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 approximate the binomial distribution.

a. 0.8577
b. 0.7967
c. 0.0606
d. 0.0823

Explanation:

Step1: Calculate the mean and standard deviation

For a binomial distribution \(X\sim B(n,p)\), where \(n = 48\) (number of trials/questions) and \(p=\frac{1}{4}=0.25\) (probability of success/correct answer).
The mean \(\mu=np\), so \(\mu = 48\times0.25=12\).
The standard deviation \(\sigma=\sqrt{np(1 - p)}\), so \(\sigma=\sqrt{48\times0.25\times(1 - 0.25)}=\sqrt{48\times0.25\times0.75}=\sqrt{9}=3\).

Step2: Apply the continuity correction

To approximate \(P(X = 15)\) using the normal distribution \(N(\mu,\sigma^{2})\), we use the continuity correction. We approximate \(P(X = 15)\) as \(P(14.5<Y<15.5)\) where \(Y\sim N(12,3^{2})\).

Step3: Standardize the values

The \(z\)-score is given by \(z=\frac{x-\mu}{\sigma}\).
For \(x = 14.5\), \(z_1=\frac{14.5 - 12}{3}=\frac{2.5}{3}\approx0.83\).
For \(x = 15.5\), \(z_2=\frac{15.5 - 12}{3}=\frac{3.5}{3}\approx1.17\).

Step4: Find the probabilities

Using the standard normal table \(P(Z<z)\), we know that \(P(Z < 0.83)=0.7967\) and \(P(Z < 1.17)=0.8790\).
Then \(P(0.83<Z<1.17)=P(Z < 1.17)-P(Z < 0.83)=0.8790 - 0.7967=0.0823\).

Answer:

D. \(0.0823\)