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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Answer:y+2.9=11, 2y+1=17.2, 28.4-y=20.3
Step-by-step explanation:
The best option is A) d - 90.
Expressions has no equal sign nor an inequality symbol. An expression has only standard or normal numbers, with variables and some basic mathematical operations.
A ratio that compares two quantities measured in different units is a conversion factor!
Answer:
a) 0.96
b) 0.016
c) 0.018
d) 0.982
e) x = 2
Step-by-step explanation:
We are given with the Probability density function f(x)= 2/x^3 where x > 1.
<em>Firstly we will calculate the general probability that of P(a < X < b) </em>
P(a < X < b) = 
= 
= ![2[ \frac{x^{-3+1} }{-3+1}]^{b}_a dx](https://tex.z-dn.net/?f=2%5B%20%5Cfrac%7Bx%5E%7B-3%2B1%7D%20%7D%7B-3%2B1%7D%5D%5E%7Bb%7D_a%20%20%20dx)
{ Because ![\int_{a}^{b} x^{n} dx = [ \frac{x^{n+1} }{n+1}]^{b}_a](https://tex.z-dn.net/?f=%5Cint_%7Ba%7D%5E%7Bb%7D%20x%5E%7Bn%7D%20dx%20%3D%20%5B%20%5Cfrac%7Bx%5E%7Bn%2B1%7D%20%7D%7Bn%2B1%7D%5D%5E%7Bb%7D_a)
}
= ![2[ \frac{x^{-2} }{-2}]^{b}_a](https://tex.z-dn.net/?f=2%5B%20%5Cfrac%7Bx%5E%7B-2%7D%20%7D%7B-2%7D%5D%5E%7Bb%7D_a)
= ![\frac{2}{-2} [ x^{-2} ]^{b}_a](https://tex.z-dn.net/?f=%5Cfrac%7B2%7D%7B-2%7D%20%5B%20x%5E%7B-2%7D%20%5D%5E%7Bb%7D_a)
= ![-1 [ b^{-2} - a^{-2} ]](https://tex.z-dn.net/?f=-1%20%5B%20b%5E%7B-2%7D%20-%20a%5E%7B-2%7D%20%20%5D)
= 
a) Now P(X < 5) = P(1 < X < 5) {because x > 1 }
Comparing with general probability we get,
P(1 < X < 5) = 
= 
= 0.96 .
b) P(X > 8) = P(8 < X < ∞) = 1/
- 1/∞ = 1/64 - 0 = 0.016
c) P(6 < X < 10) = 
= 
= 0.018 .
d) P(x < 6 or X > 10) = P(1 < X < 6) + P(10 < X < ∞)
= 
+ (1/
- 1/∞) = 1 - 1/36 + 1/100 + 0 = 0.982
e) We have to find x such that P(X < x) = 0.75 ;
⇒ P(1 < X < x) = 0.75
⇒ 
= 0.75
⇒ 
= 1 - 0.75 = 0.25
⇒ 
= 
⇒ = 4 ⇒ x = 
Therefore, value of x such that P(X < x) = 0.75 is 2.
