Problem

Psychology I can be taken as correspondence course on a Pass/Fail basis. Experience shows that about $65 \%$ of students pass. This semester, 88 students are taking Psychology. a.) To set this problem up as a Binomial Experiment, find the following values: \[ \mathrm{n}=\square \mathrm{p}=\square \mathrm{q}=\square \] b.) What is the mean and standard deviation \[ \begin{array}{l} \mu=\square \text { (Round to } 3 \text { decimal places) } \\ \sigma=\square \end{array} \] c.) Using the Normal Approximation to the Binomial distribution method, find the following probabilities: What is the probability that at least 56 pass the course? What is the probability that no more than 63 pass the course?

Solution

Step 1 :The number of trials, n, is the total number of students taking the course, which is 88.

Step 2 :The probability of success, p, is the probability that a student passes the course, which is 65% or 0.65.

Step 3 :The probability of failure, q, is the probability that a student fails the course, which is 1 - p = 1 - 0.65 = 0.35.

Step 4 :So, n = 88, p = 0.65, q = 0.35.

Step 5 :The mean and standard deviation of a binomial distribution are given by the formulas: Mean, \(\mu = np\) and Standard deviation, \(\sigma = \sqrt{npq}\)

Step 6 :Substituting the given values: \(\mu = np = 88 * 0.65 = 57.2\) and \(\sigma = \sqrt{npq} = \sqrt{88 * 0.65 * 0.35} = 7.141\) (rounded to three decimal places)

Step 7 :To find the probability that at least 56 pass the course, we need to find the z-score for 56. The z-score is given by the formula: \(z = (X - \mu) / \sigma\)

Step 8 :Substituting the given values: \(z = (56 - 57.2) / 7.141 = -0.168\) (rounded to three decimal places)

Step 9 :The probability that a z-score is at least -0.168 is 1 - P(Z < -0.168) = 1 - 0.4332 = 0.5668 or 56.68%.

Step 10 :To find the probability that no more than 63 pass the course, we need to find the z-score for 63. Substituting the given values: \(z = (63 - 57.2) / 7.141 = 0.811\) (rounded to three decimal places)

Step 11 :The probability that a z-score is no more than 0.811 is P(Z < 0.811) = 0.7910 or 79.10%.

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