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:
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
In this problem, we have that:
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
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
This probability is 1 subtracted by the pvalue of Z when X = 20.4. So
By the Central Limit Theorem
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