Answer:
a) 33% probability that a single student randomly chosen from all those taking the test scores 21 or higher.
b) 0.39% probability that the mean score for 76 students randomly selected from all who took the test nationally is 20.4 or higher
Step-by-step explanation:
To solve this question, we need to understand the normal probability distribution and the central limit theorem.
Normal probability distribution:
Problems of normally distributed samples are solved using the z-score formula.
In a set with mean
and standard deviation
, the zscore of a measure X is given by:
![Z = \frac{X - \mu}{\sigma}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7BX%20-%20%5Cmu%7D%7B%5Csigma%7D)
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.
Central limit theorem:
The Central Limit Theorem estabilishes that, for a random variable X, with mean
and standard deviation
, the sample means with size n of at least 30 can be approximated to a normal distribution with mean
and standard deviation ![s = \frac{\sigma}{\sqrt{n}}](https://tex.z-dn.net/?f=s%20%3D%20%5Cfrac%7B%5Csigma%7D%7B%5Csqrt%7Bn%7D%7D)
In this problem, we have that:
![\mu = 18.6, \sigma = 5.9](https://tex.z-dn.net/?f=%5Cmu%20%3D%2018.6%2C%20%5Csigma%20%3D%205.9)
a) What is the probability that a single student randomly chosen from all those taking the test scores 21 or higher?
This is 1 subtracted by the pvalue of Z when X = 21. So
![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{21 - 18.6}{5.4}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7B21%20-%2018.6%7D%7B5.4%7D)
![Z = 0.44](https://tex.z-dn.net/?f=Z%20%3D%200.44)
has a pvalue of 0.67
1 - 0.67 = 0.33
33% probability that a single student randomly chosen from all those taking the test scores 21 or higher.
b) The average score of the 76 students at Northside High who took the test was x =20.4. What is the probability that the mean score for 76 students randomly selected from all who took the test nationally is 20.4 or higher?
Now we have ![n = 76, s = \frac{5.9}{\sqrt{76}} = 0.6768](https://tex.z-dn.net/?f=n%20%3D%2076%2C%20s%20%3D%20%5Cfrac%7B5.9%7D%7B%5Csqrt%7B76%7D%7D%20%3D%200.6768)
This probability is 1 subtracted by the pvalue of Z when X = 20.4. So
![Z = \frac{X - \mu}{\sigma}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7BX%20-%20%5Cmu%7D%7B%5Csigma%7D)
By the Central Limit Theorem
![Z = \frac{X - \mu}{s}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7BX%20-%20%5Cmu%7D%7Bs%7D)
![Z = \frac{20.4 - 18.6}{0.6768}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7B20.4%20-%2018.6%7D%7B0.6768%7D)
![Z = 2.66](https://tex.z-dn.net/?f=Z%20%3D%202.66)
has a pvalue of 0.9961
1 - 0.9961 = 0.0039
0.39% probability that the mean score for 76 students randomly selected from all who took the test nationally is 20.4 or higher