1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
julia-pushkina [17]
3 years ago
10

PLEASEEE HELP MEEE DO THISSS♥️♥️♥️♥️​

Mathematics
1 answer:
Firlakuza [10]3 years ago
7 0
Question 5 is C ASA but I did not know question 6
You might be interested in
The dot plot shows the number of hours, to the nearest hour, that a sample of 5th- and 7th-grade students spend watching televis
SSSSS [86.1K]

Answer:5th grade range-7

7th grade range-7

The ratio variation is -1

Step-by-step explanation:just got the question right

5 0
3 years ago
Read 2 more answers
Please help if you can<br> 1. ( 6k+2) - (3k-2)
viktelen [127]
(6k+2)-(3k-2)
expand brackets  18k²-12k+6k-4
simplify 18k²-6k-4
4 0
3 years ago
Anyone know what 68010+01410 is ​
alex41 [277]
Put the first number on top then start adding 6+0, 8+1, 0+4, 1+1, 0+0 and the answer is 69420
7 0
2 years ago
Can someone pls help
Cloud [144]

Answer:

Step-by-step explanation:

3 0
2 years ago
Read 2 more answers
Let $$X_1, X_2, ...X_n$$ be uniformly distributed on the interval 0 to a. Recall that the maximum likelihood estimator of a is $
Solnce55 [7]

Answer:

a) \hat a = max(X_i)  

For this case the value for \hat a is always smaller than the value of a, assuming X_i \sim Unif[0,a] So then for this case it cannot be unbiased because an unbiased estimator satisfy this property:

E(a) - a= 0 and that's not our case.

b) E(\hat a) - a= \frac{na}{n+1} - a = \frac{na -an -a}{n+1}= \frac{-a}{n+1}

Since is a negative value we can conclude that underestimate the real value a.

\lim_{ n \to\infty} -\frac{1}{n+1}= 0

c) P(Y \leq y) = P(max(X_i) \leq y) = P(X_1 \leq y, X_2 \leq y, ..., X_n\leq y)

And assuming independence we have this:

P(Y \leq y) = P(X_1 \leq y) P(X_2 \leq y) .... P(X_n \leq y) = [P(X_1 \leq y)]^n = (\frac{y}{a})^n

f_Y (Y) = n (\frac{y}{a})^{n-1} * \frac{1}{a}= \frac{n}{a^n} y^{n-1} , y \in [0,a]

e) On this case we see that the estimator \hat a_1 is better than \hat a_2 and the reason why is because:

V(\hat a_1) > V(\hat a_2)

\frac{a^2}{3n}> \frac{a^2}{n(n+2)}

n(n+2) = n^2 + 2n > n +2n = 3n and that's satisfied for n>1.

Step-by-step explanation:

Part a

For this case we are assuming X_1, X_2 , ..., X_n \sim U(0,a)

And we are are ssuming the following estimator:

\hat a = max(X_i)  

For this case the value for \hat a is always smaller than the value of a, assuming X_i \sim Unif[0,a] So then for this case it cannot be unbiased because an unbiased estimator satisfy this property:

E(a) - a= 0 and that's not our case.

Part b

For this case we assume that the estimator is given by:

E(\hat a) = \frac{na}{n+1}

And using the definition of bias we have this:

E(\hat a) - a= \frac{na}{n+1} - a = \frac{na -an -a}{n+1}= \frac{-a}{n+1}

Since is a negative value we can conclude that underestimate the real value a.

And when we take the limit when n tend to infinity we got that the bias tend to 0.

\lim_{ n \to\infty} -\frac{1}{n+1}= 0

Part c

For this case we the followng random variable Y = max (X_i) and we can find the cumulative distribution function like this:

P(Y \leq y) = P(max(X_i) \leq y) = P(X_1 \leq y, X_2 \leq y, ..., X_n\leq y)

And assuming independence we have this:

P(Y \leq y) = P(X_1 \leq y) P(X_2 \leq y) .... P(X_n \leq y) = [P(X_1 \leq y)]^n = (\frac{y}{a})^n

Since all the random variables have the same distribution.  

Now we can find the density function derivating the distribution function like this:

f_Y (Y) = n (\frac{y}{a})^{n-1} * \frac{1}{a}= \frac{n}{a^n} y^{n-1} , y \in [0,a]

Now we can find the expected value for the random variable Y and we got this:

E(Y) = \int_{0}^a \frac{n}{a^n} y^n dy = \frac{n}{a^n} \frac{a^{n+1}}{n+1}= \frac{an}{n+1}

And the bias is given by:

E(Y)-a=\frac{an}{n+1} -a=\frac{an-an-a}{n+1}= -\frac{a}{n+1}

And again since the bias is not 0 we have a biased estimator.

Part e

For this case we have two estimators with the following variances:

V(\hat a_1) = \frac{a^2}{3n}

V(\hat a_2) = \frac{a^2}{n(n+2)}

On this case we see that the estimator \hat a_1 is better than \hat a_2 and the reason why is because:

V(\hat a_1) > V(\hat a_2)

\frac{a^2}{3n}> \frac{a^2}{n(n+2)}

n(n+2) = n^2 + 2n > n +2n = 3n and that's satisfied for n>1.

8 0
3 years ago
Other questions:
  • 5/6 - 2/6 in simplest form
    13·2 answers
  • The senior class took an end of the year trip. There were 290 total students in buses and cars. If each bus holds 45 students an
    13·1 answer
  • 4⁄6 + 2⁄4 + 3⁄5 please answer my question
    7·2 answers
  • onathan bought a new computer for $2,016, using the electronics store's finance plan. He will pay $112 a month for 18 months. Wh
    6·1 answer
  • 8 gray buttons out of 20 buttons
    9·1 answer
  • Find positive numbers a and b so that the change of variables s=ax,t=by transforms the integral ∫∫rdxdy into ∫∫t∣∣∣∂(x,y)∂(s,t)∣
    5·1 answer
  • The mean value of land and buildings per acre from a sample of farms is $1400, with a standard deviation of $100. The data set h
    8·1 answer
  • In baseball, the equation E = 9RI gives a pitcher's earned run average, E where R is the number of earned runs the player is all
    14·1 answer
  • Diameter 32 inches what is the radius?
    12·2 answers
  • RSM HW HELP!!! PLEASE HELP ME ASAP!!!! I WILL GIVE BRAINLIEST IF YOU SOLVE THEM ALL CORRECTLY!!! THANKS!
    10·2 answers
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!