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yaroslaw [1]
3 years ago
6

A company is producing christmas light strings containing 20 small bulbs. Suppose the success or failure of a bulb is independen

t of the success or failure of other bulbs and the probability that a bulb is lightning properly is 0.92.
a. When you order a christmas light string, what is the probability of at most 3 of those small bulbs are not lightning properly.
b. Find the mean and the variance of bulbs in a string that are not lightning properly.
Business
1 answer:
monitta3 years ago
6 0

Answer:

Explanation:

Number of bulbs = 20

Success of a bulb lightning properly = 0.92

Success of a bulb not lightning properly = 0.08

We have to find the probability of atmost 3 bulbs not lightning properly -

The random Variable X follows binomial distribution

= 0.929385

Mean and variance of bulbs in a string that are not lightning properly is given by

Mean of binomial distribution or E(X) = n*p = 20*0.08 = 1.6

Variance of binomial distribution or V(X) = n*p*(1-p) = 20*0.08*0.92 = 1.472

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You want to be able to withdraw the specified amount periodically from a payout annuity with the given terms. Find how much the
SpyIntel [72]

The question is incomplete. The complete question is :

You want to be able to withdraw the specified amount periodically from a payout annuity with the given terms. Find how much the account needs to hold to make this possible. Round your answer to the nearest dollar.

Regular withdrawal    $ 2200

Interest rate                        2%

Frequency                   Monthly

Time                                20 years

Solution :

Given :

Monthly withdrawal = $ 2200

Interest rate = 2%

Frequency = monthly

Time = 20 years

        = 20 x 12 = 240 months

Formula used :

$w=\frac{[PZ^{r-1}(Z-1)]}{[Z^Y-1]}$         with Z = 1 + r

where, w = monthly withdrawal

P = principal amount

r = monthly interest rate

Y = Number of months

So, w = 2200

     r = 2% = 0.02

     Z = 1 + r

        = 1 + 0.02 = 1.02

Y = 240

Therefore,

$2200=\frac{P(1.02)^{240-1}(1.02-1)}{(1.02)^{240-1}(1.02-1)}$

$P=\frac{2200(115.888-1)}{113.6164(0.02)}$

   = 111,231829

   ≈ 111,232 (rounding off)

Thus, the account balance = $ 111,232

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