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Rasek [7]
3 years ago
14

Construct a sample (with at least two different values in the set) of 4 measurements whose mean, median, and mode are equal. If

this is not possible, indicate "Cannot create sample".
Mathematics
1 answer:
KiRa [710]3 years ago
3 0

Answer:

The first set 4 , 6 , 6 , 8

The second set 2 , 6 , 6 , 10

Step-by-step explanation:

Mean is the average of the values in a set. This can be found by summing all the numbers and dividing by the total number. The median is the middle number when the values are listed in order of increasing size. The mode is the number(s) that appear the most often in the set.

The first set 4 , 6 , 6 , 8

mean = (4 + 6 + 6 + 8 )/4 = 6

median = (6+6)/2 = 6

Mode = 6

The second set 2 , 6 , 6 , 10

mean = (2 + 6 + 6 + 10)/4 = 6

median = (6+6)/2 = 6

Mode = 6

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Answer:

x = -1/7   or  -0.14285 - in decimal form

Step-by-step explanation:

8x + 5 = -6x + 3

8x + 6x = 3 - 5       { change side }

14x = -2                  { basic addition and subtraction }

x = -2/14                 { changing sides }

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What is 64 miles per hour = how much feet per hour
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Since one mile is 5280 feet,
64 miles is 337920 feet.

337920 Feet per hour. 
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3 years ago
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A shipment To a warehouse consists of 4500 MP3 players the manager chooses a random sample of 50 MP3 players and finds that 7 ar
riadik2000 [5.3K]

Answer:

<u>630</u> MP3 players are likely to be defective.

Step-by-step explanation:

Given:

A shipment to a warehouse consists of 4500 MP3 players the manager chooses a random sample of 50 MP3 players and finds that 7 are defective.

Now, to find the MP3 players which are likely to be defective.

Let the defective MP3 players be x.

Total MP3 players = 4500.

In sample of 50 MP3 players 7 are defective.

Now, setting the proportion:

If, 50 is equivalent to 7.

Then, 4500 is equivalent to x.

Now, to solve using cross multiplication method:

\frac{50}{7} =\frac{4500}{x}

<em>By cross multiplying we get:</em>

<em />50x=31500<em />

<em>Dividing both sides by 50 we get:</em>

x=630.

Therefore, 630 MP3 players are likely to be defective.

5 0
3 years ago
The mean points obtained in an aptitude examination is 159 points with a standard deviation of 13 points. What is the probabilit
Korolek [52]

Answer:

0.4514 = 45.14% probability that the mean of the sample would differ from the population mean by less than 1 point if 60 exams are sampled

Step-by-step explanation:

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

Normal probability distribution:

Problems of normally distributed samples are 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.

Central limit theorem:

The Central Limit Theorem estabilishes that, for a random variable X, with mean \mu and standard deviation \sigma, a large sample size can be approximated to a normal distribution with mean \mu and standard deviation s = \frac{\sigma}{\sqrt{n}}

In this problem, we have that:

\mu = 159, \sigma = 13, n = 60, s = \frac{13}{\sqrt{60}} = 1.68

What is the probability that the mean of the sample would differ from the population mean by less than 1 point if 60 exams are sampled?

This is the pvalue of Z when X = 159+1 = 160 subtracted by the pvalue of Z when X = 159-1 = 158. So

X = 160

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

By the Central Limit Theorem

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

Z = \frac{160 - 159}{1.68}

Z = 0.6

Z = 0.6 has a pvalue of 0.7257

X = 150

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

Z = \frac{158 - 159}{1.68}

Z = -0.6

Z = -0.6 has a pvalue of 0.2743

0.7257 - 0.2743 = 0.4514

0.4514 = 45.14% probability that the mean of the sample would differ from the population mean by less than 1 point if 60 exams are sampled

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