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3 years ago

12

A Monte Carlo experiment is a statistical analysis that uses repetitive random sampling to solve a problem.

1 answer:

zzz [600]3 years ago

8 0

Answer:

True

Step-by-step explanation:

In the Monte Carlo simulations process, the probability of outcomes is determined on the basis of repeated sampling for the outcomes in a process which can not be determined easily due to several random variables involved in the process. This techniques helps in determining the risk, uncertainty in prediction and forecasting models.

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Simplify 3m + 8 -x = -24

erastova [34]

3m+8-x=-24
3m=-24-8+x
3m=-32+x
m=(x-32)/3
3m+8-x=-24
3m+8+24=x
x=3m+32

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3 years ago

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How to simplify a30xa10

Thepotemich [5.8K]

Is the "x" a multiplication sign?

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3 years ago

The total weight of 15 containers on a freight train is 330 tonnes. What is the mean weight of the containers?

svetoff [14.1K]

Answer:

the mean weight is 22 tonnes

Step-by-step explanation:

mean=total sum/number of numbers you added

in this case the question already gave you the total so just divide right away

mean weight= 330/15 = 22 tonnes

6 0

3 years ago

Find the slope of the line that goes through (0, 2) and (4, −2).

zmey [24]

M = (y2 - y1) / (x2 - x1)

m = (-2 - 2) / (4 - 0)

m = -4 / 4

m = -1

The slope of the line that goes through (0, 2) and (4, -2) is -1.

5 0

4 years ago

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A random sample of 64 students at a university showed an average age of 20 years and a sample standard deviation of 4 years. The

adelina 88 [10]

Answer:

The 90% confidence level is $19.15< L < 20.85$

Step-by-step explanation:

From the question we are told that

The sample size is $n = 64$

The mean age is $\= x = 20 \ years$

The standard deviation is $\sigma = 4 \ years$

Generally the degree of freedom for this data set is mathematically represented as

$df = n - 1$

substituting values

$df = 64 - 1$

$df = 63$

Given that the level of confidence is 90% the significance level is mathematically evaluated as

$\alpha = 100 - 90$

$\alpha =$ 10 %

$\alpha = 0.10$

Now $\frac{\alpha }{2} = \frac{0.10}{2} = 0.05$

Since we are considering a on tail experiment

The critical value for half of this significance level at the calculated degree of freedom is obtained from the critical value table as

$t_{df, \frac{ \alpha}{2} } = t_{63, 0.05 } = 1.669$

The margin for error is mathematically represented as

$MOE = t_{df , \frac{\alpha }{2} } * \frac{\sigma}{\sqrt{n} }$

substituting values

$MOE = 1.699 * \frac{4 }{\sqrt{64} }$

$MOE = 0.85$

he 90% confidence interval for the true average age of all students in the university is evaluated as follows

$\= x - MOE < L < \= x + E$

substituting values

$20 - 0. 85 < L < 20 + 0.85$

$19.15< L < 20.85$

4 0

3 years ago