Respuesta :
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.
