Several psychology students are unprepared for a surprise true/false test with 1

Question

Several psychology students are unprepared for a surprise true/false test with 13 questions, and all of their answers are guesses. Find the mean and standard deviation for the number of correct answers for such students. Would it be unusual for a student to pass by guessing (which requires getting at least 9 correct answers)? Why or why not? mu = sigma = Choose the correct answer below. No, because 9 is within the range of usual values. Yes, because 9 is within the range of usual values.. Yes, because 9 is below the minimum usual value. Yes, because 9 is greater than the maximum usual value.

Explanation / Answer

there are total 13 question with each having the 2 options true or false

the probability of correct answer = 0.5

the probability of wrong answer = 0.5

now for getting the mean and the standard deviation we need to find the probability of the number of correct answer

p(0 correct) = 13C0*(0.5)^13 = 0.0001

P(X=1)=13C1*(0.5)^13 = 0.0013

P(X=2)=13C2*(0.5)^13 = 0.0078

P(X=3)=13C3*(0.5)^13 = 0.028

P(X=4)=13C4*(0.5)^13 = 0.071

P(X=5)=13C5*(0.5)^13 = 0.128

P(X=6)=13C6*(0.5)^13=0.171

P(X=7)=13C7*(0.5)^13=0.171

P(X=8)=13C8*(0.5)^13= 0.128

P(X=9)=13C9*(0.5)^13=0.071

P(X=10)=13C10*(0.5)^13=0.028

P(X=11)=13C11*(0.5)^13=0.0078

P(X=12)=13C12*(0.5)^13=0.0013

P(X=13)=13C13*(0.5)^13 = 0.0001

TO GET THE MEAN = E(X) = X1*P(X1)+X2*P(X2)+,,,,+XN*P(XN)

NOW E(X) = 0*0.0001+1*0.0013+2*0.0078+3*0.028+4*0.071+5*0.128+6*0.171+7*0.171+8*0.128+9*0.071+10*0.028+11*0.0078+12*0.0013+13*0.0001

= 5.293

VARIANCE = E(X^2) - E(X)^2

E(X^2) = 37.046

E(X)^2 = 28.015

THEREFORE VARIANCE = 37.046 - 28.015 = 9.031

STANDARD DEVIATION = VARIANCE^(1/2) = 9.031^(1/2)= 3.005

B) AS THE PROBABILITY OF GETTING 9 CORRECT ANSWERS IS 0.071 WHICH IS WAY LESS THEN THE USUAL VALUE OF 0.5 TEHERFORE STUDENT WILL NOT BE ABLE TO PASS HENCE OPTION B IS CORRECT.

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