SOLUTION: Several students are unprepared for multiple choice quiz with 10 questions and all their answers are guesses each question has five possible answers and only one of them is correct

Question 1128376: Several students are unprepared for multiple choice quiz with 10 questions and all their answers are guesses each question has five possible answers and only one of them is correct
A) find the mean and standard deviation for the number of correct answers for such students
B) would it be unusual such a student to pass by guessing and getting at least 7 correct answers
Answer by jim_thompson5910(35256) (Show Source): You can put this solution on YOUR website!
Part A)

n = 10
p = probability of getting a question right
p = 1/5 = 0.2

mean = n*p
mean = 10*0.2
mean = 2

standard deviation = sqrt(n*p*(1-p))
standard deviation = sqrt(10*0.2*(1-0.2))
standard deviation = sqrt(10*0.2*0.8)
standard deviation = sqrt(1.6)
standard deviation = 1.2649
This is approximate to four decimal places

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Part B)

Compute the binomial probability of getting exactly 7 questions correct
n C k = (n!)/(k!*(n-k)!)
10 C 7 = (10!)/(7!*(10-7)!)
10 C 7 = 120
P(X = k) = (n C k)*(p)^(k)*(1-p)^(n-k)
P(X = 7) = (10 C 7)*(0.2)^(7)*(1-0.2)^(10-7)
P(X = 7) = (120)*(0.2)^(7)*(0.8)^(3)
P(X = 7) = 0.000786432

Compute the binomial probability of getting exactly 8 questions correct
n C k = (n!)/(k!*(n-k)!)
10 C 8 = (10!)/(8!*(10-8)!)
10 C 8 = 45
P(X = k) = (n C k)*(p)^(k)*(1-p)^(n-k)
P(X = 8) = (10 C 8)*(0.2)^(8)*(1-0.2)^(10-8)
P(X = 8) = (45)*(0.2)^(8)*(0.8)^(2)
P(X = 8) = 0.000073728

Compute the binomial probability of getting exactly 9 questions correct
n C k = (n!)/(k!*(n-k)!)
10 C 9 = (10!)/(9!*(10-9)!)
10 C 9 = 10
P(X = k) = (n C k)*(p)^(k)*(1-p)^(n-k)
P(X = 9) = (10 C 9)*(0.2)^(9)*(1-0.2)^(10-9)
P(X = 9) = (10)*(0.2)^(9)*(0.8)^(1)
P(X = 9) = 0.000004096

Compute the binomial probability of getting exactly 10 questions correct
n C k = (n!)/(k!*(n-k)!)
10 C 10 = (10!)/(10!*(10-10)!)
10 C 10 = 1
P(X = k) = (n C k)*(p)^(k)*(1-p)^(n-k)
P(X = 10) = (10 C 10)*(0.2)^(10)*(1-0.2)^(10-10)
P(X = 10) = (1)*(0.2)^(10)*(0.8)^(0)
P(X = 10) = 0.0000001024

Add up the four probabilities found above to find the probability of getting 7 or more correct.
P(X >= 7) = P(X = 7) + P(X = 8) + P(X = 9) + P(X = 10)
P(X >= 7) = 0.000786432+0.000073728+0.000004096+0.0000001024
P(X >= 7) = 0.0008643584

The probability of getting 7 or more correct answers is 0.0008643584
This is a very small value so it's a very unusual (aka unlikely) event, which is what we expect for any student who randomly guesses at the answers.
So it would be very unusual for a student to pass if they were to randomly guess at all the questions.
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