Question

Suppose the returns on long-term government bonds are normally distributed. Assume long-term government bonds have a mean return of 5.7 percent and a standarddeviation of 9.4 percent.

Requirement 1:
What is the approximate probability that your return on these bonds will be less than ?3.7 percent in a given year? (Do not include the percent sign (%). Round youranswer to 2 decimal places (e.g., 32.16).)

1-Probability= ? %

Requirement 2:
What range of returns would you expect to see 68 percent of the time? (Do not include the percent signs (%). Negative amount should be indicated by a minus sign. Inputyour answers from lowest to highest to receive credit for your answers. Round your answers to 2 decimal places (e.g., 32.16).)

2-Expected range of returns=? % to ? %

Requirement 3:
What range would you expect to see 95 percent of the time? (Do not include the percent signs (%). Negative amount should be indicated by a minus sign. Input youranswers from lowest to highest to receive credit for your answers. Round your answers to 2 decimal places (e.g., 32.16).)

3-Expected range of returns=? % to ? %

Answer 1

Suppose the returns on long-term government bonds are normally distributed. Assume long-term government bonds have a mean return of 5.7 percent and astandard deviation of 9.4 percent.

Requirement 1:
What is the approximate probability that your return on these bonds will be less than −3.7 percent in a given year? (Do not include the percent sign(%). Round your answer to 2 decimal places (e.g., 32.16).)

z = (X − μ)/σ

z = (-3.7% − 5.7)/9.6% = -0.97

Requirement 2:
What range of returns would you expect to see 68 percent of the time? (Do not include the percent signs (%). Negative amount should be indicatedby a minus sign. Input your answers from lowest to highest to receive credit for your answers. Round your answers to 2 decimal places (e.g.,32.16).)

The range of returns you would expect to see 95 percent of the time is the mean plus or minus 2 standard deviations, or:

95% level: Rε μ ± 2σ = 5.7% ± 2(9.6%) =24.9 to -23.7 % --------range of returns

Answer 2

1. Z=[-3.7-5.7]/9.4= -1.00 from table value of Standard Normal Distribution Function at 1.00 is 0.3413 mean it is 34.13% so the area between interestpoint 50-34.13=15.87%=0.16 1-0.16=.84 = 84% of the time, bond returns will be greater -3.7%.

3.

without %age answers are .245 and -.114

2 will be same as above.