How many 1/6 cups of raisins are in 3/2 cups of raisins?
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Answer:
A score of 150.25 is necessary to reach the 75th percentile.
Step-by-step explanation:
Problems of normal distributions 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.
A set of test scores is normally distributed with a mean of 130 and a standard deviation of 30.
This means that
What score is necessary to reach the 75th percentile?
This is X when Z has a pvalue of 0.75, so X when Z = 0.675.
A score of 150.25 is necessary to reach the 75th percentile.