Can someone help please?

Suppose that 50% of all young adults prefer McDonald's to Burger King when asked to state a preference. A group of 12 young adul

ddd [48]

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

7 0

2 years ago

Please help ;-; Solve for the simultaneous equation…. 5x+y=17 X+y=3

Svetach [21]

Given Equation:-

5x+y=17 . . . . . . . . 1
x+y=3 . . . . . . . . . . 2

To find :

value of x
value of y

Solution :

Let's start with equation 2:-

x + y = 3

y = 3 - x

Put the value of y in Equation 1

5x+y=17
5x + (3 - x) = 17
5x - x + 3 = 17
4x + 3 = 17
4x = 17 - 3
4x = 14
x = 14/4
x = 7/2
x = 3.5

Now Let's find value of y:

put the value of x in equation 2:

y = 3 - x

y = 3 - 3.5

y = 0.5

4 0

2 years ago

Five kittens are sharing 6 cups of milk equally. How much milk does each kitten get?

Llana [10]

about 0.83 just solve the equation 5 divided by 6

3 0

2 years ago

Read 2 more answers

Five different books (A, B, C, D and E) are to be arranged on a shelf. Books C and D are to be arranged first and second startin

vekshin1

Your answer would be B.
HOPE THIS HELPS YOU! ^_^

5 0

3 years ago

Determine weather randomly picking a blue card from a bag containing all blue cards is impossible, unlikely, as likely as not, l

Anuta_ua [19.1K]

Answer:

The correct answer is certain with probability equal to 1.

Step-by-step explanation:

Probability is a mathematical framework which helps us to analyze chance of the outcome in a particular experiment. The value of probability is given by the ratio of the possible outcomes favorable to a certain experiment to the total outcomes.

We say an event is certain when the probability is 1 and the probability is zero when the event is uncertain.

Here the experiment is picking a blue card from a bag containing all blue cards.

Possible outcomes are all the cards colored blue in the bag.

Total outcomes are also all the blue cards in the bag.

∴ The value of probability is 1 as the event is certain because if we pick a card from the bag containing only blue cards, it would certainly give us a blue card.

5 0

2 years ago