 kg. Regulations allow to sell sugar bags as "1 kg bags" if at least 97.5% of the production is 1 kg or heavier. Answer the following question, round off the final results to 2 dp, but without rounding numbers during your calculations.

What mean weight $\mu$ should the machine be set to in order to produce 97.5% of the bags weighting at least 1kg?

10. ## Re: University Statistics Discussion Marathon

Anyway I had a go at. If someone could confirm that would be great.

11. ## Re: University Statistics Discussion Marathon

Doing Statistics on a Good Friday is insanely awesome.

12. ## Re: University Statistics Discussion Marathon

Answer the following question, keeping in mind that your answer must be a number between 0 and 1, i.e do not use percentages. Round off the final results to 2d.p but dont round numbers during your calculation.

Computer the probability that an observation 't' drawn from a normal distribution with mean $\mu$ = 24.1 and standard deviation $\sigma = 8.56$ is in the interval (15.05, 32.55) i.e the Prob (15.05 < t < 32.56) .

13. ## Re: University Statistics Discussion Marathon

Just want to confirm if this is the right answer :

14. ## Re: University Statistics Discussion Marathon

Originally Posted by davidgoes4wce
Just want to confirm if this is the right answer :

$\noindent I haven't checked your numerical calculations but yes that's the right approach, i.e. standardise the random variable using Z = \frac{X -\mu}{\sigma} so that Z becomes standard normal, correspondingly change the inequality bounds, and then use standard normal CDF values to find the required probabilities (e.g. in a table of values).$

15. ## Re: University Statistics Discussion Marathon

50 % of the meals sold at Graylands Hospital contain rice.

Answer the following question, keeping in mind that your answer must be a number between 0 and 1, i.e do not use percentages. Round off the final result to 2dp, but don't round numbers during your calculations.

What is the probability that out of 78 meals sold at Graylands Hospital, 43 or more contain rice?

16. ## Re: University Statistics Discussion Marathon

Originally Posted by davidgoes4wce
50 % of the meals sold at Graylands Hospital contain rice.

Answer the following question, keeping in mind that your answer must be a number between 0 and 1, i.e do not use percentages. Round off the final result to 2dp, but don't round numbers during your calculations.

What is the probability that out of 78 meals sold at Graylands Hospital, 43 or more contain rice?

Not sure... how to compute... (p = 0.5)

$X \sim \mathrm{Bin}(78, p) \implies P(X \geq 43) = \sum_{k=43}^{78} \binom{78}{k}p^k (1-p)^{78-k}$

17. ## Re: University Statistics Discussion Marathon

Originally Posted by He-Mann
Not sure... how to compute... (p = 0.5)

$X \sim \mathrm{Bin}(78, p) \implies P(X \geq 43) = \sum_{k=43}^{78} \binom{78}{k}p^k (1-p)^{78-k}$
what value of p did you get?

18. ## Re: University Statistics Discussion Marathon

Originally Posted by davidgoes4wce
what value of p did you get?
Do you mean the value of the sum? (The "p" He-Mann used was referring to the success probability parameter, which is 0.5.)

19. ## Re: University Statistics Discussion Marathon

Can't any quick way of doing it at the moment. Please let me know if any of this is wrong.

$P(X \geq 43) = \sum_{k = 43}^{78} \binom{78}{k} p^k (1-p)^{78 -k} = 1 - \sum_{k=0}^{42} \binom{78}{k} p^k (1-p)^{78-k}$

Since probabilities sum to 1, we have

$\left(\sum_{k=0}^{38}\binom{78}{k} p^k (1-p)^{78-k} \right) + \left(\binom{78}{39} p^{39} (1-p)^{39} \right) + \left(\sum_{k=40}^{78} \binom{78}{k}p^k (1-p)^{78-k} \right) = 1 \quad \dots (*)$

By symmetry (i.e. Pascal's Triangle), we have

$\sum_{k=0}^{38}\binom{78}{k} p^k (1-p)^{78-k} = \sum_{k=40}^{78} \binom{78}{k}p^k (1-p)^{78-k}$

Then from (*), we have

\begin{align*}\sum_{k=0}^{38}\binom{78}{k} p^k (1-p)^{78-k} &= \frac{1}{2}\binom{78}{39} p^{39} (1-p)^{39} \\&= \frac{1}{2} \binom{78}{39}(0.5)^{78} \approx 0.045 \\\\ \textbf{MISTAKE HAPPENS ABOVE. THE RHS SHOULD BE }&=\frac{1}{2} (1 - \binom{78}{39}(0.5)^{78}) \\ &\approx 0.95
\end{align*}

