Question

A student answers all 48 questions on a multiple-choice test by guessing. Each question has four possible answers, only one of which is correct. Find the probability that the student gets exactly 15 correct answers. Use the normal distribution to aproximate the binomial distribution.
0.7967
0.0606
0.0823
0.8577

Answer 1

The best option is C. 0.0823mean = np = 48*1/4 = 12 

SD = sqrt(npq) = sqrt(48 *1/4 *3/4) = 3 

aplying continuity correction for using a continuous distribution for a discrete one, 

"exactly 15" means 14.5 to 15.5 

z1= (14.5-12)/3 = 0.833 , z2 = (15.5-12)/3 = 1.167 

P(0.833 < z < 1.167) = 0.0808

Answer 2

Answer:

Correct answer is C

Step-by-step explanation:

First we need to find expected value of the event:

$E(x)=n*p=48*0.25=12$

And standard deviation of the event:

$Var(x)=\sqrt{n*p*q} =\sqrt{48*0.25*0.75} =3$

Then need to z value of normal distribution function for 15.5 and 14.5 correct answers:

$z_{1}=(15.5-12)/3=1.167\\z_{2}=(14.5-12)/3=0.833$

$P(z_{1})=0.8790\\P(z_{2})=0.7967$

P(15)=0.8790-0.7967=0.0823

Ps. Please find z table at the attachment.