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Maru [420]
2 years ago
13

In the year 2012, 12\%12%12, percent of the residents of the US moved to a different residence. Suppose that several census work

ers took random samples of n=150n=150n, equals, 150 residents from the population and computed the proportion \hat p
p
^
​
p, with, hat, on top of residents in each sample who had moved in 2012.
What would be the shape of the sampling distribution of \hat p
p
^
​
p, with, hat, on top?
Mathematics
1 answer:
yan [13]2 years ago
7 0

Answer: approximately normal (c) trust me-khan academy

Step-by-step explanation:

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In this equation, we can start by understanding that "x" has a value of 8, as given in the ordered pair. When multiplied by 5, this leads to "40 - 2y = 30". Next, we can subtract 40 from both sides of the equation. This leads us to a value of "-2y = -10". The next step would be to divide both sides by -2 as a way of isolating "y", which leads us to a final value of "y = 5". The final ordered pair would be (8,5).

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An article presents a new method for timing traffic signals in heavily traveled intersections. The effectiveness of the new meth
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With 89.9% we can say that the mean improvement is between 583.1 and 728.1 vehicles per hour.

Step-by-step explanation:

We are given that the effectiveness of the new method was evaluated in a simulation study. In 50 simulations, the mean improvement in traffic flow in a particular intersection was 655.6 vehicles per hour, with a standard deviation of 311.7 vehicles per hour.

A traffic engineer states that the mean improvement is between 583.1 and 728.1 vehicles per hour.

<em>Let </em>\bar X<em> = sample mean improvement</em>

The z-score probability distribution for sample mean is given by;

            Z =  \frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } } ~ N(0,1)

where, \mu = mean improvement = 655.6 vehicles per hour

            \sigma = standard deviation = 311.7 vehicles per hour

            n = sample of simulations = 50

The Z-score measures how many standard deviations the measure is away from the mean. After finding the Z-score, we look at the z-score table and find the p-value (area) associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X.

Now, Probability that the mean improvement is between 583.1 and 728.1 vehicles per hour is given by = P(583.1 < \bar X < 728.1) = P(\bar X < 728.1) - P(\bar X \leq 583.1)

  P(\bar X < 728.1) = P( \frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } } < \frac{728.1-655.6}{\frac{311.7}{\sqrt{50} } } ) = P(Z < 1.64) = 0.9495

  P(\bar X \leq 583.1) = P( \frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } } \leq \frac{583.1-655.6}{\frac{311.7}{\sqrt{50} } } ) = P(Z \leq -1.64) = 1 - P(Z < 1.64)

                                                            = 1 - 0.9495 = 0.0505                      

<em />

<em>So, in the z table the P(Z </em>\leq<em> x) or P(Z < x) is given. So, the above probability is calculated by looking at the value of x = 1.64 in the z table which has an area of 0.9495.</em>

Therefore, P(583.1 < \bar X < 728.1) = 0.9495 - 0.0505 = 0.899 or 89.9%

Hence, with 89.9% we can say that the mean improvement is between 583.1 and 728.1 vehicles per hour.

6 0
3 years ago
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