MHB Probability question (mean, SD)

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Scores on the examination are normally distributed with a mean of 58 and a standard deviation of 18. To find the probability of scoring higher than 72, the z-score formula is applied, converting the score into a standardized value. For students to receive an A grade, the minimum score must fall within the top 10% of the distribution, which can be calculated using z-scores. Additionally, the proportion of students failing (scoring 40 or below) can be determined, along with the probability of at most 2 failures in a sample of 10 students. Finally, the probability that the mean score of 9 randomly selected students exceeds 65 can also be calculated using the normal distribution properties.
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Scores on an examination are assumed to be normally distributed with a mean of 58 and a standard deviation of 18.

(a) What is the probability that a person taking the examination scores higher than 72?

(b) Suppose that students scoring in the top 10% of this distribution are to receive an A grade. What is the minimum score a student must achieve to earn an A grade?

(c) Suppose that students scoring 40 or below are to receive a fail grade F. What is the proportion of failure in the examination?

(d) According to (c), if 10 students are randomly selected, what is the probability that there are at most 2 failures?

(e) Find the probability that the mean score of 9 randomly selected students exceeds 65.
 
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Hello, tiffyuyu! :D

Just for future reference, we ask that people posting questions show what they have tried so far, so that those helping have an idea where you are stuck and how best to help.

Let's begin with part a).

First. we need to standardize the raw datum given, so we need to use the following formula:

$$z=\frac{x-\mu}{\sigma}$$

Can you use this to convert the value of 72 into a $z$-score?
 
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