Question

9. An investment analyst ranks a mutual fund by analyzing the type of investment made by the fund. According to the analyst's investigation of this fund, she constructs a probability distribution where X is the risk of the investment. X P(X) 0.27 0.30 0.12 2 5 0.11 a) Determine the average risk of this fund. b) Determine the population standard dev to the nearest hundredth. c) Using your results from parts a and b, how would you classify this fund's risk, using the following classifications: lowest risk is 1, medium is 3, and highest risk is 5. d) What is the probability of a risk less than 3 e) What is the probaility of a risk between 2 and 4 inclusive. 1) What is the probability of a risk greater than S.

Answer 1

We know that

ΣΡ(x) =1

In the given Probability Disribution the probability value of 4 has not given. So; we find that as follows.

ΣΡ(x) =1

→ 0.27 + 0.30 + 0.12 + 0.11 + P(4)

→ 0.80 + P(4) = 1

P(4) = 1 0.80

P(4) = 0.20

AVERAGE (i.e EXPECTATION )

Ε(α) = ΣαΡ()

VARIANCE:

VI) E(22) - (E(2)

= E(.2) = Σ2P()

STANDARD DEVIATION:

S.D = V VARIANCE

x	P(x)	1* PC	.22 * P.)
1	0.27	0.27	0.27
2	0.3	0.6	1.2
3	0.12	0.36	1.08
4	0.2	0.8	3.2
5	0.11	0.55	2.75
TOTAL	1	2.58	8.5

Therefore

Ε(α) = ΣπP(x) = 2.58

:: E(C) = 2.58

:: E(2) = H = 2.58

(E(2))2 = (2.58)

:.(E (2) 6.6564

Also we got

X.rºPC) = 8.5

V(2) = E(22) - (EL))

VOC) 8.5 – 6.6564

VC) 1.8436

S.D= VARIANCE

S.D= 1.8436

S.D= 1.3578

S.D=0= 1.3578

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(c) We know that the Standard Normal Variate is Z

- Z

Substiture X = 1

Z = 1 – 2.58 1.3578

Z = -1.1638

Therefore (1.16)S.D is Below the Mean

Substitute X = 3

Z = 3 – 2.58 1.3578

Z = 0.3093

Therefore (0.31)S.D is above from Mean

Substitute X = 5

Z 5 – 2.58 1.3578

Z = 1.7823

Therefore (1.78)S.D is above the Mean

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(d) P(X < 3) = ?

Z = 3 – 2.58 1.3578

Z=0.3093

Z= 0.31

P(X<3) = P(Z < 0.31) =?

P(Z < 0.31) =?

0.5 0.5 0.3 -3 -2 +3 -1 Z=0 +1 +2

We have to find out the Probability of Shaded Area

P(Z < 0.31) = P(-00<7 < +0.31)

→ P(Z < 0.31) = P(-00 < Z < 0) + P(0< Z < +0.31)

→ P(Z < 0.31) = 0.5 + 0.1217

P(Z < 0.31) = 0.6217

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(e) Between 2 and 4:

P(2 < X < 4) =?

Now

Z 2 – 2.58 1.3578 and z 4 – 2.58 1.3578

Z= -0.58 1.3578 and Z 1.42 1.3578

Z= -0.43 and Z= +1.05

⇒ P(-0.43<Z<+1.05) ?

0.5 0.5 +3 -0.43 Z=0 +1 +1.05 +2 -3 -2 -1

We have to find out the Probability of Shaded Area

→ P(-0.43 < 3 < +1.05) = P(-0.43 < 3 < 0) + PO<Z < +1.05

⇒ P(-0.43<Z<+1.05) = 0.1664 +0.3531

⇒ P(-0.43<Z<+1.05) = 0.5195

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(f) P(X >5) =?

Z 5 – 2.58 1.3578

Z= 1.78

P(Z > 1.78) =?

