Estimate by clustering: 56.9, 63.2, 59.3, 61.1 A. 240.0 B. 225.0 C. 200.0 D. 250.0
2 answers:
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Answer:
4.05% probability that a randomly selected adult has an IQ greater than 123.4.
Step-by-step explanation:
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.
In this question, we have that:
Probability that a randomly selected adult has an IQ greater than 123.4.
This is 1 subtracted by the pvalue of Z when X = 123.4. So
has a pvalue of 0.9595
1 - 0.9595 = 0.0405
4.05% probability that a randomly selected adult has an IQ greater than 123.4.
Using limits, it is found that the infinite sequence converges, as the limit does not go to infinity.
<h3>How do we verify if a sequence converges of diverges?</h3>Suppose an infinity sequence defined by:
Then we have to calculate the following limit:
If the <u>limit goes to infinity</u>, the sequence diverges, otherwise it converges.
In this problem, the function that defines the sequence is:
Hence the limit is:
Hence, the infinite sequence converges, as the limit does not go to infinity.
More can be learned about convergent sequences at brainly.com/question/6635869
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