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butalik [34]
1 year ago
10

We have k coins. The probability of Heads is the same for each coin and is the realized value q of a random variable Q that is u

niformly distributed on [0,1]. We assume that conditioned on Q=q, all coin tosses are independent. Let Ti be the number of tosses of the ith coin until that coin results in Heads for the first time, for i=1,2,…,k. (Ti includes the toss that results in the first Heads.)
You may find the following integral useful: For any non-negative integers k and m,

∫10qk(1−q)mdq=k!m!(k+m+1)!.
Find the PMF of T1. (Express your answer in terms of t using standard notation.)

For t=1,…, pT1(t)=- unanswered
Find the least mean squares (LMS) estimate of Q based on the observed value, t, of T1. (Express your answer in terms of t using standard notation.)

E[Q∣T1=t]=- unanswered
We flip each of the k coins until they result in Heads for the first time. Compute the maximum a posteriori (MAP) estimate q^ of Q given the number of tosses needed, T1=t1,…,Tk=tk, for each coin.
Mathematics
1 answer:
nata0808 [166]1 year ago
3 0

Answer:

Step-by-step explanation:

The question is not in correct order ; so the first thing we are required to do is to put them together in the right form to make it easier to proof; having said that. let's get started!.

From the second part " You may find the following integral useful: For any non-negative integers k and m,"

The next  equation goes thus : \int\limits^1_0 \ \  q^k (1-q)^m dq = \frac{k!m!}{(k+m+1)!}

a. Find the PMF of T_1 .   (Express your answer in terms of t using standard notation.)

For t=1,2…,

p_T_1(t)= \frac{1}{(t*(t*1))}  by using conditional probability to solve PMF

b)  Find the least mean squares (LMS) estimate of Q based on the observed value, t, of T1. (Express your answer in terms of t using standard notation.)

By using the mathematical definition of integration to solve the estimate of Q:

E|Q|T_1=t|= \frac{2}{t+2}

c)  We flip each of the k coins until they result in Heads for the first time. Compute the maximum a posteriori (MAP) estimate q^ of Q given the number of tosses needed, T1=t1,…,Tk=tk, for each coin.

The MAP estimate of  q is :

\bar q = \frac{k}{\sum^k_{i=1}}t_i

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A quality-control inspector is testing a batch of printed circuit boards to see wheater they are capable of performing in a high
avanturin [10]

Answer:

see explaination

Step-by-step explanation:

Here the null hypothesis is that the PCB survives against the alternate that the PCB 'does not survive'. The test says that the PCB will survice if it is classified as 'good'; or, it will not survive if it is classifies as 'bad'.

a. The Type II error is the error committed when a PCB which cannot actually survive is classified as 'good'.

b. Therefore P(Type II error) = P(The PCB is classified as 'good' | PCB does not survives) = 0.03.

6 0
2 years ago
? A survey of factories in five northeastern states found that 10% of the 300 workers surveyed were satisfied with the benefits
Anastasy [175]
Percent means per one-hundred

300(10/100)=30

So 30 of the 300 workers were satisfied with their benefits.
5 0
2 years ago
Two sides of an obtuse triangle measure 9 inches and 14 inches. The length of longest side is unknown. What is the smallest poss
julia-pushkina [17]

Answer:

17 inches

Step-by-step explanation:

An obtuse triangle is the triangle in which one of the side is the longest. It contains an obtuse angle and the longest side is the side that is opposite to the vertex of the obtuse angle.

Let the three sides of the obtuse triangle be a, b and c respectively with c as the longest side. Let a = 9 inches and b = 14 inches.

Now we know that for an obtuse triangle,

$c^2 > a^2 +b^2$

$c^2 > (9)^2 +(14)^2$

$c^2 > 81 +196$

$c^2 > 277$

c > 16.64

Therefore the smallest possible whole number is 17 inches.

4 0
1 year ago
Read 2 more answers
From a survey of coworkers you find that 32​% of 150 have already received this​ year's flu vaccine. An approximate 98​% confide
Artyom0805 [142]

Answer:

a) Narrower

b) Narrower

c) Wider            

Step-by-step explanation:

We are given the following in the question:

Proportion of coworker who received flu vaccine = 32​%

98​% confidence interval: (0.231​, 0.409​)

Confidence interval:

\hat{p}\pm z_{stat}\sqrt{\dfrac{\hat{p}(1-\hat{p})}{n}}

​a) Sample size had been 600 instead of 150​

If we increase the sample size, thus the standard error of the interval decreases.

Since the standard error decreases, the confidence interval become narrower.

b) Confidence level had been 90​% instead of 98​%

As the confidence level increases, the confidence interval becomes narrower. This is due to a smaller value of z-statistic at 90% confidence level.​

c) Confidence level had been 99​% instead of 98​%

As the confidence level increases, the width of the confidence interval increases and the confidence interval become wider. This is because of a larger value of z-statistic at 99% confidence interval.

4 0
1 year ago
Question: 2. Musah Stands At The Centre Of A Rectangular Field. He First Takes 50 Steps North, Then 25 Steps West And Finally 50
Iteru [2.4K]

Answer:

60.36 steps West from centre

85.36 steps North from centre

Step-by-step explanation:

<em>Refer to attached</em>

Musah start point and movement is captured in the picture.

  • 1. He moves 50 steps to North,
  • 2. Then 25 steps to West,
  • 3. Then 50 steps on a bearing of 315°. We now North is measured 0°

or 360°, so bearing of 315° is same as North-West 45°.

<em />

<em>Note. According to Pythagorean theorem, 45° right triangle with hypotenuse of a has legs equal to a/√2.</em>

<u />

<u>How far West Is Musah's final point from the centre?</u>

  • 25 + 50/√2 ≈ 60.36 steps

<u>How far North Is Musah's final point from the centre?</u>

  • 50 + 50/√2 ≈ 85.36 steps

7 0
1 year ago
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