Qualification exams for becoming a state-certified welding inspector are based on multiple-choice tests. As in any multiple-choice test, there is a possibility that someone who is simply guessing the answers to each question might pass the test. Let x denote the number of correct answers given by a person who is guessing each answer on a 25-question exam, with each question having five possible answers (for each question, assume only one of the five choices is correct).(a) What type of probability distribution does x have?(b) For the 25-question test, what are the mean and standard deviation of x?(c) The exam administrators want to make sure that there is a very small chance, say, 1%, that a person who is guessing will pass the test. What minimum passing score should they allow on the exam to meet this requirement?

(a) The probability distribution of the random variable X is Binomial.

(b) The mean and standard deviation of the 25-question test are 5 and 2 respectively.

(c) The value of x is 0.34.

Step-by-step explanation:

The random variable X is defined as the number of correct answers given by a person who is guessing each answer on a 25-question exam.

There are five possible answer for every question.

This implies that the probability of getting a correct answer is:

P (X) = 0.20.

There are a total of n = 25 questions.

Every answer is independent of the others.

(a)

The random variable X has  finite number of independent trials (i.e. 25 questions). There are only two outcomes for each trial, i.e. Success = correct answer and Failure = wrong answer. Each trial has the same probability of success (, i.e. P (X) = 0.25).

Thus, the probability distribution of the random variable X is Binomial with parameters n = 25 and p = 0.20.

(b)

Compute the mean of the random variable X as follows:

Compute the standard deviation of the random variable X as follows:

Thus, the mean and standard deviation of the 25-question test are 5 and 2 respectively.

(c)

The sample is large and the probability of success is close to 0.50.

So a Normal approximation to binomial can be applied to approximate the distribution of X if the following conditions are satisfied:

1.np ≥ 5

2.n(1 - p) ≥ 5

Check the conditions as follows:

Thus, a Normal approximation to binomial can be applied.

So,  

It is provided that the minimum passing score foe the test is such that only 1% of students who are guessing will pass the test.

That is P (X < x) = 0.01.

⇒ P (Z < z) = 0.01

The value of z is -2.33.

Compute the value of x as follows:

Thus, the value of x is 0.34.

simplifying

5(x + -6) = 2(x + 3)

reorder the terms:

5(-6 + x) = 2(x + 3)

(-6 * 5 + x * 5) = 2(x + 3)

(-30 + 5x) = 2(x + 3)

reorder the terms:

-30 + 5x = 2(3 + x)

-30 + 5x = (3 * 2 + x * 2)

-30 + 5x = (6 + 2x)

solving

-30 + 5x = 6 + 2x

solving for variable 'x'.

move all terms containing x to the left, all other terms to the right.

add '-2x' to each side of the equation.

-30 + 5x + -2x = 6 + 2x + -2x

combine like terms: 5x + -2x = 3x

-30 + 3x = 6 + 2x + -2x

combine like terms: 2x + -2x = 0

-30 + 3x = 6 + 0

-30 + 3x = 6

add '30' to each side of the equation.

-30 + 30 + 3x = 6 + 30

combine like terms: -30 + 30 = 0

0 + 3x = 6 + 30

3x = 6 + 30

combine like terms: 6 + 30 = 36

3x = 36

divide each side by '3'.

x = 12

simplifying

x = 12

this should .