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mr Goodwill [35]

3 years ago

8

A die is rolled. The set of equally-likely outcomes is {1, 2, 3, 4, 5, 6}. Find the probability of rolling a number greater than

4.

1 answer:

zalisa [80]3 years ago

6 0

Answer:

33.33 percent or 2:4

Step-by-step explanation:

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Find the distance between the two given points. Round to the nearest tenth.

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9.9

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A recent study examined the association between high blood pressure and increased risk of death from cardiovascular disease. The

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According to sources, the most probable answer to this query is that thee results seems to support a .05 probability. Meaning, this supports the hypothesis' claim. Thank you for your question. Please don't hesitate to ask in Brainly your queries.

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Gala2k [10]

A maybe I’m not sure

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Suppose 52% of the population has a college degree. If a random sample of size 563563 is selected, what is the probability that

amm1812

Answer:

The value is $P(| \^ p - p| < 0.05 ) = 0.9822$

Step-by-step explanation:

From the question we are told that

The population proportion is $p = 0.52$

The sample size is n = 563

Generally the population mean of the sampling distribution is mathematically represented as

$\mu_{x} = p = 0.52$

Generally the standard deviation of the sampling distribution is mathematically evaluated as

$\sigma = \sqrt{\frac{ p(1- p)}{n} }$

=> $\sigma = \sqrt{\frac{ 0.52 (1- 0.52 )}{563} }$

=> $\sigma = 0.02106$

Generally the probability that the proportion of persons with a college degree will differ from the population proportion by less than 5% is mathematically represented as

$P(| \^ p - p| < 0.05 ) = P( - (0.05 - 0.52 ) < \^ p < (0.05 + 0.52 ))$

Here $\^ p$ is the sample proportion of persons with a college degree.

So

$P( - (0.05 - 0.52 ) < \^ p < (0.05 + 0.52 )) = P(\frac{[[0.05 -0.52]]- 0.52}{0.02106} < \frac{[\^p - p] - p}{\sigma } < \frac{[[0.05 -0.52]] + 0.52}{0.02106} )$

Here

$\frac{[\^p - p] - p}{\sigma } = Z (The\ standardized \ value \ of\ (\^ p - p))$

=> $P( - (0.05 - 0.52 ) < \^ p < (0.05 + 0.52 )) = P[\frac{-0.47 - 0.52}{0.02106 } < Z < \frac{-0.47 + 0.52}{0.02106 }]$

=> $P( - (0.05 - 0.52 ) < \^ p < (0.05 + 0.52 )) = P[ -2.37 < Z < 2.37 ]$

=> $P( - (0.05 - 0.52 ) < \^ p < (0.05 + 0.52 )) = P(Z < 2.37 ) - P(Z < -2.37 )$

From the z-table the probability of (Z < 2.37 ) and (Z < -2.37 ) is

$P(Z < 2.37 ) = 0.9911$

and

$P(Z < - 2.37 ) = 0.0089$

So

=> $P( - (0.05 - 0.52 ) < \^ p < (0.05 + 0.52 )) =0.9911-0.0089$

=> $P( - (0.05 - 0.52 ) < \^ p < (0.05 + 0.52 )) = 0.9822$

=> $P(| \^ p - p| < 0.05 ) = 0.9822$

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3 years ago

Consider the conditional statement: "If I skip dinner, then I won't sleep well tonight."

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