If scores on an exam follow an approximately normal distribution with a mean of 76.4 and a standard deviation of 6.1 points, then the minimum score you would need to be in the top 2% is equal to 88.929.
A problem of this type in mathematics can be characterized as a normal distribution problem. We can use the z-score to solve it by using the formula;
Z = x - μ / σ
In this formula the standard score is represented by Z, the observed value is represented by x, the mean is represented by μ, and the standard deviation is represented by σ.
The p-value can be used to determine the z-score with the help of a standard table.
As we have to find the minimum score to be in the top 2%, p-value = 0.02
The z-score that is found to correspond with this p-value of 0.02 in the standard table is 2.054
Therefore,
2.054 = x - 76.4 ÷ 6.1
2.054 × 6.1 = x - 76.4
12.529 = x - 76.4
12.529 + 76.4 = x
x = 88.929
Hence 88.929 is calculated to be the lowest score required to be in the top 2%.
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Yea you did jsixidjdjiejs
Hello there!
The correct answer is option B (-3,3)
-3x - y = 6
-3(-3) - 3 = 6
9 - 3 = 6
6 = 6
Thus, the correct answer is option B
Good luck with your studied!
Answer:
the first one
Step-by-step explanation:
In the given question, there are numerous information's already provided. It is important to note then down first. With the help of those given information's the required answer can be easily reached.
Percentage of students that weighed 140 pounds = 75 percent
Then
Percentage of students that weighed more than 140 pounds = (100 - 75) percent
= 25 percent
Total number of students that were weighed = 40 students
Total number of students that weighed more than 140 pounds = (25/100) * 40
= (40/4) students
= 10 students
So the number of students that weighed more than 140 pounds is 10. So the correct option in regards to the given question is option "1".