Now, we compute the desired probability,

\begin{align*}P(X \geq 43)& = 1 - \sum_{k=0}^{42} \binom{78}{k} p^k (1-p)^{78-k} \\ & \approx 1 - \left( 0.045 + 2\times 0.045+ \sum_{k=40}^{42}\binom{78}{k} p^k (1-p)^{78-k}\right) \\ & = 1 - \left( 0.045 + 2\times 0.045 + 0.2409 \right) \\& = 0.6241 \end{align*}

In retrospect, I should have wrote p = 0.5 = 1 - p and combined the product which would make it easier to read.

20. ## Re: University Statistics Discussion Marathon

Bump! Really interested if someone has thought up of a more efficient way to do the question (or identify, if it exists, the flaws in my work).

21. ## Re: University Statistics Discussion Marathon

Mathematica 11 gives me 0.2141 from your initial sum...

22. ## Re: University Statistics Discussion Marathon

Originally Posted by He-Mann
Can't any quick way of doing it at the moment. Please let me know if any of this is wrong.

$P(X \geq 43) = \sum_{k = 43}^{78} \binom{78}{k} p^k (1-p)^{78 -k} = 1 - \sum_{k=0}^{42} \binom{78}{k} p^k (1-p)^{78-k}$

Since probabilities sum to 1, we have

$\left(\sum_{k=0}^{38}\binom{78}{k} p^k (1-p)^{78-k} \right) + \left(\binom{78}{39} p^{39} (1-p)^{39} \right) + \left(\sum_{k=40}^{78} \binom{78}{k}p^k (1-p)^{78-k} \right) = 1 \quad \dots (*)$

By symmetry (i.e. Pascal's Triangle), we have

$\sum_{k=0}^{38}\binom{78}{k} p^k (1-p)^{78-k} = \sum_{k=40}^{78} \binom{78}{k}p^k (1-p)^{78-k}$

Then from (*), we have

\begin{align*}\sum_{k=0}^{38}\binom{78}{k} p^k (1-p)^{78-k} &= \frac{1}{2}\binom{78}{39} p^{39} (1-p)^{39} \\&= \frac{1}{2} \binom{78}{39}(0.5)^{78} \approx 0.045 \end{align*}

Now, we compute the desired probability,

\begin{align*}P(X \geq 43)& = 1 - \sum_{k=0}^{42} \binom{78}{k} p^k (1-p)^{78-k} \\ & \approx 1 - \left( 0.045 + 2\times 0.045+ \sum_{k=40}^{42}\binom{78}{k} p^k (1-p)^{78-k}\right) \\ & = 1 - \left( 0.045 + 2\times 0.045 + 0.2409 \right) \\& = 0.6241 \end{align*}

In retrospect, I should have wrote p = 0.5 = 1 - p and combined the product which would make it easier to read.
Originally Posted by He-Mann
Bump! Really interested if someone has thought up of a more efficient way to do the question (or identify, if it exists, the flaws in my work).
Note that the answer to the sum cannot exceed 0.5 (in general for X ~ Bin(n, 1/2), if k > n/2, then Pr(X ≥ k) ≤ 0.5).

23. ## Re: University Statistics Discussion Marathon

I have corrected my original working. Thanks for the checks, I knew something was strange.

__________________________________________________ _________________________

Given that p = 0.5 = 1-p,

$P(X \geq 43) = \sum_{k = 43}^{78} \binom{78}{k} p^k (1-p)^{78 -k} = 1 - (0.5)^{78} \sum_{k=0}^{42} \binom{78}{k}$

By Binomial Theorem (with x = 1) and Pascal's Triangle (i.e. C(n,k) = C(n, n-k)),

\begin{align*}(1+x)^{78} = \sum_{k=0}^{78} \binom{78}{k} = 2^{78} &\iff 2\left(\sum_{k=0}^{38} \binom{78}{k} \right) + \binom{78}{39} = 2^{78} \\& \iff \sum_{k=0}^{38} \binom{78}{k} = \frac{2^{78} - \binom{78}{39}}{2}\end{align*}

Now, we compute the desired probability,

\begin{align*}P(X \geq 43)& = 1 - (0.5)^{78} \left( \left[\sum_{k=0}^{38} \binom{78}{k} \right] + \binom{78}{39} + \binom{78}{40} + \binom{78}{41} + \binom{78}{42} \right)\\&\approx 0.2141\end{align*}

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