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Zinaida [17]
3 years ago
15

Many people believe that the daily change of price of a company's stock on the stock market is a random variable with mean 0 and

variance 2. That is, if Yn represents the price of the stock on the n-th day, then Yn = Yn-1 + Xn, n > 1 where X1, X2, . . . are independent and identically distributed random variables with mean 0 and variance 2. Suppose that the stock's price today is 100. If 2 = 1, use CLT to approximate the probability that the stock's price will exceed 105 on the 10-th day?
Mathematics
1 answer:
Tomtit [17]3 years ago
8 0

Answer:

Step-by-step explanation:

We can use normal aproximation, assuming that the random variables are a lot of that means the sample size is large.

P(x>105) = P(z>\frac{105-100}{1})=P(z>5)

Using the normal distribution table,

P(z>5) = 0.00005

Hence, we can conclude that the probability that the stock’s price will exceed 105 after 10 days is very small.

Hope this helps!

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3 0
3 years ago
A rock was thrown from the top of a cliff such that its distance above sea level was given by s(t)= at² + bt + c, where t is the
frutty [35]

Answer:

<h2>a) a = -3, b = 18, c = 48; </h2><h2>s(t) = -3t²+18t+48</h2><h2>b) 48m</h2><h2>c) 8secs</h2>

Step-by-step explanation:

The question is incomplete. Here is the complete question.

'A rock was thrown from the top of a cliff such that its distance above sea level was given by s(t)=at²+bt+c, where t is the time in seconds after the rock was released. After 1 second the rock was 63 m above sea level, after 2 seconds 72 m, and after 7 seconds 27 m. a. Find a, b and c and hence an expression for s(t). b. Find the height of the cliff. c. Find the time taken for the rock to reach sea level.'

Given the equation of the distance modelled as s(t)=at²+bt+c

If after 1 second the rock was 63 m, then 63 = a+b+c

If after 2 seconds, the distance was 72 m then 72 = 4a+2b+c

Also if after 7 seconds, the distance is 27 m, then 27 = 49a+7b+c

i) Solving the 3 equations simultaneously to get a, b and c we have;

a+b+c = 63 ... 1

4a+2b+c = 72 ...2

49a+7b+c = 27 ...3

Subtracting 2 from 1 and 3 from 2 we will generate 2 new equations as shown;

eqn 2- eqn1: 3a + b = 9...4

eqn 3- eqn 2: 45a + 5b = -45

eqn 3- eqn 2: 9a+b = -9 ... 5

solving 4 and 5 simultaneously

3a + b = 9 ...4

9a+b = -9 ... 5  

Taking the difference of 4 and 5 we have

6a - -9-9

6a = -18

a = -3

substituting a = -3 into equation 4 to get b we have;

3(-3)+b = 9

-9 + b = 9

b = 9+9

b = 18

substituting a = -3 and b = 18 into equation 1 to get c we have;

-3+18+c = 63

15+c = 63

c = 48

a = -3, b= 18 and c = 48

The distance function will be s(t) = -3t²+18t+48

ii) If the height of the cliff is modelled by the equation  s(t)=at²+bt+c

The height of the cliff is at when t = 0

s(0) = -3(0)²+18(0)+48

s(0) = 48m

The height of the cliff is 48m

iii) At the sea level, the height of the rock will be 0m, substituting this into the modeled equation for the height to get the time we have;

s(t)=at²+bt+c

0 = -3t²+18t+48

3t²-18t-48 = 0

t² - 6t - 16 =0

t² - 8t+2t - 16 = 0

t(t-8)+2(t-8) = 0

(t+2)(t-8) = 0

t = -2 or 8

Taking the positive value of the time, t = 8secs

Time taken for the rock to reach sea level is 8secs

7 0
3 years ago
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