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On each trial of an experiment, a subject is presented with a constant soft noise, which is interrupted at some unpredictable time by a noticeably louder sound. The time it takes for the subject to react to this louder sound is recorded. The following list contains the reaction times (in milliseconds) for the trials of this experiment: 296, 311, 202, 217, 231, 171, 241, 164, 257, 269, 175, 273, 285, 226, 261 Find 25th and 80th percentiles for these reaction times.

Answers

Answer 1

Answer:

25th percentile = 202

80th percentile = 285

Step-by-step explanation:

Rearranging the values in increasing order

164

171

175

202

217

226

231

241

257

261

269

273

285

296

311

N = total number of variables = 15.

25th percentile = [(N + 1)/4]th variable = (15 + 1)/4 = 4th variable.

So, the 25th percentile = 4th variable = 202.

80th percentile = 0.8(N + 1) th variable = 0.8 × (15 + 1) = 12.8th variable = 13th variable = 285

Answer 2

Answer:

Final answer:

To find the 25th and 80th percentiles for the reaction times, sort the list and determine the corresponding values.

Explanation:

To find the 25th and 80th percentiles for the reaction times in this experiment, we need to first sort the list in ascending order:

The 25th percentile is the value that separates the lowest 25% of the data. In this case, it is the 4th value in the sorted list, which is 202 milliseconds. The 80th percentile is the value that separates the lowest 80% of the data. In this case, it is the 12th value in the sorted list, which is 273 milliseconds.

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If x1, x2, . . . , xn are independent and identically distributed random variables having uniform distributions over (0, 1), find (a) e[max(x1, . . . , xn)]; (b) e[min(x1, . . . , xn)].

Answers

Denote by $X_((n))$ the maximum order statistic, with $X_((n))=\max\{X_1,\ldots,X_n\}$ , and similarly denote by $X_((1))$ the minimum order statistic. Then the CDF for $X_((n))$ is

$F_{X_((n))}(x)=\mathbb P(X_((n))\le x)$

In order for there to be some $x$ that exceeds the value of $X_((n))$ , it must be true that $x$ exceeds the value of all the $X_i$ , so the above is equivalent to the joint probability

$F_{X_((n))}(x)=\mathbb P(X_1\le x,\ldots,X_n\le x)$

and since the $X_i$ are i.i.d., we have

$F_{X_((n))}(x)=\mathbb P(X_1\le x)\cdots\mathbb P(X_n\le x)=\mathbb P(X_1\le x)^n$
$\implies F_{X_((n))}(x)=F_X(x)^n$

where $X\sim\mathrm{Unif}(0,1)$ . We have

$F_X(x)=\begin{cases}0&\text{for }x<0\nx&\text{for }0\le x\le1\n1&\text{for }x>1\end{cases}$

and so

$F_{X_((n))}(x)=\begin{cases}0&\text{for }x<0\nx^n&\text{for }0\le x\le1\n1&\text{for }x>1\end{cases}$
$\implies f_{X_((n))}(x)=\begin{cases}nx^(n-1)&\text{for }0<x<1\n0&\text{otherwise}\end{cases}$
$\implies\mathbb E[X_((n))]=\displaystyle\int_0^1xnx^(n-1)\,\mathrm dx=n\int_0^1x^n\,\mathrm dx=\frac n{n+1}$

Using similar reasoning, we can find the CDF for $X_((1))$ . We have

$F_{X_((1))}(x)=\mathbb P(X_((1))\le x)=1-\mathbb P(X_((1))>x)$
$F_{X_((1))}(x)=1-\mathbb P(X_1>x,\ldots,X_n>x)=1-\mathbb P(X_1>x)^n$
$F_{X_((1))}(x)=1-(1-\mathbb P(X\le x))^n=1-(1-F_X(x))^n$
$\implies F_{X_((1))}(x)=\begin{cases}0&\text{for }x<0\n1-(1-x)^n&\text{for }0\le x\le1\n1&\text{for }x>1\end{cases}$
$\implies f_{X_((1))}(x)=\begin{cases}n(1-x)^(n-1)&\text{for }0<x<1\n0&\text{otherwise}\end{cases}$
$\implies\mathbb E[X_((1))]=\displaystyle\int_0^1xn(1-x)^(n-1)\,\mathrm dx=\frac1{n+1}$

Final answer:

The expected values of the maximum and minimum of independent and identically distributed (iid) uniform random variables, x1, x2, ..., xn, are given by E[max(x1, ..., xn)] = n / (n + 1) and E[min(x1, ..., xn)] = 1 / (n + 1) respectively.

