If interest is compounded annually, the account balance will be
.. 4400*1.0825^29 ≈ 43,839.13 . . . . dollars
Answer:
And using the probability mass function we got:
Step-by-step explanation:
Let X the random variable of interest, on this case we now that:
The probability mass function for the Binomial distribution is given as:
Where (nCx) means combinatory and it's given by this formula:
And we want to find the following probability:
And using the probability mass function we got:
5.7y-5.2=y/2.5
Add 5.2 to both sides:
5.7y = y/2.5 + 5.2
y/2.5 = 0.4y
5.7y = 0.4y + 5.2
Subtract 0.4y from both sides:
5.3y = 5.2
Divide both sides by 5.3:
y = 5.2/5.3
y = 0.98113
The number of tests that it would take for the probability of committing at least one type I error to be at least 0.7 is 118 .
In the question ,
it is given that ,
the probability of committing at least , type I error is = 0.7
we have to find the number of tests ,
let the number of test be n ,
the above mentioned situation can be written as
1 - P(no type I error is committed) ≥ P(at least type I error is committed)
which is written as ,
1 - (1 - 0.01)ⁿ ≥ 0.7
-(0.99)ⁿ ≥ 0.7 - 1
(0.99)ⁿ ≤ 0.3
On further simplification ,
we get ,
n ≈ 118 .
Therefore , the number of tests are 118 .
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Answer:
Step-by-step explanation:
5.364
