Mathematics, 18.02.2021 22:20 nattalliiee67

Consider a multiple-choice examination with 50 questions. Each question has four possible answers. Assume that a student who has done the homework and attended lectures has a 75% chance of answering any question correctly. (Round your answers to two decimal places.) (a) A student must answer 43 or more questions correctly to obtain a grade of A. What percentage of the students who have done their homework and attended lectures will obtain a grade of A on this multiple-choice examination

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0.55% of the students who have done their homework and attended lectures will obtain a grade of A on this multiple-choice examination

Step-by-step explanation:

To solve this question, we use the normal approximation to the binomial, since both np and n(1-p) >= 10.

Binomial probability distribution

Probability of exactly x sucesses 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 zscore 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 pvalue, 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 $\mu = E(X)$ , $\sigma = \sqrt{V(X)}$ .

Consider a multiple-choice examination with 50 questions.

This means that n = 50

Assume that a student who has done the homework and attended lectures has a 75% chance of answering any question correctly.

This means that p = 0.75

Mean and standard deviation:

$\mu = E(X) = np = 50*0.75 = 35$

$\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{50*0.75*0.15} = 2.96$

(a) A student must answer 43 or more questions correctly to obtain a grade of A. What percentage of the students who have done their homework and attended lectures will obtain a grade of A on this multiple-choice examination?

Using continuity correction, this is $P(X \geq 43 - 0.5) = P(X \geq 42.5)$ , which, as a proportion, is 1 subtracted by the pvalue of Z when X = 42.5. So

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

$Z = \frac{42.5 - 35}{2.96}$

Z = 2.54

Z = 2.54 has a pvalue of 0.9945

1 - 0.9945 = 0.0055 = 0.55%, which means that 0.55% of the students who have done their homework and attended lectures will obtain a grade of A on this multiple-choice examination

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