A European growth mutual fund specializes in stocks from the British Isles, cont

Question

A European growth mutual fund specializes in stocks from the British Isles, continental Europe, and Scandinavia. The fund has over175 stocks. Let x be a random variable that represents the monthly percentage return for this fund. Suppose x has mean = 1.4%and standard deviation = 1.2%.

(a) Let's consider the monthly return of the stocks in the fund to be a sample from the population of monthly returns of all European stocks. Is it reasonable to assume that x (the average monthly return on the 175 stocks in the fund) has a distribution that is approximately normal? Explain.

(b) After 9 months, what is the probability that the average monthly percentage return x will be between 1% and 2%? (Round your answer to four decimal places.)

(c) After 18 months, what is the probability that the average monthly percentage return x will be between 1% and 2%? (Round your answer to four decimal places.)

(d) Compare your answers to parts (b) and (c). Did the probability increase as n (number of months) increased? Why would this happen?

(e) If after 18 months the average monthly percentage return x is more than 2%, would that tend to shake your confidence in the statement that = 1.4%? If this happened, do you think the European stock market might be heating up? (Round your answer to four decimal places.)

Explain.

Explanation / Answer

a)

Yes, as we can have negative returns, so we can cater many standard deviations around the mean.

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b)

We first get the z score for the two values. As z = (x - u) sqrt(n) / s, then as          
x1 = lower bound =    1      
x2 = upper bound =    2      
u = mean =    1.4      
n = sample size =    9      
s = standard deviation =    1.2      
          
Thus, the two z scores are          
          
z1 = lower z score = (x1 - u) * sqrt(n) / s =    -1      
z2 = upper z score = (x2 - u) * sqrt(n) / s =    1.5      
          
Using table/technology, the left tailed areas between these z scores is          
          
P(z < z1) =    0.158655254      
P(z < z2) =    0.933192799      
          
Thus, the area between them, by subtracting these areas, is          
          
P(z1 < z < z2) =    0.774537545   [ANSWER]

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c)

We first get the z score for the two values. As z = (x - u) sqrt(n) / s, then as          
x1 = lower bound =    1      
x2 = upper bound =    2      
u = mean =    1.4      
n = sample size =    18      
s = standard deviation =    1.2      
          
Thus, the two z scores are          
          
z1 = lower z score = (x1 - u) * sqrt(n) / s =    -1.414213562      
z2 = upper z score = (x2 - u) * sqrt(n) / s =    2.121320344      
          
Using table/technology, the left tailed areas between these z scores is          
          
P(z < z1) =    0.078649604      
P(z < z2) =    0.983052573      
          
Thus, the area between them, by subtracting these areas, is          
          
P(z1 < z < z2) =    0.90440297   [ANSWER]

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d)

YES. This happens because the standard error decreased with larger sample size, so there is less variability. This made z values increase, hence, a larger area between them.

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e)

We first get the z score for the critical value. As z = (x - u) sqrt(n) / s, then as          
          
x = critical value =    2      
u = mean =    1.4      
n = sample size =    18      
s = standard deviation =    1.2      
          
Thus,          
          
z = (x - u) * sqrt(n) / s =    2.121320344      
          
Thus, using a table/technology, the right tailed area of this is          
          
P(z >   2.121320344   ) =    0.016947427 [ANSWER]

As this is a small probability (P<0.05)   , then YES, the European stock market might be heating up. [ANSWER]
  

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