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galina1969 [7]
3 years ago
8

Consider the following three bond quotes: a Treasury bond quoted at 106:30, a corporate bond quoted at 96.25, and a municipal bo

nd quoted at 100.65. If the Treasury and corporate bonds have a par value of $1,000 and the municipal bond has a par value of $5,000, what is the price of these three bonds in dollars?
2 decimal places

Treasury ?

Corporate bond?

Municipal Bond?
Mathematics
1 answer:
boyakko [2]3 years ago
5 0

Answer:

The value of Treasury bond in dollars = $1,069.38

The value of Corporate bond in dollars = $962.50

The value of Municipal bond in dollars = $5,032.50

Step-by-step explanation:

Data provided in the question:

Treasury bond quoted at 106:30

corporate bond quoted at 96.25

municipal bond quoted at 100.65

Treasury bonds have a par value = $1,000

corporate bonds have a par value = $1,000

municipal bond has a par value = $5,000

Now,

The value of Treasury bond in dollars = (106+\frac{30}{32})\% × $1,000

= 106.9375% × $1,000

= $1,069.38

The value of Corporate bond in dollars = 96.25% × $1,000

= 0.9625 × $1,000

= $962.50

The value of Municipal bond in dollars = 100.65% × $5,000

= 1.0065 × $5,000

= $5,032.50

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yan [13]

Answer:

The 99% confidence interval for the scores of men is (64.406,75.594).

The 99% confidence interval for the scores of women is (73.609,90.391).

Step-by-step explanation:

We have the standard deviations for the sample, which means that the t-distribution will be used to solve this question.

The first step to solve this problem is finding how many degrees of freedom, we have. This is the sample size subtracted by 1. So

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99% confidence interval

Now, we have to find a value of T, which is found looking at the t table, with 24 degrees of freedom(y-axis) and a confidence level of 1 - \frac{1 - 0.99}{2} = 0.995. So we have T = 2.797.

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The margin of error is:

M = T\frac{s}{\sqrt{n}} = 2.797\frac{10}{\sqrt{25}} = 5.594

In which s is the standard deviation of the sample and n is the size of the sample.

The lower end of the interval is the sample mean subtracted by M. So it is 70 - 5.594 = 64.406

The upper end of the interval is the sample mean added to M. So it is 70 + 5.594 = 75.594

The 99% confidence interval for the scores of men is (64.406,75.594).

For women:

Standard deviation of 15.

The margin of error is:

M = T\frac{s}{\sqrt{n}} = 2.797\frac{15}{\sqrt{25}} = 8.391

In which s is the standard deviation of the sample and n is the size of the sample.

The lower end of the interval is the sample mean subtracted by M. So it is 82 - 8.391 = 73.609

The upper end of the interval is the sample mean added to M. So it is 82 + 8.391 = 90.391

The 99% confidence interval for the scores of women is (73.609,90.391).

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3 years ago
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