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Sati [7]
3 years ago
11

g If the economy improves, a certain stock stock will have a return of 23.4 percent. If the economy declines, the stock will hav

e a loss of 11.9 percent. The probability of the economy improving is 67 percent while the probability of a recession is 33 percent. What is the standard deviation of the returns on the stock
Mathematics
1 answer:
dusya [7]3 years ago
7 0

Answer:

E(X) = 23.4* 0.67 -11.9*0.33= 11.759 \%

Now we can find the second central moment with this formula:

E(X^2) = \sum_{i=1}^n X^2_i P(X_i)

And replacing we got:

E(X^2) = (23.4)^2* 0.67 +(-11.9)^2*0.33= 413.5965

And the variance is given by:

Var(X) = E(X^2) - [E(X)]^2

And replacing we got:

Var(X) = 413.5965 -(11.759)^2 =275.5105

And finally the deviation would be:

Sd(X) = \sqrt{275.5105}= 16.599 \%

Step-by-step explanation:

We can define the random variable of interest X as the return from a stock and we know the following conditions:

X_1 = 23.4 , P(X_1) =0.67 represent the result if the economy improves

X_2 = -11.9 , P(X_1) =0.33 represent the result if we have a recession

We want to find the standard deviation for the returns on the stock. We need to begin finding the mean with this formula:

E(X) = \sum_{i=1}^n X_i P(X_i)

And replacing the data given we got:

E(X) = 23.4* 0.67 -11.9*0.33= 11.759 \%

Now we can find the second central moment with this formula:

E(X^2) = \sum_{i=1}^n X^2_i P(X_i)

And replacing we got:

E(X^2) = (23.4)^2* 0.67 +(-11.9)^2*0.33= 413.5965

And the variance is given by:

Var(X) = E(X^2) - [E(X)]^2

And replacing we got:

Var(X) = 413.5965 -(11.759)^2 =275.5105

And finally the deviation would be:

Sd(X) = \sqrt{275.5105}= 16.599 \%

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3 years ago
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3 years ago
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4 0
4 years ago
Do you tailgate the car in front of you? About 35% of all drivers will tailgate before passing, thinking they can make the car i
Vesna [10]

Answer:

(a) The histogram is shown below.

(b) E (X) = 4.2

(c) SD (X) = 2.73

Step-by-step explanation:

Let <em>X</em> = <em>r</em><em> </em>= a driver will tailgate the car in front of him before passing.

The probability that a driver will tailgate the car in front of him before passing is, P (X) = <em>p</em> = 0.35.

The sample selected is of size <em>n</em> = 12.

The random variable <em>X</em> follows a Binomial distribution with parameters <em>n</em> = 12 and <em>p</em> = 0.35.

The probability function of a binomial random variable is:

P(X=x)={n\choose x}p^{x}(1-p)^{n-x}

(a)

For <em>X</em> = 0 the probability is:

P(X=0)={12\choose 0}(0.35)^{0}(1-0.35)^{12-0}=0.006

For <em>X</em> = 1 the probability is:

P(X=1)={12\choose 1}(0.35)^{1}(1-0.35)^{12-1}=0.037

For <em>X</em> = 2 the probability is:

P(X=2)={12\choose 2}(0.35)^{2}(1-0.35)^{12-2}=0.109

Similarly the remaining probabilities will be computed.

The probability distribution table is shown below.

The histogram is also shown below.

(b)

The expected value of a Binomial distribution is:

E(X)=np

The expected number of vehicles out of 12 that will tailgate is:

E(X)=np=12\times0.35=4.2

Thus, the expected number of vehicles out of 12 that will tailgate is 4.2.

(c)

The standard deviation of a Binomial distribution is:

SD(X)=np(1-p)

The standard deviation of vehicles out of 12 that will tailgate is:

SD(X)=np(1-p)=12\times0.35\times(1-0.35)=2.73\\

Thus, the standard deviation of vehicles out of 12 that will tailgate is 2.73.

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3 years ago
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