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Drupady
10 days ago
13

In a multiple choice exam, there are 5 questions and 4 choices for each question (a, b, c, d). Nancy has not studied for the exa

m at all and decides to randomly guess the answers. What is the probability that:
(a) the first question she gets right is the 5th question?
(b) she gets all of the questions right?
(c) she gets at least one question right?
What is the area under the standard normal distribution for each region?
(a) Z< -1.65
(b) Z>1.5
(c) -1.1 (d) |Z|>1.3
Mathematics
1 answer:
zzz [9K]10 days ago
6 0

Answer:

Part 1

a) 0.0791

b) 0.000977

c) 0.7627

Part 2

a) 0.049

b) 0.067

c) 0.864 or 0.136 (depending on the exact question wording)

d) 0.806

Step-by-step explanation:

Part 1

Since each question presents 4 possibilities, and only one answer is correct per question.

The likelihood of answering a question correctly = (1/4) = 0.25

The chance of getting a question incorrect = 1 - 0.25 = 0.75

a) The probability of correctly answering the first question as the 5th implies answering the first four incorrectly, and then getting the last one correct.

0.75 × 0.75 × 0.75 × 0.75 × 0.25 = 0.0791

b) The probability that she answers all questions correctly

0.25 × 0.25 × 0.25 × 0.25 × 0.25 = 0.000977

c) Probability that she answers at least one question correctly = 1 - (probability that none are answered correctly) = 1 - (0.75⁵) = 1 - 0.2373 = 0.7627

Part 2

We will utilize the standard normal distribution for this

a) Area under (Z< -1.65) = P(z < -1.65) = 1 - P(z ≥ -1.65) = 1 - P(z ≤ 1.65) = 1 - 0.951 = 0.049

b) Area under (Z > 1.5) = P(z > 1.5) = 1 - P(z ≤ 1.5) = 1 - 0.933 = 0.067

c) P(z > -1.1) or P(z < -1.1)

P(z > - 1.1) = 1 - P(z ≤ -1.1) = 1 - 0.136 = 0.864

P(z < - 1.1) = 1 - P(z ≥ - 1.1) = 1 - P(z ≤ 1.1) = 1 - 0.864 = 0.136

d) |Z|>1.3 = P(-1.3 < z < 1.3) = P(z < 1.3) - P(z < -1.3)

P(z < 1.3) = 1 - P(z ≥ 1.3) = 1 - P(z ≤ -1.3) = 1 - 0.097 = 0.903

P(z < -1.3) = 1 - P(z ≥ -1.3) = 1 - P(z ≤ 1.3) = 1 - 0.903 = 0.097

P(z < 1.3) - P(z < -1.3) = 0.903 - 0.097 = 0.806

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