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Answer:
0.1% probability that the average score of a random sample of 20 students exceeds 59.5.
Step-by-step explanation:
The Central Limit Theorem estabilishes that, for a random variable X, with mean and standard deviation
, a large sample size can be approximated to a normal distribution with mean
and standard deviation
Normal probability distribution
Problems of normally distributed samples 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.
In this problem, we have that:
What is the probability that the average score of a random sample of 20 students exceeds 59.5?
This is 1 subtracted by the pvalue of Z when . So
has a pvalue of 0.9990.
So there is a 1-0.9990 = 0.001 = 0.1% probability that the average score of a random sample of 20 students exceeds 59.5.
Answer:
y = - 13, y = 13
Step-by-step explanation:
Given
y² = 169 ( take the square root of both sides )
y = ± = ± 13
Since 13 × 13 = 169 and - 13 × - 13 = 169