Suppose that the rate of return for a particular stock during the past two years

Question

Suppose that the rate of return for a particular stock during the past two years was 10% and 30%. Compute the geometric rate of return per year. (Note: A rate of return of 10% is recorded as 0.10, and a rate of return of 30% is recorded as 0.30.) 3. Consider a population of 1,024 mutual funds that primarily invest in large companies. You have determined that m, the mean one-year total percentage return achieved by all the funds, is 8.20 and that s, the standard deviation, is 2.75. a. According to the empirical rule, what percentage of these funds are expected to be within -/+1 standard deviation of the mean? b. According to the empirical rule, what percentage of these funds are expected to be within -/+2 standard deviations of the mean? c. According to the Chebyshev rule, what percentage of these funds are expected to be within -/+1, -/+2, or -/+3 standard deviations of the mean? d. According to the Chebyshev rule, at least 93.75% of these funds are expected to have one-year total returns between what two amounts?

Explanation / Answer

NORMAL DISTRIBUTION
the PDF of normal distribution is = 1/ * 2 * e ^ -(x-u)^2/ 2^2
standard normal distribution is a normal distribution with a,
mean of 0,
standard deviation of 1
equation of the normal curve is ( Z )= x - u / sd ~ N(0,1)
mean ( u ) = 8.2
standard Deviation ( sd )= 2.75
68% OF DATA
About 68% of the area under the normal curve is within one standard deviation of the mean. i.e. (u ± 1s.d)
So to the given normal distribution about 68% of the observations lie in between
= [8.2 ± 2.75]
= [ 8.2 - 2.75 , 8.2 + 2.75]
= [ 5.45 , 10.95 ]
95% OF DATA
About 95% of the area under the normal curve is within two standard deviation of the mean. i.e. (u ± 2s.d)
So to the given normal distribution about 95% of the observations lie in between
= [8.2 ± 2 * 2.75]
= [ 8.2 - 2 * 2.75 , 8.2 + 2* 2.75]
= [ 2.7 , 13.7 ]
99.7% OF DATA
About 99.7% of the area under the normal curve is within two standard deviation of the mean. i.e. (u ± 3s.d)
So to the given normal distribution about 99.7% of the observations lie in between
= [8.2 ± 3 * 2.75]
= [ 8.2 - 3 * 2.75 , 8.2 + 3* 2.75]
= [ -0.0500000000000007 , 16.45 ]

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