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svetoff [14.1K]
4 years ago
13

You have just completed an experiment measuring the length (in mm) of 12 soybean plants after 3 days of sprouting. The measureme

nts are listed below:
53 47 51 54 43 39 61 57 55 46 44 43

If you doubled every measurement in this dataset (e.g., 106, 94, 102, ...), what would the new IQR equal? (Think about the SD too!)
Mathematics
1 answer:
sweet-ann [11.9K]4 years ago
4 0

Answer:

The new IQR is 22.    

Step-by-step explanation:

We are given the following data of  length (in mm) of 12 soybean plants after 3 days of sprouting.

53, 47, 51, 54, 43, 39, 61, 57, 55, 46, 44, 43

Sorted data:

39, 43, 43, 44, 46, 47, 51, 53, 54, 55, 57, 61

Formula:

IQR = Q_3 - Q_1\\Q_3 = \text{upper median},\\Q_1 = \text{ lower median}

Median:\\\text{If n is odd, then}\\\\Median = \displaystyle\frac{n+1}{2}th ~term \\\\\text{If n is even, then}\\\\Median = \displaystyle\frac{\frac{n}{2}th~term + (\frac{n}{2}+1)th~term}{2}  

Median ==\dfrac{6^{th}+7^{th}}{2} \dfrac{47+51}{2} = 49

Q_1 =\dfrac{3^{rd}+4^{th}}{2} = \dfrac{43 + 44}{2} = 43.5\\\\Q_3 =\dfrac{9^{th}+10^{th}}{2}= \frac{54 + 55}{2} = 54.5

IQR = Q_3 -Q_1 =54.5-43.5= 11

If every measurement is doubled, then, the IQR will also double itself.

Thus,

New IQR =

2\times \text{IQR}\\=2\TIMES 11\\=22

Thus, the new IQR is 22.

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Answer:

Step-by-step explanation:

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Step-by-step explanation:

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The full answer is below.

Step-by-step explanation:

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3 years ago
A fitness center is interested in finding a 90% confidence interval for the mean number of days per week that Americans who are
Blababa [14]

Answer:

A. Normal

B. Between 2.078 and 2.722 visits.

C. About 90 percent of these confidence intervals will contain the true population mean number of visits per week and about 10 percent will not contain the true population mean number of visits per week.

Step-by-step explanation:

Central Limit Theorem

The Central Limit Theorem estabilishes 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}}

x% confidence interval:

A confidence interval is built from a sample, has bounds a and b, and has a confidence level of x%. It means that we are x% confident that the population mean is between a and b.

Question A:

By the Central Limit Theorem, normal.

Question B:

We have that to find our \alpha level, that is the subtraction of 1 by the confidence interval divided by 2. So:

\alpha = \frac{1 - 0.9}{2} = 0.05

Now, we have to find z in the Ztable as such z has a pvalue of 1 - \alpha.

That is z with a pvalue of 1 - 0.05 = 0.95, so Z = 1.645.

Now, find the margin of error M as such

M = z\frac{\sigma}{\sqrt{n}}

In which \sigma is the standard deviation of the population and n is the size of the sample.

M = 1.645\frac{2.9}{\sqrt{219}} = 0.322

The lower end of the interval is the sample mean subtracted by M. So it is 2.4 - 0.322 = 2.078 visits

The upper end of the interval is the sample mean added to M. So it is 2.4 + 0.322 = 2.722 visits.

Between 2.078 and 2.722 visits.

Question c:

90% confidence level, so 90% will contain the true population mean, 100 - 90 = 10% wont.

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