Consider the probability that no less than 92 out of 159 students will pass their college placement exams. Assume the probability that a given student will pass their college placement exam is 55%. Approximate the probability using the normal distribution. Round your answer to four decimal places.

0.2843 = 28.43% probability that no less than 92 out of 159 students will pass their college placement exams.

Step-by-step explanation:

I am going to use the binomial approximation to the normal to solve this question.

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:

The standard deviation of the binomial distribution is:

Normal probability distribution

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

In a set with mean and standard deviation , the zscore of a measure X is given by:

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 , .

In this problem, we have that:

. So

Probability that no less than 92 out of 159 students will pass their college placement exams.

No less than 92 is more than 91, which is 1 subtracted by the pvalue of Z when X = 91. So

has a pvalue of 0.7157

1 - 0.7157 = 0.2843

0.2843 = 28.43% probability that no less than 92 out of 159 students will pass their college placement exams.