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Vikentia [17]
3 years ago
12

The points scored by a basketball team in its last 6 gamesare shown. Use these data for 7 and 8.Points Scored73 77 85 84 37 1157

. Find the mean score and the median score.MeanMedians. Which measure better describes the typical number of points scored?Explain.

Mathematics
1 answer:
Mrac [35]3 years ago
4 0
The median is 73 and the mean is 60.6 (rounded to the nearest tenth). Mean better describes the typical number of points scored.
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What is 694 divided by 41 and what is the remainder please help thank you
Studentka2010 [4]
16... Remainder of 38

41 x 16 = 656 therefore 38 left over
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What is the <img src="https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7B7%7D" id="TexFormula1" title="\sqrt[3]{7}" alt="\sqrt[3]{7}" alig
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The answer you want will be 1.91293\dots

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The ratio of lollipops to pieces of bubble gum at the party is 2 to 3. There are 24 lollipops. ​ ​What is the total number of lo
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Answer: 40

Step-by-step explanation: first make the ratio into a fraction, 2/3

then make 24 into a fraction with the numerator as a variable and the denominator as 24, x/24

then cross multiply 2 and 24.

2x24=48

then divide 48 by 3, 16.

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3 years ago
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What are the factors of 30 that have a sum of -11
Fed [463]

Answer:

Why is 42 the sum of all factors of 30, excluding 30? The factors of the number 30 are: 1, 2, 3, 5, 6, 10, 15, 30.

Step-by-step explanation:

8 0
3 years ago
The mean number of words per minute (WPM) read by sixth graders is 8888 with a standard deviation of 1414 WPM. If 137137 sixth g
Bingel [31]

Noticing that there is a pattern of repetition in the question (the numbers are repeated twice), we are assuming that the mean number of words per minute is 88, the standard deviation is of 14 WPM, as well as the number of sixth graders is 137, and that there is a need to estimate the probability that the sample mean would be greater than 89.87.

Answer:

"The probability that the sample mean would be greater than 89.87 WPM" is about \\ P(z>1.56) = 0.0594.

Step-by-step explanation:

This is a problem of the <em>distribution of sample means</em>. Roughly speaking, we have the probability distribution of samples obtained from the same population. Each sample mean is an estimation of the population mean, and we know that this distribution behaves <em>normally</em> for samples sizes equal or greater than 30 \\ n \geq 30. Mathematically

\\ \overline{X} \sim N(\mu, \frac{\sigma}{\sqrt{n}}) [1]

In words, the latter distribution has a mean that equals the population mean, and a standard deviation that also equals the population standard deviation divided by the square root of the sample size.

Moreover, we know that the variable Z follows a <em>normal standard distribution</em>, i.e., a normal distribution that has a population mean \\ \mu = 0 and a population standard deviation \\ \sigma = 1.

\\ Z = \frac{\overline{X} - \mu}{\frac{\sigma}{\sqrt{n}}} [2]

From the question, we know that

  • The population mean is \\ \mu = 88 WPM
  • The population standard deviation is \\ \sigma = 14 WPM

We also know the size of the sample for this case: \\ n = 137 sixth graders.

We need to estimate the probability that a sample mean being greater than \\ \overline{X} = 89.87 WPM in the <em>distribution of sample means</em>. We can use the formula [2] to find this question.

The probability that the sample mean would be greater than 89.87 WPM

\\ Z = \frac{\overline{X} - \mu}{\frac{\sigma}{\sqrt{n}}}

\\ Z = \frac{89.87 - 88}{\frac{14}{\sqrt{137}}}

\\ Z = \frac{1.87}{\frac{14}{\sqrt{137}}}

\\ Z = 1.5634 \approx 1.56

This is a <em>standardized value </em> and it tells us that the sample with mean 89.87 is 1.56<em> standard deviations</em> <em>above</em> the mean of the sampling distribution.

We can consult the probability of P(z<1.56) in any <em>cumulative</em> <em>standard normal table</em> available in Statistics books or on the Internet. Of course, this probability is the same that \\ P(\overline{X} < 89.87). Then

\\ P(z

However, we are looking for P(z>1.56), which is the <em>complement probability</em> of the previous probability. Therefore

\\ P(z>1.56) = 1 - P(z

\\ P(z>1.56) = P(\overline{X}>89.87) = 0.0594

Thus, "The probability that the sample mean would be greater than 89.87 WPM" is about \\ P(z>1.56) = 0.0594.

5 0
3 years ago
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