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olchik [2.2K]
3 years ago
8

A 78​% of U.S. adults think that political correctness is a problem in America today. You randomly select six U.S. adults and as

k them whether they think that political correctness is a problem in America today. The random variable represents the number of U.S. adults who think that political correctness is a problem in America today. Answer the questions below.Find the mean of the binomial distribution ​(Round to the nearest tenth as​ needed.)Find the variance of the binomial distribution. (Round to the nearest tenth as​ needed.)Find the standard deviation of the binomial distribution. (Round to the nearest tenth as​ needed.)Most samples of 6 adults would differ from the mean by no more than nothing. ​(Type integers or decimals rounded to the nearest tenth as​ needed.)
Mathematics
1 answer:
Contact [7]3 years ago
8 0

Answer:

Step-by-step explanation:

Let x be a random variable representing the number of U.S. adults who think that political correctness is a problem in America today. Since it is a binomial probability distribution, the probability if success, p = 78/100 = 0.78

The probability of failure, q = 1 - p = 1 - 0.78 = 0.22

Number of samples = 6

Mean = np = 6 × 0.78 = 4.7

Variance = npq = 6 × 0.78 × 0.22 = 1.0

Standard deviation = √variance = 1.0

Most samples of 6 adults would differ from the mean by no more than 1

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Find the area [ a = bh ] of both then subtract.

16 * 20 = 320

11 * 14 = 154

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Best of Luck!

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50 pts!!!! Luciana's laptop has 3,000 pictures. The size of the pictures is skewed to the right, with a mean of 3.7MB and a stan
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Answer:

<u></u>

  • <u>Part A: No, you cannot.</u>

<u></u>

  • <u>Part B: 0.4491</u>

Explanation:

<u>Part A:</u>

The mean of samples of symmetrical (bell shaped) distributions follow a normal distribution pattern.

Thus, for symmetrical distributions you can use the z-score tables to make calculations that permit calculate probabilities for particular values.

For <em>skewed </em>distributions, in general, the samples do not follow a normal distribution pattern, except that the samples are large.

Since <em>a sample of 20 pictures</em> is not large enough, the answer to this question is negative: <em>you cannot accurately calculate the probability that the mean picture size is more than 3.8MB for an SRS (skewed right sample) of 20 pictures.</em>

<u>Part B.</u>

For large samples, the<em> Central Limit Theorem</em> will let you work with samples from skewed distributions.

Although the distribution of a population is skewed, the <em>Central Limit Theorem</em> states that  large samples follow a normal distribution shape.

It is accepted that samples of 30 data is large enough to use the <em>Central Limit Theorem.</em>

Hence, you can use the z-core tables for standard normal distributions to calculate the probabilities for a <em>random sample of 60 pictures</em> instead of 20.

The z-score is calculated with the formula:

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Here, mean = 3.7MB, value = 3.8MB, and standard deviation = 0.78MB.

Thus:

         z=(3.7M-3.8MB)/(0.78MB)=-0.128

Then, you must use the z-score table to find the probability that the z-score is greater than - 0.128.

There are tables that show the cumulative probability in the right end and tables that show the cumulative probability in the left end of a normal standard distribution .

The probability that a z-score is greater than -0.128 is taken directly from a table with the cumulative probability in the left end. It is 0.4491.

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Step-by-step explanation:

<em>Hi</em><em>!</em><em>!</em>

<em> </em><em>It's</em><em> </em><em>simple</em><em>, </em><em>you</em><em> </em><em>can</em><em> </em><em>do</em><em> </em><em>it</em><em> </em><em>easily</em><em>. </em>

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<em>now</em><em>,</em>

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<em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em>=</em><em> </em><em>1</em><em>0</em><em> </em><em>/</em><em>5</em><em>5</em>

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