Question

Consider the probability that at least 91 out of 155 students will pass their college placement exams. Assume the probability that a given student will pass their college placement exam is 59%.Approximate the probability using the normal distribution. Round your answer to four decimal places.

Answer 1

0.5616 = 56.16% probability that at least 91 out of 155 students will pass their college placement exams.

Explanation:

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:

`√(V(X)) = √(np(1-p))`

Normal probability distribution

Problems of normally distributed samples 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 = (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 = √(V(X))`.

In this problem, we have that:

`n = 155, p = 0.59`

So

`\mu = E(X) = np = 155*0.59 = 91.45`

`√(V(X)) = √(np(1-p)) = √(155*0.59*0.41) = 6.12`

Probability that at least 91 out of 155 students will pass their college placement exams.

Using continuity correction, this is
`P(X \geq 91 - 0.5) = P(X \geq 90.5)`, which is 1 subtracted by the pvalue of Z when X = 90.5. So

`Z = (X - \mu)/(\sigma)`

`Z = (90.5 - 91.45)/(6.12)`

`Z = -0.155`

`Z = -0.155` has a pvalue of 0.4384

1 - 0.4384 = 0.5616

0.5616 = 56.16% probability that at least 91 out of 155 students will pass their college placement exams.