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 exam 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

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 what the question truly says)

d) 0.806

Step-by-step explanation:

Part 1

Since there are 4 choices per question, and only one correct answer per question.

The probability of getting a question right = (1/4) = 0.25

Probability of getting a question wrong = 1 - 0.25 = 0.75

a) Probability that the first question she gets right is the 5th question means she gets the first 4 questions wrong, and gets the last question.

0.75 * 0.75 * 0.75 * 0.75 * 0.25 = 0.0791

b) Probability that she gets all of the questions right

0.25 * 0.25 * 0.25 * 0.25 * 0.25 = 0.000977

c) Probability that she gets at least one question right = 1 - (probability that she doesn't get any question right) = 1 - (0.75⁵) = 1 - 0.2373 = 0.7627

Part 2

We use 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