Question

A really tough trivia quiz has 50 multiple-choice questions, each with four possible answers.
a. List the probability of success & failure for one question, success mean and standard deviation to two decimal
places. [T 3]
b. Determine whether a binomial distribution using a normal distribution model is reasonable. If so, explain
your reasoning. [C 2]
c. Calculate the probability of randomly guessing 15 to 20 (inclusive) questions correctly. [T 3]

Answer 1

For each question, there are only two possible outcomes. Either the correct answer is guessed, or it is not. The probability of the correct answer being guessed on a question is independent of any other question, which means that the binomial probability distribution can be used to solve this question.

Since it is a large sample, we test if the approximation to the normal distribution can be used, and thus, first concepts of the binomial distribution and it's approximation to the normal are presented.

After this, we get that:

• For one question: 0.25 probability of success and 1 - 0.25 = 0.75 probability of failure. The mean number of successes is of 12.5, with a standard deviation of 3.06.
• Since both $np \geq 10$ and $n(1-p) \geq 10$, it is reasonable to approximate the binomial distribution using a normal distribution model.
• 0.2533 = 25.33% probability of randomly guessing 15 to 20 (inclusive) questions correctly.

Binomial probability distribution

Probability of exactly x successes on n repeated trials, with p probability.

Can be approximated to a normal distribution, using the expected value and the standard deviation.

The expected value of the binomial distribution is:

$E(X) = np$

The standard deviation of the binomial distribution is:

$\sqrt{V(X)} = \sqrt{np(1-p)}$

Normal probability distribution

Problems of normally distributed distributions can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$, the z-score of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.

When we are approximating a binomial distribution to a normal one, we have that $\mu = E(X)$, $\sigma = \sqrt{V(X)}$, if $np \geq 10$ and $n(1-p) \geq 10$.

Question a:

Each question has four possible answers, one of which is correct, meaning that:

$p = \frac{1}{4} = 0.25$

Thus, for one question: 0.25 probability of success and 1 - 0.25 = 0.75 probability of failure.

50 questions means that $n = 50$, and thus:

$E(X) = np = 50(0.25) = 12.5$

$\sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{50(0.25)(0.75)} = 3.06$

The mean number of successes is of 12.5, with a standard deviation of 3.06.

Question b:

$np = 50(0.25) = 12.5$

$n(1-p) = 50(0.75) = 37.5$

Since both $np \geq 10$ and $n(1-p) \geq 10$, it is reasonable to approximate the binomial distribution using a normal distribution model.

Question c:

Using the approximation, with $\mu = 12.5, \sigma = 3.06$, and continuity correction(as the normal distribution is continuous and the binomial is discrete), this is:

$P(15 - 0.5 \leq X \leq 20 + 0.5) = P(14.5 \leq X \leq 20.5)$

Which is the p-value of Z when X = 20.5 subtracted by the p-value of Z when X = 14.5.

X = 20.5

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{20.5 - 12.5}{3.06}$

$Z = 2.61$

$Z = 2.61$ has a p-value of 0.9955.

X = 14.5

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{14.5 - 12.5}{3.06}$

$Z = 0.65$

$Z = 0.65$ has a p-value of 0.7422.

0.9955 - 0.7422 = 0.2533

0.2533 = 25.33% probability of randomly guessing 15 to 20 (inclusive) questions correctly.

A similar question is found at https://brainly.com/question/24261244