A set of exam scores is normally distributed and has a mean of 79.4 and a standard deviation of 11. What is the probability that a randomly selected score will be greater than 54.1? Answer = (round to four decimal places) Note: Be careful...only use the Z Table here and round z scores to two places since that’s what the table uses...do not use technology or the 68-95-99.7 Rule.

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 [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the zscore of a measure X is given by:[tex]Z = \frac{X - \mu}{\sigma}[/tex]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 [tex]\mu = 79.4, \sigma = 11[/tex].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 [tex]X = 54.1[/tex].[tex]Z = \frac{X - \mu}{\sigma}[/tex][tex]Z = \frac{54.1 - 79.4}{11}[/tex][tex]Z = -2.3[/tex][tex]Z = -2.3[/tex] 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.