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ruslelena [56]
3 years ago
11

In a test for extrasensory perception, the experimenter looks at cards containing either a star, circle, wave, or square. (The s

ubject cannot see the cards.) As the experimenter looks at each of 20 cards, the subject names the shape on the card. Assuming that any success guessing shapes is due purely to chance, what is the probability a subject correctly guesses at least 10 of the 20 shapes? (Round to 3 decimal places)
Mathematics
1 answer:
babymother [125]3 years ago
4 0

Answer:

The probability of a subject correctly guesses at least 10 of the 20 shapes is 0.014

Step-by-step explanation:

Looking at the experiment, we can find the probability thinking it as a Bernoulli experiment (dicotomic variable), as the subject either guess the shape right or not, and each guess is independent from the others (as are due to chance).

We are going to use a binomial distribution (useful to model successes in n   Bernoulli experiments), as we want to know the probability of at least 10 right guesses (number of sucess) in 20 cards (number of independent experiments). The probability of guessing right a shape is:

\mbox{Probability of guessing right}=\frac{\mbox{favorable cases}}{\mbox{total cases}}=\frac{1}{4}

The probability of k sucesses in n experiments, each with a p probability of sucess is given by:

P(X = k) = \binom{n}{k}p^k(1-p)^{n-k}

As we want to know the probability of getting at least 10 correct guesses, we have to add the odds for 10, 11, 12, ..., 20 right guesses:

P(X \ge 10) = \sum_{i=10}^{20} {20\choose i}0.25^i(1-0.25)^{20-i}=0.014

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Pls help me solve this problem!!
ddd [48]

Answer:

49.85

Step-by-step explanation:

41 is the mean, so half of the daily requests are above that number, half below, the question is about numbers above the mean, so the lower numbers won't be included in the percentage.

The difference between 59 and 41 is 18. The standard deviation given is 6.

18 ÷ 6 = 3  So that is 3 standard deviations. The empirical rule states that 99.7 of all values are within 3 standard deviations from the mean. But we are looking at only the upper half of those values so 99.7 ÷ 2 = 49.85 %

If the answer asks for approximate, you could round to 49.9 or 50.

And tell the math people to learn the correct spelling of fluorescent!

4 0
3 years ago
Jason deposited $18 000 in a bank that offers an interest rate of 5% per annum compounded
8_murik_8 [283]

Answer:

$1995.97.

Step-by-step explanation:

Amount in bank after 20 days =

18000(1 + 0.05/365)^20

= $18049.38

So H = 18049.38 - 16093.41

        =  $1995.97.

3 0
2 years ago
PLEASE HELP!!!!!!!!!<br> Make sure you show all your work for full points. What is o?
Pie

Answer:

  o = 54

Step-by-step explanation:

The angle sum theorem tells you the sum of angles in a triangle is 180°. The definition of a linear pair tells you the two angles of a linear pair total 180°. Together, these relations tell you that an exterior angle of a triangle is equal to the sum of the remote interior angles.

In this geometry, the angle marked 78° is exterior to the left-side triangle. That means ...

  78° = o° +24°

  o° = 78° -24° = 54°

The value of 'o' is 54.

__

<em>Additional comment</em>

n° is the supplement of 78°, so is 102°.

m° is the difference between 102° and 22°, so is 80°.

6 0
3 years ago
Read 2 more answers
Roger's flight took off at 9:27 a.m. The flight is scheduled to land at 1:05 p.m. If the flight lands on schedule, how long is t
arsen [322]
So Roger flight left at exactly 9.27:am and it will land at 1:05pm

So count on to 12:27 will be 3 hours and subtract from 5 because 65 is 05 3hours and 38mins
5 0
3 years ago
Read 2 more answers
The youth group is going on a trip to the State Fair trip cost 63 dollars included in that price is 13 dollars for a concert tic
Marina CMI [18]

Answer:

The cost of one pass is 25$

Step-by-step explanation:

Call the cost of one of the passes, P.   So we have

63 =  13 + 2P - subtract 13 from both sides

50 = 2P - divide both sides by 2

25 = P

So the cost of one pass = $25

4 0
2 years ago
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