Jerry has taken a random sample of students and determined the number of electives that each student in his sample took last yea
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Using the normal distribution, it is found that the percentages are given as follows:
- 3.59% of the grades will be A.
- 9.98% of the grades will be B.
- 74.92% of the grades will be C.
- 8.64% of the grades will be D.
- 2.87% of the grades will be F.
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 prorportion of students who receive an A is <u>one subtracted by the p-value of Z = 1.8</u>.
Looking at the z-table, Z = 1.8 has a p-value of 0.9641.
1 - 0.9641 = 0.0359.
3.59% of the grades will be A.
For B, it is the <u>p-value of Z = 1.8 subtracted by the p-value of Z = 1.1</u>, hence:
0.9641 - 0.8643 = 0.0998.
9.98% of the grades will be B.
For C, it is the <u>p-value of Z = 1.1 subtracted by the p-value of Z = -1.2</u>, hence:
0.8643 - 0.1151 = 0.7492.
74.92% of the grades will be C.
For D, it is the <u>p-value of Z = -1.2 subtracted by the p-value of Z = -1.9</u>, hence:
0.1151 - 0.0287 = 0.0864.
8.64% of the grades will be D.
For F, it is the <u>p-value of Z = -1.9</u>, hence 2.87% of the grades will be F.
More can be learned about the normal distribution at brainly.com/question/15181104
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