A professor at a local university finds that the mean quiz grade for her student

Question

A professor at a local university finds that the mean quiz grade for her students is 72 with a standard deviation of 9. It is known that the grades are normally distributed. Complete the following question using Excel.

In blank #1 enter the probability that a student received a grade less than 65 on the quiz . (Round your answer to 4 decimal places and enter your answer as 0.1234 and not .1234.)

In blank #2 enter the probability that a student received a grade of at least 87 on the quiz. (Round your answer to 4 decimal places and enter your answer as 0.1234 and not .1234.)

In blank #3 enter the probability that a student received a grade between 60 and 80 on the quiz. (Round your answer to 4 decimal places and enter your answer as 0.1234 and not .1234.)

25% of the students received a grade more than what number? Enter your answer in blank #4.  (Round your answer to 2 decimal place.)

Question 11 options:

Explanation / Answer

= 72 = 9

Z = (X - ) /

1. X = 65

=> Z = (65 - 72) / 9

=> X = -7/9 = -0.7778

From the table, P(X < 65) = 0.2183

2. X = 87

Z = (87 - 72) / 9

= 15/9 = 1.6667

From the Z table, P(X < 87) = 0.9522

=> P(X >= 87) = 1 - 0.9522 = 0.0478.

3. X = 60

Z = (60 - 72) / 9

= -12/9 = -1.3333

From the table, P(Z < 60) = 0.0912

X = 80

Z = (80 - 72) / 9

= 8/9 = 0.8889

From the table P(Z < 80) = 0.8889

=> P(60 < Z < 80) = 0.8889 - 0.0912 = 0.7977.

4. Percentile = 0.75

Z value from table = -1.4395

X = 72 - 1.4395 * 9

= 59.0445

= 59.

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