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Scorpion4ik [409]
2 years ago
13

He asked the class to write a number that is greater than R, and less than S.

Mathematics
2 answers:
Annette [7]2 years ago
7 0

Answer:

0.45

Step-by-step explanation:

<em>R = .34</em>

<em>S = .66</em>

<em>.30 is </em><em>less than R</em><em> and less than S</em>

<em>.70 is greater than R and </em><em>greater than S</em>

<em>.75 is greater than R and </em><em>greater than S</em>

<em>.45 is the only one that is greater than R and less than S</em>

<u>Hope this helped <3 brainliest please :)</u>

Goryan [66]2 years ago
3 0

number greater than R and less than S is 0,45

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Answer:

a) 0.194 = 19.4% probability that more than 7 preferred McDonald's

b) 0.787 = 78.7% probability that between 3 and 7 (inclusive) preferred McDonald's

c) 0.787 = 78.7% probability that between 3 and 7 (inclusive) preferred Burger King

Step-by-step explanation:

For each young adult, there are only two possible outcomes. Either they prefer McDonalds, or they prefer burger king. The probability of an adult prefering McDonalds is independent from other adults. So we use the binomial probability distribution to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

In which C_{n,x} is the number of different combinations of x objects from a set of n elements, given by the following formula.

C_{n,x} = \frac{n!}{x!(n-x)!}

And p is the probability of X happening.

50% of all young adults prefer McDonald's to Burger King when asked to state a preference.

This means that p = 0.5

12 young adults were randomly selected

This means that n = 12

(a) What is the probability that more than 7 preferred McDonald's?

P(X > 7) = P(X = 8) + P(X = 9) + P(X = 10) + P(X = 11) + P(X = 12)

In which

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

P(X = 8) = C_{12,8}.(0.5)^{8}.(0.5)^{4} = 0.121

P(X = 9) = C_{12,9}.(0.5)^{9}.(0.5)^{3} = 0.054

P(X = 10) = C_{12,10}.(0.5)^{10}.(0.5)^{2} = 0.016

P(X = 11) = C_{12,11}.(0.5)^{11}.(0.5)^{1} = 0.003

P(X = 12) = C_{12,12}.(0.5)^{12}.(0.5)^{0} = 0.000

P(X > 7) = P(X = 8) + P(X = 9) + P(X = 10) + P(X = 11) + P(X = 12) = 0.121 + 0.054 + 0.016 + 0.003 + 0.000 = 0.194

0.194 = 19.4% probability that more than 7 preferred McDonald's

(b) What is the probability that between 3 and 7 (inclusive) preferred McDonald's?

P(3 \leq X \leq 7) = P(X = 3) + P(X = 4) + P(X = 5) + P(X = 6) + P(X = 7)

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

P(X = 3) = C_{12,3}.(0.5)^{3}.(0.5)^{9} = 0.054

P(X = 4) = C_{12,4}.(0.5)^{4}.(0.5)^{8} = 0.121

P(X = 5) = C_{12,5}.(0.5)^{5}.(0.5)^{7} = 0.193

P(X = 6) = C_{12,6}.(0.5)^{6}.(0.5)^{6} = 0.226

P(X = 7) = C_{12,7}.(0.5)^{7}.(0.5)^{5} = 0.193

P(3 \leq X \leq 7) = P(X = 3) + P(X = 4) + P(X = 5) + P(X = 6) + P(X = 7) = 0.054 + 0.121 + 0.193 + 0.226 + 0.193 = 0.787

0.787 = 78.7% probability that between 3 and 7 (inclusive) preferred McDonald's

(c) What is the probability that between 3 and 7 (inclusive) preferred Burger King?

Since p = 1-p = 0.5, this is the same as b) above.

So

0.787 = 78.7% probability that between 3 and 7 (inclusive) preferred Burger King

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