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Fantom [35]
3 years ago
9

The annual rainfall in a certain region is approximately normally distributed with mean 41.4 inches and standard deviation 5.7 i

nches. Round answers to the nearest tenth of a percent.
Mathematics
1 answer:
stich3 [128]3 years ago
6 0

Complete question :

The annual rainfall in a certain region is approximately normally distributed with mean 41.4 inches and standard deviation 5.7 inches. Round answers to the nearest tenth of a percent. a) What percentage of years will have an annual rainfall of less than 43 inches? b) What percentage of years will have an annual rainfall of more than 39 inches? c) What percentage of years will have an annual rainfall of between 38 inches and 42 inches?

Answer:

0.61053

0.66314

0.26647

Step-by-step explanation:

Given that :

Mean (m) = 41.4 inches

Standard deviation (s) = 5.7 inches

a) What percentage of years will have an annual rainfall of less than 43 inches?

P(x < 43)

USing the relation to obtain the standardized score (Z) :

Z = (x - m) / s

Z = (43 - 41.4) / 5.7 = 0.2807017

p(Z < 0.2807) = 0.61053 ( Z probability calculator)

b) What percentage of years will have an annual rainfall of more than 39 inches? c)

P(x > 39)

USing the relation to obtain the standardized score (Z) :

Z = (x - m) / s

Z = (39 - 41.4) / 5.7 = −0.421052

p(Z > −0.421052) = 0.66314 ( Z probability calculator)

What percentage of years will have an annual rainfall of between 38 inches and 42 inches?

P(x < 38)

USing the relation to obtain the standardized score (Z) :

Z = (x - m) / s

Z = (38 - 41.4) / 5.7 = −0.596491

p(Z < −0.596491) = 0.27545 ( Z probability calculator)

P(x < 42)

USing the relation to obtain the standardized score (Z) :

Z = (x - m) / s

Z = (42 - 41.4) / 5.7 = 0.1052631

p(Z < 0.1052631) = 0.54192 ( Z probability calculator)

0.54192 - 0.27545 = 0.26647

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The market and Stock J have the following probability distributions:
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Answer:

1) E(M) = 14*0.3 + 10*0.4 + 19*0.3 = 13.9 \%

2) E(J)= 22*0.3 + 4*0.4 + 12*0.3 = 11.8 \%

3) E(M^2) = 14^2*0.3 + 10^2*0.4 + 19^2*0.3 = 207.1

And the variance would be given by:

Var (M)= E(M^2) -[E(M)]^2 = 207.1 -(13.9^2)= 13.89

And the deviation would be:

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And the variance would be given by:

Var (J)= E(J^2) -[E(J)]^2 = 194.8 -(11.8^2)= 55.56

And the deviation would be:

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Step-by-step explanation:

For this case we have the following distributions given:

Probability  M   J

0.3           14%  22%

0.4           10%    4%

0.3           19%    12%

Part 1

The expected value is given by this formula:

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

And replacing we got:

E(M) = 14*0.3 + 10*0.4 + 19*0.3 = 13.9 \%

Part 2

E(J)= 22*0.3 + 4*0.4 + 12*0.3 = 11.8 \%

Part 3

We can calculate the second moment first with the following formula:

E(M^2) = 14^2*0.3 + 10^2*0.4 + 19^2*0.3 = 207.1

And the variance would be given by:

Var (M)= E(M^2) -[E(M)]^2 = 207.1 -(13.9^2)= 13.89

And the deviation would be:

Sd(M) = \sqrt{13.89}= 3.73

Part 4

We can calculate the second moment first with the following formula:

E(J^2) = 22^2*0.3 + 4^2*0.4 + 12^2*0.3 =194.8

And the variance would be given by:

Var (J)= E(J^2) -[E(J)]^2 = 194.8 -(11.8^2)= 55.56

And the deviation would be:

Sd(M) = \sqrt{55.56}= 7.45

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