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Lemur [1.5K]
3 years ago
6

Sam takes out a $167,000 mortgage for 20 years. He makes monthly payments and at the end calculates

Mathematics
1 answer:
cupoosta [38]3 years ago
8 0

Answer:

The Annual rate of interest for the mortgage is 1.8%

Step-by-step explanation:

Given as :

The mortgage principal = p = $167,000

The time period of mortgage = t = 20 years

The Amount paid towards mortgage in 20 years = A = $240,141

Let the Annual percentage rate on interest = r % compounded annually

Now, <u>From Compound Interest method</u>

Amount = Principal × (1+\dfrac{\textrm rate}{100})^{\textrm time}

Or, A = p × (1+\dfrac{\textrm r}{100})^{\textrm t}

Or, $240,141 = $167,000 × (1+\dfrac{\textrm r}{100})^{\textrm 20}

or, \dfrac{240,141}{167,000} =  (1+\dfrac{\textrm r}{100})^{\textrm 20}

Or , 1.437 =  (1+\dfrac{\textrm r}{100})^{\textrm 20}

Or, (1.437)^{\frac{1}{20}} = (1+\dfrac{r}{100})

or, 1.018 = (1+\dfrac{r}{100})

Or, \dfrac{r}{100} =  1.018 - 1

Or, \dfrac{r}{100} = 0.018

∴ r = 0.018 × 100

i.e r = 1.8

So, The rate of interest applied = r = 1.8 %

Hence, The Annual rate of interest for the mortgage is 1.8%  Answer

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A random sample of 10 parking meters in a resort community showed the following incomes for a day. Assume the incomes are normal
GenaCL600 [577]

Answer:

A 95% confidence interval for the true mean is [$3.39, $6.01].

Step-by-step explanation:

We are given that a random sample of 10 parking meters in a resort community showed the following incomes for a day;

Incomes (X): $3.60, $4.50, $2.80, $6.30, $2.60, $5.20, $6.75, $4.25, $8.00, $3.00.

Firstly, the pivotal quantity for finding the confidence interval for the population mean is given by;

                         P.Q.  =  \frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }  ~ t_n_-_1

where, \bar X = sample mean income = \frac{\sum X}{n} = $4.70

            s = sample standard deviation = \sqrt{\frac{\sum (X-\bar X)^{2} }{n-1} }  = $1.83

            n = sample of parking meters = 10

            \mu = population mean

<em>Here for constructing a 95% confidence interval we have used a One-sample t-test statistics because we don't know about population standard deviation.</em>

<u>So, 95% confidence interval for the population mean, </u>\mu<u> is ;</u>

P(-2.262 < t_9 < 2.262) = 0.95  {As the critical value of t at 9 degrees of

                                            freedom are -2.262 & 2.262 with P = 2.5%}  

P(-2.262 < \frac{\bar X-\mu}{\frac{s}{\sqrt{n} } } < 2.262) = 0.95

P( -2.262 \times {\frac{s}{\sqrt{n} } } < {\bar X-\mu < 2.262 \times {\frac{s}{\sqrt{n} } } ) = 0.95

P( \bar X-2.262 \times {\frac{s}{\sqrt{n} } } < \mu < \bar X+2.262 \times {\frac{s}{\sqrt{n} } } ) = 0.95

<u>95% confidence interval for</u> \mu = [ \bar X-2.262 \times {\frac{s}{\sqrt{n} } } , \bar X+2.262 \times {\frac{s}{\sqrt{n} } } ]

                                         = [ 4.70-2.262 \times {\frac{1.83}{\sqrt{10} } } , 4.70+ 2.262 \times {\frac{1.83}{\sqrt{10} } } ]

                                         = [$3.39, $6.01]

Therefore, a 95% confidence interval for the true mean is [$3.39, $6.01].

The interpretation of the above result is that we are 95% confident that the true mean will lie between incomes of $3.39 and $6.01.

Also, the margin of error  =  2.262 \times {\frac{s}{\sqrt{n} } }

                                          =  2.262 \times {\frac{1.83}{\sqrt{10} } }  = <u>1.31</u>

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Answer:

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Step-by-step explanation:

We are given the following data in the question:

10, 11, 12, 13, 13, 13, 14, 15, 16, 16, 17, 18, 18, 19, 20

Formula:

\text{Standard Deviation} = \sqrt{\displaystyle\frac{\sum (x_i -\bar{x})^2}{n-1}}  

where x_i are data points, \bar{x} is the mean and n is the number of observations.  

Mean = \displaystyle\frac{\text{Sum of all observations}}{\text{Total number of observation}}

Mean =\displaystyle\frac{225}{15} = 15

Sum of squares of differences = 25 + 16 + 9 + 4 + 4+ 4 + 1 + 0+ 1+ 1 + 4 + 9 + 9+ 16 + 25 = 128

S.D = \sqrt{\dfrac{128}{14}} = 3.02

Empirical rule:

  • According to this rule almost all the data lies within three standard deviation of the mean for a normal distribution.
  • About 68% of data lies within one standard deviation of the mean.
  • About 95% of data lies within two standard deviations of mean.
  • Arround 99.7% of data lies within three standard deviation of mean.

Thus, by empirical rule,

\mu \pm 1\sigma = 15\pm (3.02) = (11.98,18.02)

According to the Empirical Rule, 68% of the data should fall between 11.98 and 18.02

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Answer:

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2 years ago
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