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1 year ago

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103-2-5 11. Pond the coordinates of the point. Pohance that lonel make (b) : (a) equidistant from (4.-6) and the origin; as well

as on (b) equidistant from (3, 8) and (-2,5) on the v-axis (c) /*+ 3y = 7

1 answer:

Bad White [126]1 year ago

6 0

Since the point is equidistant from (3,8) and (-2,5) on the y-axis, we are talking about the middle point, so:

$\begin{gathered} Let \\ (x1,y1)=(3,8) \\ (x2,y2)=(-2,5) \\ mp=(xm,ym) \\ xm=\frac{x1+x2}{2}=\frac{3-2}{2}=\frac{1}{2}=0.5 \\ ym=\frac{y1+y2}{2}=\frac{8+5}{2}=\frac{13}{2}=6.5 \end{gathered}$

Therefore the coordinates are:

(0.5,6.5)

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hichkok12 [17]

Answer:

0.007 = 0.7% probability that the proportion of Rolls Royce owners in a sample of 595 Americans would differ from the population proportion by more than 3%

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal Probability Distribution:

Problems of normal distributions can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the z-score of a measure X is given by:

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The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value 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. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem establishes that, for a normally distributed random variable X, with mean $\mu$ and standard deviation $\sigma$ , the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$ .

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

For a proportion p in a sample of size n, the sampling distribution of the sample proportion will be approximately normal with mean $\mu = p$ and standard deviation $s = \sqrt{\frac{p(1-p)}{n}}$

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This means that $p = 0.08$

Sample of 595:

This means that $n = 595$

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$\mu = p = 0.08$

$s = \sqrt{\frac{p(1-p)}{n}} = \sqrt{\frac{0.08*0.92}{595}} = 0.0111$

What is the probability that the proportion of Rolls Royce owners in a sample of 595 Americans would differ from the population proportion by more than 3%?

Proportion above 8% + 3% = 11% or below 8% - 3% = 5%. Since the normal distribution is symmetric, these probabilities are equal, and so we find one of them and multiply by 2.

Probability the proportion is less than 5%:

P-value of Z when X = 0.05. So

$Z = \frac{X - \mu}{\sigma}$

By the Central Limit Theorem

$Z = \frac{X - \mu}{s}$

$Z = \frac{0.05 - 0.08}{0.0111}$

$Z = -2.7$

$Z = -2.7$ has a p-value of 0.0035

2*0.0035 = 0.0070

0.007 = 0.7% probability that the proportion of Rolls Royce owners in a sample of 595 Americans would differ from the population proportion by more than 3%

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3 years ago

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Answer:

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