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noname [10]
3 years ago
15

To find the 90% confidence interval for the population standard deviation, you randomly sample with replacement from the origina

l sample thousands of times. From each new sample, you compute the sample standard deviation. Using the bootstrap method, how can you find the 90% confidence interval for the population standard deviation from these values?
Mathematics
1 answer:
vitfil [10]3 years ago
8 0

Answer:

use 5th and the 95th percentile of these values.

Step-by-step explanation:

To find the 90%  confidence interval for the population standard deviation, randomly sample with replacement from the original sample thousands of times. From each new sample, you compute the sample standard deviation.  Using the bootstrap method, we can find the 90%  confidence interval for the population standard deviation,from the values of 5th and the 95th percentile of these values.

So we need to use 5th and the 95th percentile of these values.

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19. Use Gauss-Jordan elimination to solve the following system of equations. 3x + 5y = 7 6x − y = −8
MrRa [10]

\left[\begin{array}{cc}3&5\\6&-1\end{array}\right] \left[\begin{array}{c}x\\y\end{array}\right] = \left[\begin{array}{c}7\\-8\end{array}\right]\\\left[\begin{array}{cc|c}3&5 &7\\ 6& -1 & -8\end{array}\right] \rightarrow \left[\begin{array}{cc|c}3&5& 7\\0&-11 & -22\end{array}\right] \rightarrow \left[\begin{array}{cc|c}3&5& 7\\0&1 & 2\end{array}\right]\\3x +5y = 7\\\\y = 2\\\therefore x = -1

6 0
3 years ago
john spent half of his weekly allowance at the batting cages. to earn more money his parents let him vacuum the house for $6. wh
KATRIN_1 [288]

Answer:

$28

Step-by-step explanation:

First to solve this get the original amount before his parents gave him $6

20-6=14

So he had $14 after spending it at the batting cages.

It is stated he spent half of his money there so multiply 14 by 2

14 x 2=28

So he has a weekly allowance of $28


4 0
3 years ago
Lindsey earns 5% commission off all sales in the clothing boutique she works at. If a customer bought $124.50 in merchandise, ho
goldenfox [79]

Answer:

Choice C: 6.23

Step-by-step explanation:

\frac{5}{100} = \frac{x}{124.50} -----> cross multiply

5(124.50) = 100x

622.5 = 100x

\frac{622.5}{100} = \frac{100x}{100}

6.225 = x

6.225 rounded to the nearest hundredth is 6.23

She earned 6.23 commission.

4 0
3 years ago
Can someone answer this please
yanalaym [24]

Answer:

x=7

Step-by-step explanation:

8x-5y=16

8x-40=16

8x=56

x=7

8 0
3 years ago
You know that in a specific population of rainbow trout 15% of the individuals carry intestinal parasites. Assume you obtain a r
kicyunya [14]

Answer:

a) 0.2316 = 23.16% probability that 0 carry intestinal parasites.

b) 0.4005 = 40.05% probability that at least two individuals carry intestinal parasites.

Step-by-step explanation:

For each trout, there are only two possible outcomes. Either they carry intestinal parasites, or they do not. Trouts are independent. This means that we use the binomial probability distribution to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

In which C_{n,x} is the number of different combinations of x objects from a set of n elements, given by the following formula.

C_{n,x} = \frac{n!}{x!(n-x)!}

And p is the probability of X happening.

You know that in a specific population of rainbow trout 15% of the individuals carry intestinal parasites.

This means that p = 0.15

Assume you obtain a random sample of 9 individuals from this population:

This means that n = 9

a. Calculate the probability that __ (last digit of your ID number) carry intestinal parasites.

Last digit is 0, so:

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

P(X = 0) = C_{9,0}.(0.15)^{0}.(0.85)^{9} = 0.2316

0.2316 = 23.16% probability that 0 carry intestinal parasites.

b. Calculate the probability that at least two individuals carry intestinal parasites.

This is

P(X \geq 2) = 1 - P(X < 2)

In which

P(X < 2) = P(X = 0) + P(X = 1)

So

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

P(X = 0) = C_{9,0}.(0.15)^{0}.(0.85)^{9} = 0.2316

P(X = 1) = C_{9,1}.(0.15)^{1}.(0.85)^{8} = 0.3679

P(X < 2) = P(X = 0) + P(X = 1) = 0.2316 + 0.3679 = 0.5995

P(X \geq 2) = 1 - P(X < 2) = 1 - 0.5995 = 0.4005

0.4005 = 40.05% probability that at least two individuals carry intestinal parasites.

5 0
3 years ago
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