Using the normal distribution, it is found that:
a) The rating would have to be of at least 9.78.
b) The z-score for a rating of 5 is of -1.53.
<h3>Normal Probability Distribution</h3>The z-score of a measure X of a normally distributed variable with mean and standard deviation
is given by:
- The z-score measures how many standard deviations the measure is above or below the mean.
- Looking at the z-score table, the p-value associated with this z-score is found, which is the percentile of X.
The mean and the standard deviation are given by:
For item a, we have to find the 90th percentile, which is <u>X when Z = 1.28</u>, hence:
1.28 = (X - 7.6)/1.7
X - 7.6 = 1.28 x 1.7
X = 9.78.
The rating would have to be of at least 9.78.
For item b, we have to find <u>Z when X = 5</u>, hence:
Z = (5 - 7.6)/1.7
Z = -1.53.
The z-score for a rating of 5 is of -1.53.
More can be learned about the normal distribution at brainly.com/question/24808124
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