Question

Nate and Gabriel are two friends who say they have no previous knowledge of statistics. They are going to take a placement test to see if they should start at the introductory level or not.

The placement test is a multiple-choice test with 85 questions. Each question has 4 multiple-choice options with exactly one correct answer.

What is the expected number (mean) of correct answers if a student were to randomly guess?  (Do not round)

What is the standard deviation for the number of correct answers if a student were to randomly guess?  (Round to two decimal places)

The usual number of correct answers, if a student were to randomly guess, would be  to  . (Round to the nearest whole number that would be within the interval.)

What is the probability of getting 24 or more questions correct if you are randomly guessing?  (Round to four decimal places.)

What is the probability of getting 53 or more questions correct if you are randomly guessing?  (Round to four decimal places.)

Nate got 24 questions correct.

Gabriel got 53 questions correct.

Is it reasonable to believe that Nate had no prior knowledge of statistics?

• yes
• no

Why or why not?

Is it reasonable to believe that Gabriel had no prior knowledge of statistics?

• yes
• no

Why or why not?

Answer 1

Test have 85 multiple choice questions with 4 choices, among 1 choice is correct.

Probability of correct answer where student randomly guess is 1 / 4 since there are total 4 choices and only one is correct

That is random variable X: Number of questions guessing correct have binomial distribution with n = 85 and probability of success is 1/4 that is 0.25

Expected number of correct answer were student randomly guess = n * p = 85 * (1/4) = 21.25

Standard deviation for the number of correct answers

=

The range of usual values is

Where

round to the whole number as 13

round to the whole number as 29

The usual number of correct answers is (13, 29)

P(Getting 24 or more) that is

Here n = 85, so 24 or more means x takes values from 24 to onwards.

So here use compliment rule to find the required probability

X < 24 means X takes values from 0 to 23

Using excel find the probability,

Use binomdist function with number_s = 23, trials = 85, probability_s = 0.25, cumulative = 1 to get the sum of probabilities from 0 to 23

P(X < 24) = 0.718216

Therefore, probability of getting 24 or more questions correctly if you randomly guessing is 0.2818

P(Getting 53 or more) that is

53 or more means X takes values from 53 to 85

Similarly using excel first find the less than probability that is use compliment rule

P(X < 53) that means X takes values from 0 to 52

Use binomdist function with number_s = 52, trials = 85, probability_s = 0.25, cumulative = 1 to get the sum of probabilities from 0 to 52

P(X < 53) = 1

Therefore, probability of getting 53 or more questions correctly if you randomly guessing is 1

Nate got 24 questions correct, 24 falls in the range of usual values so it is possible.

No, it is not reasonable to believe that Nate had no prior knowledge of statistics.

Gabriel got 53 questions correct, 53 doesn't fall in the range of usual values so it is not possible.

Yes, it is reasonable to believe that Gabriel had no prior knowledge of statistics.