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Ksju [112]
3 years ago
14

A company that sells annuities must base the annual payout on the probability distribution of the length of life of the particip

ants in the plan. Suppose the probability distribution of the lifetimes of the participants is approximately a normal distribution with a mean of 68 years and a standard deviation of 3.5 years (Show your workings clearly).
a) What proportion of the plan recipients would receive payments beyond age 75?

b) What proportion of the plan recipients die before they reach the standard retirement age of 65?

c) Find the age at which payments have ceased for approximately 86% of the plan participants.
Mathematics
1 answer:
Svetlanka [38]3 years ago
5 0

Answer:

Step-by-step explanation:

Let X be length of life of the participants in the plan.

Given that X is N(68,3.5)

We convert this to standard normal score z using

z=\frac{x-68}{3.5}

a)  proportion of the plan recipients that would receive payments beyond age 75=P(X\geq 75) = P(Z\geq 2)\\= 0.025

b) proportion of the plan recipients die before they reach the standard retirement age of 65=P(X\leq 65) = P(z\leq -0.86)\\=0.5-0.2764\\=0.2236

c) x for 86% ceased

P(Z

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Answer:

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6 0
3 years ago
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What is tan(θ/2), when tan(θ) = 4/3 and when π &lt; θ &lt; 3π/2?
Tamiku [17]
Tan (Ф/2)=⁺₋√[(1-cosФ)/(1+cosФ)]
if π<Ф<3π/2;
 then,  Where is Ф/2??
π/2<Ф/2<3π/4; therefore Ф/2 is in the second quadrant; then tan (Ф/2) will have a negative value.

tan(Ф/2)=-√[(1-cosФ)/(1+cosФ)]

Now, we have to find the value of cos Ф.
tan (Ф)=4/3
1+tan²Ф=sec²Ф
1+(4/3)²=sec²Ф
sec²Ф=1+16/9
sec²Ф=(9+16)/9
sec²Ф=25/9
sec Ф=-√(25/9)        (sec²Ф will have a negative value, because Ф is in the        sec Ф=-5/3                                 third quadrant).

cos Ф=1/sec Ф
cos Ф=1/(-5/3)
cos Ф=-3/5

Therefore:

tan(Ф/2)=-√[(1-cosФ)/(1+cosФ)]
tan(Ф/2)=-√[(1+3/5)/(1-3/5)]
tan(Ф/2)=-√[(8/5)/(2/5)]
tan(Ф/2)=-√4
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4 years ago
Which equation gives the line shown on the graph?
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QUESTION 1 The driving distance for the top 100 golfers on the PGA tour is between 284.7 and 310.6 yards.Assume that the driving
GarryVolchara [31]

Answer:

P(X>300)

And we can find this probability with the complement rule:

P(X>300)= 1-P(X

Step-by-step explanation:

For this case we define the random variable X ="driving distance for the top 100 golfers on the PGA tour" and we know that:

X \sim Unif (a=284.7, b=310.6)

And for this case the probability density function is given by:

f(x) =\frac{1}{310.6 -284.7} =0.0386 , 284.7 \leq X \leq 310.6

And the cumulative distribution function is given by:

F(x) =\frac{x-284.7}{310.6-284.7} , 284.7 \leq X \leq 310.6

And we want to find this probability:

P(X>300)

And we can find this probability with the complement rule:

P(X>300)= 1-P(X

6 0
3 years ago
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