Answer:
(a) The probability distribution of the random variable <em>X</em> is Binomial.
(b) The mean and standard deviation of the 25-question test are 5 and 2 respectively.
(c) The value of <em>x</em> is 0.34.
Step-by-step explanation:
The random variable <em>X </em>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:
<em>P</em> (X) = 0.20.
There are a total of <em>n</em> = 25 questions.
Every answer is independent of the others.
(a)
The random variable <em>X</em> 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 <em>X</em> is Binomial with parameters <em>n</em> = 25 and <em>p</em> = 0.20.
(b)
Compute the mean of the random variable <em>X</em> as follows:
![E(X)=np\\=25\times 0.20\\=5](https://tex.z-dn.net/?f=E%28X%29%3Dnp%5C%5C%3D25%5Ctimes%200.20%5C%5C%3D5)
Compute the standard deviation of the random variable <em>X</em> as follows:
![SD(X)=\sqrt{np(1-p)}\\=\sqrt{25\times 0.20\times (1-0.20)}\\=2](https://tex.z-dn.net/?f=SD%28X%29%3D%5Csqrt%7Bnp%281-p%29%7D%5C%5C%3D%5Csqrt%7B25%5Ctimes%200.20%5Ctimes%20%281-0.20%29%7D%5C%5C%3D2)
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, ![X\sim N(5, 2^{2})](https://tex.z-dn.net/?f=X%5Csim%20N%285%2C%202%5E%7B2%7D%29)
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 <em>z</em> is -2.33.
Compute the value of <em>x</em> as follows:
![z=\frac{x-\mu}{\sigma}\\\\-2.33=\frac{x-5}{2}\\\\x=5-(2.33\times 2)\\\\x=0.34](https://tex.z-dn.net/?f=z%3D%5Cfrac%7Bx-%5Cmu%7D%7B%5Csigma%7D%5C%5C%5C%5C-2.33%3D%5Cfrac%7Bx-5%7D%7B2%7D%5C%5C%5C%5Cx%3D5-%282.33%5Ctimes%202%29%5C%5C%5C%5Cx%3D0.34)
Thus, the value of <em>x</em> is 0.34.