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Andreas93 [3]
2 years ago
9

Amy, Bess, Cassie, Diana, and Ellen are a basketball team. Two of them will be guards. Assuming they all have an equal chance of

becoming guards, what is the probability that Diana will be a guards. which outcome (or outcomes) of the sample space compose the event? Describe the probability Diana becoming a guard as impossiable, unlikey, neither likey nor unlikey, likey or certain. Justify your answ
Mathematics
2 answers:
Vinvika [58]2 years ago
5 0
There are 5 people who can be guards, 2 will be guards. Since each person has the same chance of becoming a guard, each person has a 2/5 chance of being a guard which makes it unlikely for Diana to become a guard.
alexgriva [62]2 years ago
5 0

Answer: 1: The sample space for this problem is {Amy, Bess, Cassie, Diana, Ellen} 2: 10 outcomes 3: 2/5 or 0.4 4: The probability of Diana becoming a guard is possible since her chance of being chosen is 40%.

Step-by-step explanation: The explanations are on every problem except for 2 so here it is. The outcomes of the sample problem are Diana and Ellen, Diana and Cassie, Diana and Amy, Diana and Bess, Amy and Bess, Amy and Cassie, Amy and Ellen, Cassie and Bess, Cassie and Ellen, Ellen and Bess which adds up to ten outcomes since two people can be chosen from the group of five.

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You are dealt two cards successively without replacement from a standard deck of 52 playing cards. Find the probability that the
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Answer:

(D) 0.006

Step-by-step explanation:

Total number of cards :52

Please note that, all cards have a the possibility of appearing 4 times.

Hence total possible number of a '2' is 4 cards and so it is also for a '10'

Having this Understanding, let's solve the question properly.

The probability that the FIRST CARD is 2 = 4/52

Probability that the second card without replacement is a 10 = 4 / 51

P( 1st two and 2nd four)

4/52 * 4/51 = 4/663

= 0.0060332

Rounding to 3 decimal places = 0.006

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3 years ago
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Rita has a loan of 40,000. This loan has a simple interest rate of 6% per year. What is the amount of interest that rita will be
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Step-by-step explanation:

7 0
3 years ago
A researcher examines 27 water samples for iron concentration. The mean iron concentration for the sample data is 0.802 cc/cubic
Delvig [45]

Answer:

90% confidence interval for the population mean iron concentration is [0.771 , 0.832].

Step-by-step explanation:

We are given that a researcher examines 27 water samples for iron concentration. The mean iron concentration for the sample data is 0.802 cc/cubic meter with a standard deviation of 0.093.

Firstly, the pivotal quantity for 90% confidence interval for the population mean iron concentration is given by;

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

where, \bar X = sample mean iron concentration = 0.802 cc/cubic meter

             s = sample standard deviation = 0.093

             n = number of water samples = 27

             \mu = population mean

<em>Here for constructing 90% confidence interval we have used t statistics because we don't know about population standard deviation.</em>

So, 90% confidence interval for the population mean, \mu is ;

P(-1.706 < t_2_6 < 1.706) = 0.90  {As the critical value of t at 26 degree of

                                                 freedom are -1.706 & 1.706 with P = 5%}

P(-1.706 < \frac{\bar X - \mu}{\frac{s}{\sqrt{n} } } < 1.706) = 0.90

P( -1.706 \times {\frac{s}{\sqrt{n} } } < {\bar X - \mu} < 1.706 \times {\frac{s}{\sqrt{n} } } ) = 0.90

P( \bar X -1.706 \times {\frac{s}{\sqrt{n} } < \mu < \bar X +1.706 \times {\frac{s}{\sqrt{n} } ) = 0.90

<u>90% confidence interval for</u> \mu = [ \bar X -1.706 \times {\frac{s}{\sqrt{n} } , \bar X +1.706 \times {\frac{s}{\sqrt{n} } ]

                                                 = [ 0.802 -1.706 \times {\frac{0.093}{\sqrt{27} } , 0.802 +1.706 \times {\frac{0.093}{\sqrt{27} } ]

                                                 = [0.771 , 0.832]

Therefore, 90% confidence interval for the population mean iron concentration is [0.771 , 0.832].

4 0
3 years ago
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