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3 years ago

Find the SUm of the infinite series.

Mathematics

1 answer:

givi [52]3 years ago

7 0

I'll do problem 13 to get you started.

The expression $4\left(\frac{2^n}{7^n}\right)$ is the same as $4\left(\frac{2}{7}\right)^n$

Then we can do a bit of algebra like so to change that n into n-1

$4\left(\frac{2}{7}\right)^n\\\\4\left(\frac{2}{7}\right)^n*1\\\\4\left(\frac{2}{7}\right)^n*\left(\frac{2}{7}\right)^{0}\\\\4\left(\frac{2}{7}\right)^n*\left(\frac{2}{7}\right)^{1-1}\\\\4\left(\frac{2}{7}\right)^n*\left(\frac{2}{7}\right)^{1}*\left(\frac{2}{7}\right)^{-1}\\\\4*\left(\frac{2}{7}\right)^{1}\left(\frac{2}{7}\right)^n*\left(\frac{2}{7}\right)^{-1}\\\\\frac{8}{7}\left(\frac{2}{7}\right)^{n-1}\\\\$

This is so we can get the expression in a(r)^(n-1) form

a = 8/7 is the first term of the geometric sequence
r = 2/7 is the common ratio

Note that -1 < 2/7 < 1, which satisfies the condition that -1 < r < 1. This means the infinite sum converges to some single finite value (rather than diverge to positive or negative infinity).

We'll plug those a and r values into the infinite geometric sum formula below

S = a/(1-r)

S = (8/7)/(1 - 2/7)

S = (8/7)/(5/7)

S = (8/7)*(7/5)

S = 8/5

S = 1.6

------------------------

Answer in fraction form = 8/5

Answer in decimal form = 1.6

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Assume the average height for American women is 64 inches with a standard deviation of 2 inches. A sorority on campus wonders if

satela [25.4K]

Answer:

a) s = 0.4

b) Z = 2.5

c) 99th percentile.

d) The larger sample size would lead to a smaller margin of error, which would lead to a higher z-score and a increased percentile.

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.

Assume the average height for American women is 64 inches with a standard deviation of 2 inches

This means that $\mu = 64, \sigma = 2$

A) Calculate the standard error for the distribution of means.

Sample of 25 means that $n = 25$ , so $s = \frac{2}{\sqrt{25}} = 0.4$ .

B) Calculate the z statistic for the sorority group.

Sample mean of 65 means that $X = 65$ .

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

By the Central Limit Theorem

$Z = \frac{X - \mu}{0.4}$

$Z = \frac{65 - 64}{0.4}$

$Z = 2.5$

C) What is the approximate percentile for this sample? Enter as a whole number.

Z = 2.5 has a p-value of 0.9938, so 0.99*100 = 99th percentile.

D) If the sorority actually had 36 members (still with an average of 65 inches), would you expect the percentile value to increase or decrease? why?

The larger sample size would lead to a smaller margin of error, which would lead to a higher z-score and a increased percentile.

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3 years ago

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posledela

Answer:

No Solutions

Step-by-step explanation:

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3 years ago

A survey of 76 commercial airline flights of under 2 hours resulted in a sample average late time for a flight of 2.55 minutes.

nika2105 [10]

Answer:

The best point of estimate for the true mean is:

$\hat \mu = \bar X = 2.55$

$2.55-1.96\frac{12}{\sqrt{76}}=-0.148$

$2.55+1.96\frac{12}{\sqrt{76}}=5.248$

Since the time can't be negative a good approximation for the confidence interval would be (0,5.248) minutes. The interval are tellling to us that at 95% of confidence the average late time is lower than 5.248 minutes.

Step-by-step explanation:

Information given

$\bar X=2.55$ represent the sample mean for the late time for a flight

$\mu$ population mean

$\sigma=12$ represent the population deviation

n=76 represent the sample size

Confidence interval

The best point of estimate for the true mean is:

$\hat \mu = \bar X = 2.55$

The confidence interval for the true mean is given by:

$\bar X \pm z_{\alpha/2}\frac{\sigma}{\sqrt{n}}$ (1)

The Confidence level given is 0.95 or 95%, th significance would be $\alpha=0.05$ and $\alpha/2 =0.025$ . If we look in the normal distribution a quantile that accumulates 0.025 of the area on each tail we got $z_{\alpha/2}=1.96$

Replacing we got:

$2.55-1.96\frac{12}{\sqrt{76}}=-0.148$

$2.55+1.96\frac{12}{\sqrt{76}}=5.248$

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Scilla [17]

Try making a substitution:

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