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Leno4ka [110]
3 years ago
10

The sum of the first n positive odd numbers is square.

Mathematics
1 answer:
ololo11 [35]3 years ago
3 0

Answer:

See explanation below

Step-by-step explanation:

All the odd numbers can be written as 2n-1 where n is an integer.

The sum (S) of the first n positive odd numbers can be written as:

S=1+3+5+.........+(2n-5)+(2n-3)+(2n-1)

If we change the order to a decreasing order we would write it this way:

S=(2n-1)+(2n-3)+(2n-5)+.........+5+3+1

Let's sum up both equations, we are going to pair the first number of our first equation with the first number of our second equation, we can see that each pair sums up 2n ( let's see this: (2n-1 + 1 =2n, 2n-3 + 3 = 2n, 2n-5 + 5 = 2n...)) therefore we would get n pairs that sum up 2n.

2S=n(2n)\\2S=2n^{2} \\S=n^{2}

Therefore, the sum of the first n positive odd numbers is square.

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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:

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

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}}

A statistician calculates that 8% of Americans own a Rolls Royce.

This means that p = 0.08

Sample of 595:

This means that n = 595

Mean and standard deviation:

\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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