D. Mrs. Buchanan decided to throw her daughter a sweet sixteen party. She rented a room at a local venue for $500 a night plus $
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Answer:
57.49% probability that a randomly selected individual has an IQ between 81 and 109
Step-by-step explanation:
When the distribution is normal, we use 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:
Find the probability that a randomly selected individual has an IQ between 81 and 109
This is the pvalue of Z when X = 109 subtracted by the pvalue of Z when X = 81. So
X = 109
has a pvalue of 0.67
X = 81
has a pvalue of 0.0951
0.67 - 0.0951 = 0.5749
57.49% probability that a randomly selected individual has an IQ between 81 and 109
