On 2 June 2017, an analyst estimated that possible returns for the S&P500 for Ju

Question

On 2 June 2017, an analyst estimated that possible returns for the S&P500 for June 2017 had the following distribution:

  4.3% of the possible returns for June 2017 were –5.0% or less

29.8% of the possible returns for June 2017 were between –5.0% and –1.0%

[see https://seekingalpha.com/article/4078283-odds-favor-1-percent-5-percent-change-s-and-p-500-june-direction ]

If the possible returns for the S&P500 for June 2017 follow a normal curve, what percentage of returns fall between +1.0% and +5.0%?

A. 29.8%

B. 31.0%

C. 34.3%

D. 36.9%

E. 39.5%

2. The actual return for the S&P500 for June 2017 was 0.32%. Based on the assumption that possible returns follow a normal distribution, did the market perform about as well as the analyst predicted, better than the analyst predicted, or worse than the analyst predicted?

A. Much better than the analyst predicted: 0.32% was more than two SD’s above the mean predicted by the analyst

B. Better than the analyst predicted: 0.32% was more than one but less than two SD’s above the mean predicted by the analyst

C. About as well as predicted: 0.32% was within one SD of the mean predicted by the analyst

D. Worse than the analyst predicted: 0.32% was more than one but less than two SD’s below the mean predicted by the analyst

E. Much worse than the analyst predicted: 0.32% was more than two SD’s below the mean predicted by the analyst

A. 29.8%

B. 31.0%

C. 34.3%

D. 36.9%

E. 39.5%

Explanation / Answer

1) let the mean and sd of possible returns be m and s respectively .

z = (X - m)/s

P(X < -5) = 0.043

P(Z< z*) = 0.043

z* = -1.717

-1.717 = (-5 - m)/s       (1)

P(-5 < X< -1) = 0.298

P(X< -1) - P(X < -5) = 0.298

P(X< -1) = 0.341

P(Z < z*) = 0.341

z* = -0.409

-0.409 = (-1 - m)/s    (2)

hence

m = 0.250765 , s = 3.0581

hence P(1 <X< 5) =  

P ( 0.25<Z<1.55 )=0.3407

option C) 34.3 % is most close

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