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Yakvenalex [24]

3 years ago

15

A random sample of n = 36 scores is selected from a population. Which of the following distributions definitely will be normal?

The scores in the sample will form a normal distribution. The scores in the population will form a normal distribution. The distribution of sample means will form a normal distribution. Neither the sample, the population, nor the distribution of sample means will definitely be normal.

1 answer:

Alika [10]3 years ago

4 0

Answer:

The distribution of sample means will form a normal distribution

Step-by-step explanation:

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This test will not impact your grade.

ycow [4]

Answer:

im pretty sure the distance is across the y and x axis

Step-by-step explanation:

5 0

3 years ago

A group of 33 people attended a ball game. There were twice as many children as adults in the group. Set up a system of equation

hoa [83]

Answer:

1.2a+a=33

2.3a=33

3.a=11

Step-by-step explanation:

4 0

3 years ago

Determine whether the sequences converge.

Alik [6]

$a_n=\sqrt{\dfrac{(2n-1)!}{(2n+1)!}}$

Notice that

$\dfrac{(2n-1)!}{(2n+1)!}=\dfrac{(2n-1)!}{(2n+1)(2n)(2n-1)!}=\dfrac1{2n(2n+1)}$

So as $n\to\infty$ you have $a_n\to0$ . Clearly $a_n$ must converge.

The second sequence requires a bit more work.

$\begin{cases}a_1=\sqrt2\\a_n=\sqrt{2a_{n-1}}&\text{for }n\ge2\end{cases}$

The monotone convergence theorem will help here; if we can show that the sequence is monotonic and bounded, then $a_n$ will converge.

Monotonicity is often easier to establish IMO. You can do so by induction. When $n=2$ , you have

$a_2=\sqrt{2a_1}=\sqrt{2\sqrt2}=2^{3/4}>2^{1/2}=a_1$

Assume $a_k\ge a_{k-1}$ , i.e. that $a_k=\sqrt{2a_{k-1}}\ge a_{k-1}$ . Then for $n=k+1$ , you have

$a_{k+1}=\sqrt{2a_k}=\sqrt{2\sqrt{2a_{k-1}}\ge\sqrt{2a_{k-1}}=a_k$

which suggests that for all $n$ , you have $a_n\ge a_{n-1}$ , so the sequence is increasing monotonically.

Next, based on the fact that both $a_1=\sqrt2=2^{1/2}$ and $a_2=2^{3/4}$ , a reasonable guess for an upper bound may be 2. Let's convince ourselves that this is the case first by example, then by proof.

We have

$a_3=\sqrt{2\times2^{3/4}}=\sqrt{2^{7/4}}=2^{7/8}$
$a_4=\sqrt{2\times2^{7/8}}=\sqrt{2^{15/8}}=2^{15/16}$

and so on. We're getting an inkling that the explicit closed form for the sequence may be $a_n=2^{(2^n-1)/2^n}$ , but that's not what's asked for here. At any rate, it appears reasonable that the exponent will steadily approach 1. Let's prove this.

Clearly, $a_1=2^{1/2}$ . Let's assume this is the case for $n=k$ , i.e. that $a_k$ . Now for $n=k+1$ , we have

$a_{k+1}=\sqrt{2a_k}$

and so by induction, it follows that $a_n$ for all $n\ge1$ .

Therefore the second sequence must also converge (to 2).

4 0

3 years ago

Student council hosted a bake sale of the 40 items brought to sale 18 were brownies. what percent of the bake sale were Brownies

LuckyWell [14K]

Answer:

45 percent were brownies :)

7 0

4 years ago

The slope of the line whose equation is 2x + 5y + 1 = 0 is<br> 0-2/5<br> 05/2<br> 0-1/5

Leto [7]

Answer:

0-2/5

I think this is it ;-;

7 0

3 years ago