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. Approximate the probability that the student gets at least 22 correct answers by using the normal distribution. The probability is:

Answer 1

Solution:

Each question has 4 choices.

So ,

probability of correct answer = 1/4 = 0.25

n = 48

Let X be the number of correct answered questions in this sample.

X follows Binomial(n = 48 , p = 0.25)

p = 0.25

q = 1 - p = 1 - 0.25 = 0.75

n*p = 48 * 0.25 = 12

n*q  = 48 * 0.75 = 36

Both np and nq are > 10

So , we use normal approximation to binomial.

According to normal approximation binomial,

X $\rightarrow$ Normal with

Mean = $\mu$ = n*p = 12

Standard deviation = $\sigma$ =$\sqrt{}$[n*p*q] = $\sqrt{}$ [48*0.25*0.75]  = 3

Now ,

P[getting at least 22 correct answers]

= P[X  $\geq$ 22]

= P[(X - $url=https://latex.codecogs.com/gif.latex?%5Cmu&t=1596163916&ymreqid=f4764063-0d40-e2b0-1cde-f00053012f00&sig=PrkMvSl1tOa6cBMo7G19mg--~D$ )/$url=https://latex.codecogs.com/gif.latex?%5Csigma&t=1596163916&ymreqid=f4764063-0d40-e2b0-1cde-f00053012f00&sig=UUwRWH102YhAW8Ogu4PJHg--~D$$\geq$ (22 - $url=https://latex.codecogs.com/gif.latex?%5Cmu&t=1596163916&ymreqid=f4764063-0d40-e2b0-1cde-f00053012f00&sig=PrkMvSl1tOa6cBMo7G19mg--~D$ )/$url=https://latex.codecogs.com/gif.latex?%5Csigma&t=1596163916&ymreqid=f4764063-0d40-e2b0-1cde-f00053012f00&sig=UUwRWH102YhAW8Ogu4PJHg--~D$]

= P[Z $\geq$ (22 - 12)/3]

= P[Z $\geq$ 3.33]

= 1 - P[Z < 3.33]

= 1 - 0.9996 ( use z table)

= 0.0004

The probability is 0.0004