Mathematics, 06.05.2020 02:00 iamjenng5028

Suppose a student takes a multiple-choice exam composed of 8 questions with 5 possible answers to each question, exactly one of which is correct. The student has no clue as to the correct answer to any of the question! and merely guesses. [You can use binompdf or binomcdf for a), b), and c).] {5.4)

a) What is the probability that the student will get exactly 6 answers correct, which is necessary for a grade of 751 (5 points)
b) What is the probability of the student getting 3 or less answers correct? [5 points)
c) What is the probability of the student getting 4 or more answers correct? [Hint: the complement of this event is getting 3 or less answers correct.] [4 point

Answers: 3

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Answers

Answer from: mirmir62

a) 0.11% probability that the student will get exactly 6 answers correct

b) 94.37% probability of the student getting 3 or less answers correct

c) 5.63% probability of the student getting 4 or more answers correct

Step-by-step explanation:

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.

Suppose a student takes a multiple-choice exam composed of 8 questions with 5 possible answers to each question, exactly one of which is correct.

This means that $n = 8, p = \frac{1}{5} = 0.2$

a) What is the probability that the student will get exactly 6 answers correct, which is necessary for a grade of 751 (5 points)

This is P(X = 6).

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

$P(X = 6) = C_{8,6}.(0.2)^{6}.(0.8)^{2} = 0.0011$

0.11% probability that the student will get exactly 6 answers correct

b) What is the probability of the student getting 3 or less answers correct? [5 points)

$P(X \leq 3) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)$

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

$P(X = 0) = C_{8,0}.(0.2)^{0}.(0.8)^{8} = 0.1678$

$P(X = 1) = C_{8,1}.(0.2)^{1}.(0.8)^{7} = 0.3355$

$P(X = 2) = C_{8,2}.(0.2)^{2}.(0.8)^{6} = 0.2936$

$P(X = 3) = C_{8,3}.(0.2)^{3}.(0.8)^{5} = 0.1468$

$P(X \leq 3) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) = 0.1678 + 0.3355 + 0.2936 + 0.1468 = 0.9437$

94.37% probability of the student getting 3 or less answers correct

c) What is the probability of the student getting 4 or more answers correct? [Hint: the complement of this event is getting 3 or less answers

$P(X \leq 3) + P(X \geq 4) = 1$

We want $P(X \geq 4)$

$P(X \geq 4) = 1 - P(X \leq 3) = 1 - 0.9437 = 0.0563$

5.63% probability of the student getting 4 or more answers correct

Answer from: Quest

the discriminant is 64 + 4*1*12 = 112

this is prime.

step-by-step explanation:

Answer from: Quest

the camry sales+accord sale=828,000. to make it so the two numbers are 18 thousand apart and still equal 828,000, you first need to find out what half of 828,000 is. 828,000/2=414,000. for the two numbers to equal 828,000, you have to add the same amount as you subtract from the middle. in this case, the number is 9,000 because it's half of 18,000. to equal 828,000, you have to add 9,000 to 414,000 for the camry and subtract 9,000 from 414,000 for the accord. 414,000+9,000=423,000.

414,000-9,000=405,000.

423,000+405,000=828,000.

423,000-405,000=18,

hope it