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Homework 9 problem 5

jcastel1

I am having trouble figuring out how to do problem 5 on homework 9:

A multiple-choice test consists of 26 questions with possible answers of a, b, c, d. Estimate the probability that with random guessing, the number of correct answers is at least 10.

I really can’t figure out how the web work wants the answer either.

Any help is appreciated!

mark

The probability of getting any particular question right with random guessing is 0.25. Thus, if

X = \begin{cases} 1 & \text{ if we get a specific question right} \\ 0 & \text{ if we get a specific question wrong} \end{cases}

is a random variable that tallies our score on any single question, we have

\begin{aligned} \mu(X) &= 0.25 \times 1 + 0.75 \times 0 = 0.25, \text{ and } \\ \sigma^2(X) &= (1-0.25)^2 \times 0.25 + (1-0.75)^2 \times 0.75 = 0.1875. \end{aligned}

Now suppose that T is a random variable representing our total score. Then multiplying X by the number of problems (i.e 26), we get

\begin{aligned} \mu(T) &= 26 \times 0.25 = 6.5, \text{ and } \\ \sigma^2(T) &= 26 \times 0.1875 = 4.875 \end{aligned}

Thus, you should approximate with a normal distribution using \mu = 6.5 and \sigma = \sqrt{4.875}. I do think that the +0.5 adjustment factor will be important here. Thus, your integral might look like

\frac{1}{\sqrt{2\pi}} \int_{9.5}^{\infty} e^{(x-\mu)^2/(2\sigma^2)} \, dx.

Of course, that needs to be evaluated numerically.