Explanation:

In mathematics, particularly in probability theory and statistics, the question is related to independent and identically distributed (iid) random variables with a uniform distribution. The expected value or mean (E) of the maximum (max) and minimum (min) of these random variables is sought.

(a) The expected value of the max of 'n' iid uniform random variables, x1, x2, ..., xn, is calculated by integrating the nth power of x from 0 to 1. It can be found via the equation E[max(x1, ..., xn)] = n / (n + 1).

(b) Similarly, the expected value of the min of 'n' iid uniform random variables is acquired by doing (1 / (n + 1)). Hence, E[min(x1, ..., xn)] = 1 / (n + 1).

By understanding these, you could visualize the various outcomes of the random variables and their distributions, demonstrating how likely each outcome could occur.

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Stephanie claims that 26540is greater than 50. Do you agree or disagree? Explain.

Answers

Agree. 50 is a smaller

In a recent Super Bowl, a TV network predicted that 31 % of the audience would express an interest in seeing one of its forthcoming television shows. The network ran commercials for these shows during the Super Bowl. The day after the Super Bowl, and Advertising Group sampled 120 people who saw the commercials and found that 40 of them said they would watch one of the television shows. Suppose you have the following null and alternative hypotheses for a test you are running:1. H0: p = 0.53 Ha: p ≠ 0.53
2. H0: p = 0.53 Ha: p ≠ 0.53
Calculate the test statistic, rounded to 3 decimal places.

Answers

Answer:

-4.317

Step-by-step explanation:

The z test statistic for testing of 1-proportion can be computed as

$z=\frac{phat-p}{\sqrt{(pq)/(n) } }$

We know that

phat=x/n.

We know that x=40 and n=120.

Thus,

phat=40/120=0.3333

p=hypothesized proportion=0.53

q=1-p=1-0.53=0.47

So, required z-statistic is

$z=\frac{0.3333-0.53}{\sqrt{(0.53(0.47))/(120) } }$

$z=(-0.1967)/( 0.04556 )$

z=-4.317.

Thus, the required test statistic value for given hypothesis is z=-4.317.

A sample of 8 new models of automobiles provides the following data on highway miles per gallon. Highway Miles Model Per Gallon 1 33.6 2 26.8 3 20.2 4 38.7 5 35.1 6 28.0 7 26.2 8 27.6 a. What is the point estimate for the average highway miles per gallon for all new models of autos? b. Determine the point estimate for the standard deviation of the population.

Answers

Answer:

The point estimate for the mean is 29.5 miles per gallon and the standard deviation is 5.9

Step-by-step explanation:

The formula por the point estimate of the mean is:

$mean(x) = (x1+x2+x3+...+xn)/(n)$

And for the standard deviation:

$desv(x)=((1)/(n-1)sum(xi^(2) - n*x^(2) ))^(1/2)$

So for the mean:

$mean(x)=(33.6+26.8+20.2+38.7+35.1+28.0+26.2+27.6 )/(8) \n\nmean(x)=(236.2)/(8) \n\nmean(x)=29.5$

And for the standard deviation:

$desv(x)=((1)/(8-1) ((33.6-29.5)^(2) +(26.8-29.5)^(2) +(20.2-29.5)^(2) +(38.7-29.5)^(2) +(35.1-29.5)^(2) +(28-29.5)^(2) +(26.2-29.5)^(2) +(27.6-29.5)^(2))^(1/2) \n\ndesv(x)=((1)/(8-1) ((4.1)^(2) +(-2.7)^(2) +(-9.3)^(2) +(9.2)^(2) +(5.6)^(2) +(-1.5)^(2) +(-3.3)^(2) +(-1.9)^(2))^(1/2) \n\ndesv(x)=\sqrt{(243.3)/(7)} \n\ndesv(x)=√(34.7) \n\ndesv(x)=5.9$

emanuel played a game where he got a point if he drew s red marble out of a bag and flipped a coin that landed on heads. what is the probability that he will get a point on his first turn