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2 years ago

Help me plsssssssssssssssssssssssssss

Mathematics

1 answer:

strojnjashka [21]2 years ago

4 0

Answer:

This answer would be c

Step-by-step explanation:

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"Large Sample Proportion Problem. Surveys were conducted in multiple countries and respondents were asked if they felt political

melomori [17]

Answer: 0.0158

Step-by-step explanation:

Given : The data for the United States is that out of 1,000 sampled, 470 indicated yes, they felt political news was reported fairly.

According to the given information we have,

Sample size : n= 1000

Sample proportion: $\hat{p}=\dfrac{470}{1000}=0.47$

The standard error for proportion is given by :-

$S.E.=\sqrt{\dfrac{\hat{p}(1-\hat{p})}{n}}$

$S.E.=\sqrt{\dfrac{0.47(1-0.47)}{1000}}$

$S.E.=\sqrt{0.0002491}$

$S.E.=0.0157829021412\approx0.0158$

[Rounded to the nearest four decimal places.]

Hence, the standard error for the confidence interval = 0.0158

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3 years ago

I love !!!!!!!!!!!!!!!!

Phantasy [73]

Answer:

tym

Step-by-step explanation:

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3 years ago

Please help with this question

yarga [219]

5654.87 Hope this helps!

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2 years ago

If you know that 1% of a population has a certain disease. A test was made. This test is positive in 95% of the people with the

LenKa [72]

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Step-by-step explanation:

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3 years ago

Determine whether a probability distribution is given. If a probability distribution is given, find its mean and standard deviat

drek231 [11]

Answer:

$E(X) = \sum_{i=1}^n X_i P(X_i) = 0*0.031 +1*0.156+ 2*0.313+3*0.313+ 4*0.156+ 5*0.031 = 2.5$

We can find the second moment given by:

$E(X^2) = \sum_{i=1}^n X^2_i P(X_i) = 0^2*0.031 +1^2*0.156+ 2^2*0.313+3^2*0.313+ 4^2*0.156+ 5^2*0.031 =7.496$

And we can calculate the variance with this formula:

$Var(X) =E(X^2) -[E(X)]^2 = 7.496 -(2.5)^2 = 1.246$

And the deviation is:

$Sd(X) = \sqrt{1.246}= 1.116$

Step-by-step explanation:

For this case we have the following probability distribution given:

X 0 1 2 3 4 5

P(X) 0.031 0.156 0.313 0.313 0.156 0.031

The expected value of a random variable X is the n-th moment about zero of a probability density function f(x) if X is continuous, or the weighted average for a discrete probability distribution, if X is discrete.

The variance of a random variable X represent the spread of the possible values of the variable. The variance of X is written as Var(X).

We can verify that:

$\sum_{i=1}^n P(X_i) = 1$

And $P(X_i) \geq 0, \forall x_i$

So then we have a probability distribution

We can calculate the expected value with the following formula:

$E(X) = \sum_{i=1}^n X_i P(X_i) = 0*0.031 +1*0.156+ 2*0.313+3*0.313+ 4*0.156+ 5*0.031 = 2.5$

We can find the second moment given by:

$E(X^2) = \sum_{i=1}^n X^2_i P(X_i) = 0^2*0.031 +1^2*0.156+ 2^2*0.313+3^2*0.313+ 4^2*0.156+ 5^2*0.031 =7.496$

And we can calculate the variance with this formula:

$Var(X) =E(X^2) -[E(X)]^2 = 7.496 -(2.5)^2 = 1.246$

And the deviation is:

$Sd(X) = \sqrt{1.246}= 1.116$

6 0

3 years ago