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 ![p = 0.64](https://tex.z-dn.net/?f=p%20%3D%200.64)
Sample of 148 students:
This means that ![n = 148](https://tex.z-dn.net/?f=n%20%3D%20148)
Mean and standard deviation:
![\mu = E(X) = np = 148(0.64) = 94.72](https://tex.z-dn.net/?f=%5Cmu%20%3D%20E%28X%29%20%3D%20np%20%3D%20148%280.64%29%20%3D%2094.72)
![\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{148*0.64*0.36} = 5.84](https://tex.z-dn.net/?f=%5Csigma%20%3D%20%5Csqrt%7BV%28X%29%7D%20%3D%20%5Csqrt%7Bnp%281-p%29%7D%20%3D%20%5Csqrt%7B148%2A0.64%2A0.36%7D%20%3D%205.84)
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}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7BX%20-%20%5Cmu%7D%7B%5Csigma%7D)
![Z = \frac{90.5 - 94.72}{5.84}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7B90.5%20-%2094.72%7D%7B5.84%7D)
![Z = -0.72](https://tex.z-dn.net/?f=Z%20%3D%20-0.72)
has a p-value of 0.2358.
X = 89.5
![Z = \frac{X - \mu}{\sigma}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7BX%20-%20%5Cmu%7D%7B%5Csigma%7D)
![Z = \frac{89.5 - 94.72}{5.84}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7B89.5%20-%2094.72%7D%7B5.84%7D)
![Z = -0.89](https://tex.z-dn.net/?f=Z%20%3D%20-0.89)
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.