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:
P( top two horses are predicted incorrectly in incorrect order)
= 
Step-by-step explanation:
In the horse race the outcome can be predicted in 5! = 120 ways.
Now suppose the top two horses were predicted incorrectly in incorrect order. Now, the top horse can be predicted incorrectly in 4 ways.
Suppose the top horse was predicted to be in k-th position where k = 2, 3 ,4,5
so the second horse can be predicted to be in place from 1 to (k - 1)
So, the top two horses can be predicted incorrectly in incorrect order
in 
= 10 ways
and for each prediction of the two the remaining horses may be predicted in 3! = 6 ways.
Hence ,
P( top two horses are predicted incorrectly in incorrect order)
= 
=
Answer:
- 3, - 11, - 35, - 107
Step-by-step explanation:
Using the recursive formula with f(1) = - 3 , then
f(2) = - 2 + 3f(1) = - 2 + 3(- 3) = - 2 - 9 = - 11
f(3) = - 2 + 3f(2) = - 2 + 3(- 11) = - 2 - 33 = - 35
f(4) = - 2 + 3f(3) = - 2 + 3(- 35) = - 2 - 105 = - 107
Thus the first four terms are
- 3, - 11, - 35, - 107
45-9=36/18=2 I hope this helps
