4^2 -6 (4 to the second power minus 6) *
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Answer:
(a) The probability that <em>Y</em>₂₀ exceeds 1000 is 3.91 × 10⁻⁶.
(b) <em>n</em> = 28.09
Step-by-step explanation:
The random variable <em>Y</em>ₙ is defined as the total numbers of dollars paid in <em>n</em> years.
It is provided that <em>Y</em>ₙ can be approximated by a Gaussian distribution, also known as Normal distribution.
The mean and standard deviation of <em>Y</em>ₙ are:
(a)
For <em>n</em> = 20 the mean and standard deviation of <em>Y</em>₂₀ are:
Compute the probability that <em>Y</em>₂₀ exceeds 1000 as follows:
**Use a <em>z </em>table for probability.
Thus, the probability that <em>Y</em>₂₀ exceeds 1000 is 3.91 × 10⁻⁶.
(b)
It is provided that P (<em>Y</em>ₙ > 1000) > 0.99.
The value of <em>z</em> for which P (Z < z) = 0.01 is 2.33.
Compute the value of <em>n</em> as follows:
The last equation is a quadratic equation.
The roots of a quadratic equation are:
a = 16
b = -805.4289
c = 10000
On solving the last equation the value of <em>n</em> = 28.09.