Question 1128376
<font color="black" face="times" size="3">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.</font>