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vladimir2022 [97]
2 years ago
15

Juan and his friends played a game in which sticky darts were thrown onto a playing board. To start the game, Juan threw a dart

randomly onto the board. Describe the likelihood of getting each score. 8 points: 16 points: more than 0 points:
Mathematics
2 answers:
Inessa05 [86]2 years ago
6 0

Answer:

8 points:  

✔ unlikely

16 points:  

✔ impossible

more than 0 points:  

✔ likely

i did the test hope this helped

evablogger [386]2 years ago
4 0

Answer:

8 points:  unlikely

16 points:   impossible

more than 0 points:   likely

Step-by-step explanation:

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Step-by-step explanation:

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2 years ago
A random sample of 250 students at a university finds that these students take a mean of 15.2 credit hours per quarter with a st
Orlov [11]

Answer:

95% confidence interval for the mean credit hours taken by a student each quarter is [14.915 hours , 15.485 hours].

Step-by-step explanation:

We are given that a random sample of 250 students at a university finds that these students take a mean of 15.2 credit hours per quarter with a standard deviation of 2.3 credit hours.

Firstly, the pivotal quantity for 95% confidence interval for the population mean is given by;

                          P.Q. = \frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }  ~ t_n_-_1

where, \bar X = sample credit hours per quarter = 15.2 credit hours

             s = sample standard deviation = 2.3 credit hours

             n = sample of students = 250

             \mu = population mean credit hours per quarter

<em>Here for constructing 95% confidence interval we have used One-sample t test statistics as we know don't about population standard deviation.</em>

So, 95% confidence interval for the population mean, \mu is ;

P(-1.96 < t_2_4_9 < 1.96) = 0.95  {As the critical value of t at 249 degree of

                                        freedom are -1.96 & 1.96 with P = 2.5%}  

P(-1.96 < \frac{\bar X-\mu}{\frac{s}{\sqrt{n} } } < 1.96) = 0.95

P( -1.96 \times {\frac{s}{\sqrt{n} } } < {\bar X-\mu} < 1.96 \times {\frac{s}{\sqrt{n} } } ) = 0.95

P( \bar X-1.96 \times {\frac{s}{\sqrt{n} } } < \mu < \bar X+1.96 \times {\frac{s}{\sqrt{n} } } ) = 0.95

<u><em>95% confidence interval for</em></u> \mu = [ \bar X-1.96 \times {\frac{s}{\sqrt{n} } } , \bar X+1.96 \times {\frac{s}{\sqrt{n} } } ]

                  = [ 15.2-1.96 \times {\frac{2.3}{\sqrt{250} } } , 15.2+1.96 \times {\frac{2.3}{\sqrt{250} } } ]

                  = [14.915 hours , 15.485 hours]

Therefore, 95% confidence interval for the mean credit hours taken by a student each quarter is [14.915 hours , 15.485 hours].

<em>The interpretation of the above confidence interval is that we are 95% confident that the true mean credit hours taken by a student each quarter will be between 14.915 credit hours and 15.485 credit hours.</em>

8 0
1 year ago
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