Question

It is estimated that 9% of those taking the quantitative methods portion of the certified public accountant (CPA) examination fail that section. Sixty four students are taking the examination this Saturday.

  

a-1. How many would you expect to fail? (Round the final answer to 2 decimal places.)

Number of students           

a-2. What is the standard deviation? (Round the final answer to 2 decimal places.)

Standard deviation           

b. What is the probability that exactly two students will fail? (Round the final answer to 4 decimal places.)

Probability           

c. What is the probability at least two students will fail? (Round the final answer to 4 decimal places.)

Probability           

Answer 1

Solution:

We are given n = 64, p = 0.09

a-1. How many would you expect to fail?

Expected number of students = n*p = 64*0.09 = 5.76

Expected number of students = 5.76

a-2. What is the standard deviation?

Standard deviation = sqrt(np(1 – p)) = sqrt(64*0.09*(1 - 0.09)) = 2.289454

Standard deviation = 2.29

b. What is the probability that exactly two students will fail?

We have to find P(X=2)

Here, we have to use normal approximation to binomial distribution.

np = 5.76 and nq = 64*0.91 = 58.24

np and nq > 5

So, we can use normal approximation.

P(X=2) = P(1.5<X<2.5)

(by using continuity correction)

P(1.5<X<2.5) = P(X<2.5) – P(X<1.5)

Find P(X<2.5)

Z = (X – mean) / SD

Mean = np = 5.76

SD = 2.29

Z = (2.5 - 5.76) /2.29

Z = -1.42358

P(Z<-1.42358) = P(X<2.5) = 0.077284

(by using z-table)

Now, find P(X<1.5)

Z = (1.5 - 5.76) /2.29

Z = -1.86026

P(Z<-1.86026) = P(X<1.5) = 0.031424

(by using z-table)

P(1.5<X<2.5) = P(X<2.5) – P(X<1.5)

P(1.5<X<2.5) = 0.077284 - 0.031424

P(1.5<X<2.5) = 0.04586

Required probability = 0.0459

c. What is the probability at least two students will fail?

Here, we have to find P(X≥2)

P(X≥2) = P(X>1.5)

(by using continuity correction)

P(X>1.5) = 1 – P(X<1.5)

Z = (1.5 - 5.76) /2.29

Z = -1.86026

P(Z<-1.86026) = P(X<1.5) = 0.031424

(by using z-table)

P(X>1.5) = 1 – P(X<1.5)

P(X>1.5) = 1 - 0.031424

P(X>1.5) = 0.968576

Required probability = 0.9686