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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If you are graphing quadratic equations of the type
ax^2 + bx + c = 0
The equation will look like a "U" <span>if "a" is positive </span>or it will look like an upside-down "U" <span>if "a" is negative </span>
Since the population is going to increase by a certain percentage, we have to add 
to the percentage to keep the original population.
Now, convert the percentage into a decimal. You can do this by moving the decimal place two places to the left.
Now, just multiply that by 
.
X=y-2
2y=5x-17
Substituting for x into the second equation, we find
2y=5(y-2)-17
2y=5y-10-17
-3y=-27
y=9
So x=7
Thus, the two numbers are 7 and 9.
Since
, we know that
follows a Poisson distribution with parameter
.
Now assuming
denote the mean and standard deviation of
, respectively, then we know right away that
and
.
So,
