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Alexus [3.1K]
2 years ago
8

A recent study of 50 U.S. chess players details such things as the number of years the players have been active and the chess ra

tings of the 50 players. (A chess rating is a number, with a higher number indicating greater expertise.) The chess rating data for the sample of 50 players are summarized in the following histogram.
Mathematics
1 answer:
Anvisha [2.4K]2 years ago
5 0

Using the given histogram, we have that 90% of the players in the sample have scores that are less than 2300.

<h3>What is the missing information?</h3>

This problem is incomplete, but researching it on a search engine, we have that the histogram states as follows:

  • 6 of these players have scores between 300 and 700.
  • 6 of these players have scores between 700 and 1100.
  • 13 of these players have scores between 1100 and 1500.
  • 14 of these players have scores between 1500 and 1900.
  • 6 of these players have scores between 1900 and 1300.
  • 5 of these players have scores between 2300 and 2700.

It asks the proportion of players in the sample with <u>scores that are less than 2300</u>.

<h3>What is an histogram?</h3>

An histogram is a graph that shows the <u>number of times each element of x was observed.</u>

Above, we already gave the amounts presented by histogram, and we have that 45 out of 50 players have <u>scores that are less than 2300</u>, hence the proportion is given by:

p = 45/50 = 0.9.

More can be learned about histograms at brainly.com/question/25836450

#SPJ1

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10.
Kay [80]

Answer:

\frac{(82)(2.4)^2}{104.139} \leq \sigma^2 \leq \frac{(82)(2.4)^2}{62.132}

4.535 \leq \sigma^2 \leq 7.602

Now we just take square root on both sides of the interval and we got:

2.2 \leq \sigma \leq 2.8

And the best option would be:

A.  2.2 < σ < 2.8

Step-by-step explanation:

Information provided

\bar X=32.1 represent the sample mean

\mu population mean  

s=2.4 represent the sample standard deviation

n=83 represent the sample size  

Confidence interval

The confidence interval for the population variance is given by the following formula:

\frac{(n-1)s^2}{\chi^2_{\alpha/2}} \leq \sigma^2 \leq \frac{(n-1)s^2}{\chi^2_{1-\alpha/2}}

The degrees of freedom are given by:

df=n-1=83-1=82

The Confidence is given by 0.90 or 90%, the value of \alpha=0.1 and \alpha/2 =0.05, the critical values for this case are:

\chi^2_{\alpha/2}=62.132

\chi^2_{1- \alpha/2}=104.139

And replacing into the formula for the interval we got:

\frac{(82)(2.4)^2}{104.139} \leq \sigma^2 \leq \frac{(82)(2.4)^2}{62.132}

4.535 \leq \sigma^2 \leq 7.602

Now we just take square root on both sides of the interval and we got:

2.2 \leq \sigma \leq 2.8

And the best option would be:

A.  2.2 < σ < 2.8

3 0
3 years ago
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Free_Kalibri [48]

Answer:

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3 years ago
16^2+8^2= <br> Show what you did
Slav-nsk [51]

Answer:

Step-by-step explanation:

(16*16)+(8*8)

16. 8

16. 8

—————-

96. 64

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—————-

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3 years ago
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Answer:

Step-by-step explanation:

Because 48 and 89 have the same sign, that is, the negative sign, their sum takes on that sign:

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