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madreJ [45]
2 years ago
6

10. The continuously compounded annual return on a stock is normally distributed with a mean of 20% and standard deviation of 30

%. With 95.44% confidence, we should expect its actual return in any particular year to be between which pair of values
Mathematics
1 answer:
ehidna [41]2 years ago
7 0

Answer: We should expect its actual return in any particular year to be between<u> -40%</u> and<u> 80%</u>.

Step-by-step explanation:

Given : The continuously compounded annual return on a stock is normally distributed with a mean 20% and standard deviation of 30%.

From normal z-table, the z-value corresponds to 95.44 confidence is 2.

Therefore , the interval limits for 95.44 confidence level will be :

Lower limit = Mean -2(Standard deviation) = 20% -2(30%)= 20%-60%=-40%

Upper limit =  Mean +2(Standard deviation)=20% +2(30%)= 20%+60%=80%

Hence, we should expect its actual return in any particular year to be between<u> -40%</u> and<u> 80%</u>.

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Answer:

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Using the condition given:

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With K a constant. For this case the period of a pendulumn is given by this general expression:

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If we square both sides of the equation we got:

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And solving for L we got:

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3) T = 2\pi \sqrt{\frac{0.98m}{9.8\frac{m}{s^2}}}= 1.987 s

Step-by-step explanation:

Part 1

For this case we know the following info: The length, l cm, of a simple pendulum is directly proportional to the square of its period (time taken to complete one oscillation), T seconds.

L \propto T^2

Using the condition given:

2.205 m = K (3)^2

K = 0.245 \approx \frac{g}{4\pi^2}

So then if we want to create an equation we need to do this:

L = K T^2

With K a constant. For this case the period of a pendulumn is given by this general expression:

T = 2\pi \sqrt{\frac{L}{g}}

Where L is the length in m and g the gravity g = 9.8 \frac{m}{s^2}.

Part 2

For this case using the function in part a we got:

T = 2\pi \sqrt{\frac{L}{g}}

If we square both sides of the equation we got:

T^2 = 4 \pi^2 \frac{L}{g}

And solving for L we got:

L = \frac{g T^2}{4 \pi^2}

Replacing we got:

L =\frac{9.8 \frac{m}{s^2} (5s)^2}{4 \pi^2} = 6.206m

Part 3

For this case using the function in part a we got:

T = 2\pi \sqrt{\frac{L}{g}}

Replacing we got:

T = 2\pi \sqrt{\frac{0.98m}{9.8\frac{m}{s^2}}}= 1.987 s

8 0
2 years ago
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