Answer:
(a) The probability that an applicant guesses randomly and passes is 0.0577.
(b) To reduce the probability of passing by guessing to 10 percent or less the minimum score required would be 14.
Step-by-step explanation:
Let X = number of correctly marked questions.
The probability of correctly marking a question is, p = 0.50.
The test consists of n = 20 questions.
The participant marks each question independently.
The random variable X follows a Binomial distribution with parameters n and p.
The probability mass function of X is given as follows:
[tex]P(X=x)={20\choose x}0.50^{x}(1-0.50)^{20-x};\ x=0,1,2,3...[/tex]
It is provided that the company accepts candidates who scores 14 or more.
(a)
Compute the probability of the event (X ≥ 14) as follows:
P (X ≥ 14) = P (X = 14) + P (X = 15) + P (X = 16) +... +P (X = 20)
[tex]=\sum\limits^{20}_{x=14} {{20\choose x}0.50^{x}(1-0.50)^{20-x}}\\=0.037+0.0148+0.0046+0.0011+0.0002+0.0000+0.0000\\=0.0577[/tex]
Thus, the probability that an applicant guesses randomly and passes is 0.0577.
(b)
Consider the probability distribution table attached below.
The probability of getting 15 or more correct answer by guessing is,
P (X ≥ 15) = P (X = 15) + P (X = 16) +... +P (X = 20) = 0.0207
And the probability of getting 14 or more correct answer by guessing is 0.0577.
The probability of getting 13 or more correct answer by guessing is:
P (X ≥ 13) = P (X = 13) + P (X = 14) +... +P (X = 20) = 0.1316
So, to reduce the probability of passing by guessing to 10 percent or less the minimum score required would be 14.
