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tankabanditka [31]

3 years ago

10

Pls only answer if you know it:) The numbers are 19, 36, 24, 40, 17, 43, 30, 31, 37, 28, 45, 32, 16. What are the first quartile

, median and third quartile to this data set?
a) 21.5, 31, 37
b) 21.5, 31, 38.5
c) 24, 31, 38.5
d) 24, 31, 37

2 answers:

charle [14.2K]3 years ago

6 0

Answer:

The answer is B

Step-by-step explanation:

You order them from least to greatest- 16, 17, 19, 24, 28, 30, 31, 32, 36, 37, 40, 43, 45.

Then you find the middle number or the median- 31

Then you go to the left of the medain and get the first quartile- 19, 24 (You have to find the mean of these two numbers- 21.5)

Then you go to the right of the medain and get the third quartile- 37, 40 (Again you must find the mean of these two numbers- 38.5)

Last you would put them in the order needed.

Hope this helps! :)

NNADVOKAT [17]3 years ago

5 0

Answer:

The answer is b

Step-by-step explanation:

The reason being that I just did it myself on a lesson and got it right.

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2 years ago

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Suppose the weights of Farmer Carl's potatoes are normally distributed with a mean of 8.0 ounces and a standard deviation of 1.1

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a) 0.9959 = 99.59% probability that the mean weight is less than 9.3 ounces

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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 zscore 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 pvalue, we get the probability that the value of the measure is greater than X.

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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.

Mean of 8.0 ounces and a standard deviation of 1.1 ounces.

This means that $\mu = 8, \sigma = 1.1$

(a) If 5 potatoes are randomly selected, find the probability that the mean weight is less than 9.3 ounces?

$n = 5$ means that $s = \frac{1.1}{\sqrt{5}} = 0.4919$

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

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

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$Z = 2.64$

$Z = 2.64$ has a pvalue of 0.9959

0.9959 = 99.59% probability that the mean weight is less than 9.3 ounces

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Answer:

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Given Jenny has saved $\$3,800$ over the last 12 months.

She saved either $\$250$ or $\$350$ a month.

Let the number of months to save $\$250$ be x

Let the number of months to save $\$350$ be y

Total period is 12 month.

$x+y=12 \ \ \ \ equation\ 1$

Jenny has saved $3,800 over the last 12 months.

$250x+350y=3800$

Dividing both side from 50 we get,

$\frac{250}{50}x+\frac{350}{50}y= \frac{3800}{50}$

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Now Multiplying equation 1 by 5 we get,

$5\times(x+y)=5\times 12\\5x+5y=60\ \ \ \ equation \ 3$

Now Subtracting Equation 3 by equation 2 we get,

$(5x+7y=76)-(5x+5y=60)\\2y=16\\y=\frac{16}{2}= 8$

Substituting the value of y in equation 1 we get,

$x+y=12\\x+8=12\\x=12-8\\x=4$

∴ Jenny save 4 Months of $\$ 250$ and 8 Months of $\$ 350$ in a 12 month duration

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