0.5 0.5 -3 -2 -1 Z=0 +1 1.78 +2 +3

We have to find the Probability of Shaded area

→ P(Z > 1.78) = P(+1.78 <7< +00)

$\Rightarrow P(Z > 1.78) = 0.5 - 0.4625$

$\Rightarrow P(Z > 1.78) = 0.0375$

NOTE: The Probability value of Z is extracted from the Standard Normal Tabulated values; which is posted below

Normal Distribution Table a 0.00 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09 0.0 0.0000 0.0040 0.0080 0.0120 (0.0160 0.0199 0.0239 0.0279 0.0319 0.0359 0.1 0.0398 0.0438 0.0478 0.0517 0.0557 0.0596 0.0636 0.0675 0.0714 0.0753 0.2 0.0793 0.0832 0.0871 0.0910 0.0948 0.0987 0.1026 0.1064 0.1103 0.1141 0.3 0.1179 0.1217 0.1255 0.1293 0.1331 0.1368 0.1406 0.1443 0.1480 0.1517 0.4 0.1554 0.1591 0.1628 0.1664 0.1700 0.1736 0.1772 0.1808 0.1844 0.1879 0.5 0.1915 0.1950 0.1985 0.2019 0.2054 0.2088 0.2123 0.2157 0.2190 0.2224 0.6 0.2257 0.2291 0.2324 0.2357 0.2389 0.2422 0.2454 0.2486 0.2517 0.2549 0.7 0.2580 0.2611 0.2642 0.2673 0.2704 0.2734 0.2764 0.2794 0.2823 0.2852 0.8 0.2881 0.2910 0.2939 0.2967 0.2995 0.3023 0.3051 0.3078 0.3106 0.3133 0.9 0.3159 0.3186 0.3212 0.3238 0.3264 0.3289 0.3315 0.3340 0.3365 0.3389 1.0 0.3413 0.3438 0.3461 0.3485 0.3508 0.3531 0.3554 0.3577 0.3599 0.3621 1.1 0.3643 0.3665 0.3686 0.3708 0.3729 0.3749 0.3770 0.3790 0.3810 0.3830 1.2 0.3849 0.3869 0.3888 0.3907 0.3925 0.3944 0.3962 0.3980 0.3997 0.4015 1.3 0.4032 0.4049 0.4066 0.4082 0.4099 0.4115 0.4131 0.4147 0.4162 0.4177 1.4 0.4192 0.4207 0.4222 0.4236 0.4251 0.4265 0.4279 0.4292 0.4306 0.4319 1.5 0.4332 0.4345 0.4357 0.4370 0.4382 0.4394 0.4406 0.4418 0.4429 0.4441 1.6 0.4452 0.4463 0.4474 0.4484 0.4495 0.4505 0.4515 0.4525 0.4535 0.4545 1.7 0.4554 0.4564 0.4573 0.4582 0.4591 0.4599 0.4608 0.4616 0.4625 0.4633 1.8 0.4641 0.4649 0.4656 0.4664 0.4671 0.4678 0.4686 0.4693 0.4699 0.4706 1.9 0.4713 0.4719 0.4726 0.4732 0.4738 0.4744 0.4750 0.4756 0.4761 0.4767 2.0 0.4772 0.4778 0.4783 0.4788 0.4793 0.4798 0.4803 0.4808 0.4812 0.4817 2.1 0.4821 0.4826 0.4830 0.4834 0.4838 0.4842 0.4846 0.4850 0.4854 0.4857 2.2 0.4861 0.4864 0.4868 0.4871 0.4875 0.4878 0.4881 0.4884 0.4887 0.4890 2.3 0.4893 0.4896 0.4898 0.4901 0.4904 0.4906 0.4909 0.4911 0.4913 0.4916 2.4 0.4918 0.4920 0.4922 0.4925 0.4927 0.4929 0.4931 0.4932 0.4934 0.4936 2.5 0.4938 0.4940 0.4941 0.4943 0.4945 0.4946 0.4948 0.4949 0.4951 0.4952 2.6 0.4953 0.4955 0.4956 0.4957 0.4959 0.4960 0.4961 0.4962 0.4963 0.4964 2.7 0.4965 0.4966 0.4967 0.4968 0.4969 0.4970 0.4971 0.4972 0.4973 0.4974 2.8 0.4974 0.4975 0.4976 0.4977 0.4977 0.4978 0.4979 0.4979 0.4980 0.4981 2.9 0.4981 0.4982 0.4982 0.4983 0.4984 0.4984 0.4985 0.4985 0.4986 0.4986 3.0 0.4987 0.4987 0.4987 0.4988 0.4988 0.4989 0.4989 0.4989 0.4990 0.4990
REMARK:
P(-00 <7< 0) = P(O<Z< +00) = 0.5

Whatever may be the symbol either less than or greater than or equal to ; you just search the probability for equal value.

For example If P(0 <Z<1.78) then search for P(Z =1.78) i.e you search the value which is intersected between 1.7 and 0.08