Question

Consider the probability that exactly 90 out of 148 students will pass their college placement exams. Assume the probability that a given student will pass their college placement exam is 64%. Approximate the probability using the normal distribution. Round your answer to four decimal places.

Answer 1

Answer:

0.0491 = 4.91% probability that exactly 90 out of 148 students will pass their college placement exams.

Step-by-step explanation:

Binomial probability distribution

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

Assume the probability that a given student will pass their college placement exam is 64%.

This means that $p = 0.64$

Sample of 148 students:

This means that $n = 148$

Mean and standard deviation:

$\mu = E(X) = np = 148(0.64) = 94.72$

$\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{148*0.64*0.36} = 5.84$

Consider the probability that exactly 90 out of 148 students will pass their college placement exams.

Due to continuity correction, 90 corresponds to values between 90 - 0.5 = 89.5 and 90 + 0.5 = 90.5, which means that this probability is the p-value of Z when X = 90.5 subtracted by the p-value of Z when X = 89.5.

X = 90.5

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

$Z = \frac{90.5 - 94.72}{5.84}$

$Z = -0.72$

$Z = -0.72$ has a p-value of 0.2358.

X = 89.5

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

$Z = \frac{89.5 - 94.72}{5.84}$

$Z = -0.89$

$Z = -0.89$ has a p-value of 0.1867.

0.2358 - 0.1867 = 0.0491.

0.0491 = 4.91% probability that exactly 90 out of 148 students will pass their college placement exams.