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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:
The standard deviation of the binomial distribution is:
Normal probability distribution
Problems of normally distributed distributions 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 ,
.
Assume the probability that a given student will pass their college placement exam is 64%.
This means that
Sample of 148 students:
This means that
Mean and standard deviation:
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
has a p-value of 0.2358.
X = 89.5
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.