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VARVARA [1.3K]
3 years ago
15

Charlotte purchased a pool for $7,680 using a six month deferred payment plan with an interest rate of 20.45%. she did not make

any payments during the deferment period. what will Charlotte's monthly payment be if she must pay off the pool within three years after the deferment period?
Mathematics
1 answer:
sveta [45]3 years ago
6 0

There are many ways to solve this question. The basic idea is to assume that Charlotte makes a monthly payment of x and equate the amount she pays to the amount she owes, after adjusting both amounts with interest to be at the same point of time. In the following, we equate the amounts at the end of the term. 
 
 Initially, Charlotte owes $7680. She finishes her payments after a total of 6 + 36 = 42 months. Using a simple compounding formula, the amount she owes is worth P at the end of 42 months, where P is: 
 
 P = 7680 * (1 + .2045/12)^42 = 15616.67379

 Now, the first installment she pays (at the end of six months) is paid 35 months in advance of the end, so it is worth x * (1 + .2375/12)^35 at the end of her loan period. 

 Similarly, the second installment is worth x * (1 + .2375/12)^34 at the end of the loan period. 

 Continuing, this way, the last installment is worth exactly x at the end of the loan period. 

 So, the total amount she paid equals: 
 x [(1 + .2375/12)^35 + (1 + .2375/12)^34 + ... + (1 + .2375/12)^0] 

 To calculate this, assume that 1+.2045/12 = a. Then the amount Charlotte pays is: 
 x (a^35 + a^34 + ... + a^0) = x (a^36 - 1)/(a - 1) 
 Clearly, this value must equal P, so we have: 
 x (a^36 - 1)/(a - 1) = P = 15616.67379
 Substituting, a = 1 + .2045/12 and solving, we get 
 x = 317.82

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The average life of a bread-making machine is 7 years, with a standard deviation of 1 year. Assuming that the lives of these mac
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Answer:

a) P(6.4

b) a=7 +1.036*0.333=7.345

So the value of bread-making machine that separates the bottom 85% of data from the top 15% is 7.345.

Step-by-step explanation:

Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".  

The central limit theorem states that "if we have a population with mean μ and standard deviation σ and take sufficiently large random samples from the population with replacement, then the distribution of the sample means will be approximately normally distributed. This will hold true regardless of whether the source population is normal or skewed, provided the sample size is sufficiently large".

Let X the random variable life of a bread making machine. We know from the problem that the distribution for the random variable X is given by:

X\sim N(\mu =7,\sigma =1)

We take a sample of n=9 . That represent the sample size.

From the central limit theorem we know that the distribution for the sample mean \bar X is also normal and is given by:

\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})

\bar X \sim N(\mu=7, \frac{1}{\sqrt{9}})

Solution to the problem

Part a

(a) the probability that the mean life of a random sample  of 9 such machines falls between 6.4 and 7.2

In order to answer this question we can use the z score in order to find the probabilities, the formula given by:

z=\frac{\bar X- \mu}{\frac{\sigma}{\sqrt{n}}}

The standard error is given by this formula:

Se=\frac{\sigma}{\sqrt{n}}=\frac{1}{\sqrt{9}}=0.333

We want this probability:

P(6.4

Part b

b) The value of x to the right of which 15% of the  means computed from random samples of size 9 would fall.

For this part we want to find a value a, such that we satisfy this condition:

P(\bar X>a)=0.15   (a)

P(\bar X   (b)

Both conditions are equivalent on this case. We can use the z score again in order to find the value a.  

As we can see on the figure attached the z value that satisfy the condition with 0.85 of the area on the left and 0.15 of the area on the right it's z=1.036. On this case P(Z<1.036)=0.85 and P(Z>1.036)=0.15

If we use condition (b) from previous we have this:

P(X  

P(z

But we know which value of z satisfy the previous equation so then we can do this:

z=1.036

And if we solve for a we got

a=7 +1.036*0.333=7.345

So the value of bread-making machine that separates the bottom 85% of data from the top 15% is 7.345.

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