Someone answer this please!
1 answer:
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Answer:
There is a 98.93% probability that a randomly selected score will be greater than 54.1.
Step-by-step explanation:
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:
A set of exam scores is normally distributed and has a mean of 79.4 and a standard deviation of 11. This means that .
What is the probability that a randomly selected score will be greater than 54.1?
This probability is 1 subtracted by the pvalue of Z when .
has a pvalue of 0.0107.
This means that there is a 1-0.0107 = 0.9893 = 98.93% probability that a randomly selected score will be greater than 54.1.